Stochastic Calculus Solution Manual

Stochastic Calculus for Finance, Volume I and II by Yan Zeng Last updated: August 20, 2007 This is a solution manual for the two-volume textbook Stochastic calculus for ? nance, by Steven Shreve. If you have any comments or ? nd any typos/errors, please email me at [email  protected] edu. The current version omits the following problems. Volume I: 1. 5, 3. 3, 3. 4, 5. 7; Volume II: 3. 9, 7. 1, 7. 2, 7. 5–7. 9, 10. 8, 10. 9, 10. 10. Acknowledgment I thank Hua Li (a graduate student at Brown University) for reading through this solution manual and communicating to me several mistakes/typos. 1. 1. Stochastic Calculus for Finance I: The Binomial Asset Pricing Model 1. The Binomial No-Arbitrage Pricing Model Proof. If we get the up sate, then X1 = X1 (H) = ? 0 uS0 + (1 + r)(X0 ? ?0 S0 ); if we get the down state, then X1 = X1 (T ) = ? 0 dS0 + (1 + r)(X0 ? ?0 S0 ). If X1 has a positive probability of being strictly positive, then we must either have X1 (H) > 0 or X1 (T ) > 0. (i) If X1 (H) > 0, then ? 0 uS0 + (1 + r)(X0 ? ?0 S0 ) > 0. Plug in X0 = 0, we get u? 0 > (1 + r)? 0 . By condition d < 1 + r < u, we conclude ? 0 > 0.In this case, X1 (T ) = ? 0 dS0 + (1 + r)(X0 ? ?0 S0 ) = ? 0 S0 [d ? (1 + r)] < 0. (ii) If X1 (T ) > 0, then we can similarly deduce ? 0 < 0 and hence X1 (H) < 0. So we cannot have X1 strictly positive with positive probability unless X1 is strictly negative with positive probability as well, regardless the choice of the number ? 0 . Remark: Here the condition X0 = 0 is not essential, as far as a property de? nition of arbitrage for arbitrary X0 can be given. Indeed, for the one-period binomial model, we can de? ne arbitrage as a trading strategy such that P (X1 ?X0 (1 + r)) = 1 and P (X1 > X0 (1 + r)) > 0. First, this is a generalization of the case X0 = 0; second, it is â€Å"proper† because it is comparing the result of an arbitrary investment involving money and stock markets with that of a safe investment involving only money market. This can also be seen by regarding X0 as borrowed from money market account. Then at time 1, we have to pay back X0 (1 + r) to the money market account. In summary, arbitrage is a trading strategy that beats â€Å"safe† investment. Accordingly, we revise the proof of Exercise 1. 1. as follows.If X1 has a positive probability of being strictly larger than X0 (1 + r), the either X1 (H) > X0 (1 + r) or X1 (T ) > X0 (1 + r). The ? rst case yields ? 0 S0 (u ? 1 ? r) > 0, i. e. ?0 > 0. So X1 (T ) = (1 + r)X0 + ? 0 S0 (d ? 1 ? r) < (1 + r)X0 . The second case can be similarly analyzed. Hence we cannot have X1 strictly greater than X0 (1 + r) with positive probability unless X1 is strictly smaller than X0 (1 + r) with positive probability as well. Finally, we comment that the above formulation of arbitrage is equivalent to the one in the textbook. For details, see Shreve [7], Exercise 5. . 1. 2. 1 5 Proof. X1 (u) = ? 0 ? 8 + ? 0 ? 3 ? 5 (4? 0 + 1. 20? 0 ) = 3? 0 + 1. 5? 0 , a nd X1 (d) = ? 0 ? 2 ? 4 (4? 0 + 1. 20? 0 ) = 4 ? 3? 0 ? 1. 5? 0 . That is, X1 (u) = ? X1 (d). So if there is a positive probability that X1 is positive, then there is a positive probability that X1 is negative. Remark: Note the above relation X1 (u) = ? X1 (d) is not a coincidence. In general, let V1 denote the ? ? payo? of the derivative security at time 1. Suppose X0 and ? 0 are chosen in such a way that V1 can be ? 0 ? ?0 S0 ) + ? 0 S1 = V1 . Using the notation of the problem, suppose an agent begins ? replicated: (1 + r)(X with 0 wealth and at time zero buys ? 0 shares of stock and ? 0 options. He then puts his cash position ? 0 S0 ? ?0 X0 in a money market account. At time one, the value of the agent’s portfolio of stock, option and money market assets is ? X1 = ? 0 S1 + ? 0 V1 ? (1 + r)(? 0 S0 + ? 0 X0 ). Plug in the expression of V1 and sort out terms, we have ? X1 = S0 (? 0 + ? 0 ? 0 )( S1 ? (1 + r)). S0 ? Since d < (1 + r) < u, X1 (u) and X1 (d) have opposite signs. So if the price of the option at time zero is X0 , then there will no arbitrage. 1. 3. S0 1 Proof. V0 = 1+r 1+r? d S1 (H) + u? ? r S1 (T ) = 1+r 1+r? d u + u? 1? r d = S0 . This is not surprising, since u? d u? d u? d u? d this is exactly the cost of replicating S1 . Remark: This illustrates an important point. The â€Å"fair price† of a stock cannot be determined by the risk-neutral pricing, as seen below. Suppose S1 (H) and S1 (T ) are given, we could have two current prices, S0 and S0 . Correspondingly, we can get u, d and u , d . Because they are determined by S0 and S0 , respectively, it’s not surprising that risk-neutral pricing formula always holds, in both cases. That is, 1+r? d u? d S1 (H) S0 = + u? 1? r u? d S1 (T ) 1+r S0 = 1+r? d u ? d S1 (H) + u ? 1? r u ? d S1 (T ) 1+r . Essentially, this is because risk-neutral pricing relies on fair price=replication cost. Stock as a replicating component cannot determine its own â€Å"fair† price via the risk-n eutral pricing formula. 1. 4. Proof. Xn+1 (T ) = = ? n dSn + (1 + r)(Xn ? ?n Sn ) ?n Sn (d ? 1 ? r) + (1 + r)Vn pVn+1 (H) + q Vn+1 (T ) ? ? Vn+1 (H) ? Vn+1 (T ) (d ? 1 ? r) + (1 + r) = u? d 1+r = p(Vn+1 (T ) ? Vn+1 (H)) + pVn+1 (H) + q Vn+1 (T ) ? ? ? = pVn+1 (T ) + q Vn+1 (T ) ? ? = Vn+1 (T ). 1. 6. 2 Proof. The bank’s trader should set up a replicating portfolio whose payo? s the opposite of the option’s payo?. More precisely, we solve the equation (1 + r)(X0 ? ?0 S0 ) + ? 0 S1 = ? (S1 ? K)+ . 1 Then X0 = ? 1. 20 and ? 0 = ? 2 . This means the trader should sell short 0. 5 share of stock, put the income 2 into a money market account, and then transfer 1. 20 into a separate money market account. At time one, the portfolio consisting of a short position in stock and 0. 8(1 + r) in money market account will cancel out with the option’s payo?. Therefore we end up with 1. 20(1 + r) in the separate money market account. Remark: This problem illustrates why we are in terested in hedging a long position.In case the stock price goes down at time one, the option will expire without any payo?. The initial money 1. 20 we paid at time zero will be wasted. By hedging, we convert the option back into liquid assets (cash and stock) which guarantees a sure payo? at time one. Also, cf. page 7, paragraph 2. As to why we hedge a short position (as a writer), see Wilmott [8], page 11-13. 1. 7. Proof. The idea is the same as Problem 1. 6. The bank’s trader only needs to set up the reverse of the replicating trading strategy described in Example 1. 2. 4. More precisely, he should short sell 0. 1733 share of stock, invest the income 0. 933 into money market account, and transfer 1. 376 into a separate money market account. The portfolio consisting a short position in stock and 0. 6933-1. 376 in money market account will replicate the opposite of the option’s payo?. After they cancel out, we end up with 1. 376(1 + r)3 in the separate money market ac count. 1. 8. (i) 2 s s Proof. vn (s, y) = 5 (vn+1 (2s, y + 2s) + vn+1 ( 2 , y + 2 )). (ii) Proof. 1. 696. (iii) Proof. ?n (s, y) = vn+1 (us, y + us) ? vn+1 (ds, y + ds) . (u ? d)s 1. 9. (i) Proof. Similar to Theorem 1. 2. 2, but replace r, u and d everywhere with rn , un and dn .More precisely, set pn = 1+rn ? dn and qn = 1 ? pn . Then un ? dn Vn = pn Vn+1 (H) + qn Vn+1 (T ) . 1 + rn (ii) Proof. ?n = (iii) 3 Vn+1 (H)? Vn+1 (T ) Sn+1 (H)? Sn+1 (T ) = Vn+1 (H)? Vn+1 (T ) . (un ? dn )Sn 10 10 Proof. un = Sn+1 (H) = Sn +10 = 1+ Sn and dn = Sn+1 (T ) = Sn ? 10 = 1? Sn . So the risk-neutral probabilities Sn Sn Sn Sn at time n are pn = u1? dnn = 1 and qn = 1 . Risk-neutral pricing implies the price of this call at time zero is ? ? 2 2 n ? d 9. 375. 2. Probability Theory on Coin Toss Space 2. 1. (i) Proof. P (Ac ) + P (A) = (ii) Proof. By induction, it su? ces to work on the case N = 2.When A1 and A2 are disjoint, P (A1 ? A2 ) = A1 ? A2 P (? ) = A1 P (? ) + A2 P (? ) = P (A1 ) + P (A2 ). When A1 and A2 are arbitrary, using the result when they are disjoint, we have P (A1 ? A2 ) = P ((A1 ? A2 ) ? A2 ) = P (A1 ? A2 ) + P (A2 ) ? P (A1 ) + P (A2 ). 2. 2. (i) 1 3 1 Proof. P (S3 = 32) = p3 = 8 , P (S3 = 8) = 3p2 q = 3 , P (S3 = 2) = 3pq 2 = 8 , and P (S3 = 0. 5) = q 3 = 8 . 8 Ac P (? ) + A P (? ) = P (? ) = 1. (ii) Proof. E[S1 ] = 8P (S1 = 8) + 2P (S1 = 2) = 8p + 2q = 5, E[S2 ] = 16p2 + 4  · 2pq + 1  · q 2 = 6. 25, and 3 1 E[S3 ] = 32  · 1 + 8  · 8 + 2  · 3 + 0.  · 8 = 7. 8125. So the average rates of growth of the stock price under P 8 8 5 are, respectively: r0 = 4 ? 1 = 0. 25, r1 = 6. 25 ? 1 = 0. 25 and r2 = 7. 8125 ? 1 = 0. 25. 5 6. 25 (iii) 8 1 Proof. P (S3 = 32) = ( 2 )3 = 27 , P (S3 = 8) = 3  · ( 2 )2  · 1 = 4 , P (S3 = 2) = 2  · 1 = 2 , and P (S3 = 0. 5) = 27 . 3 3 3 9 9 9 Accordingly, E[S1 ] = 6, E[S2 ] = 9 and E[S3 ] = 13. 5. So the average rates of growth of the stock price 9 6 under P are, respectively: r0 = 4 ? 1 = 0. 5, r1 = 6 ? 1 = 0. 5, and r2 = 13. 5 ? 1 = 0. 5. 9 2. 3. Proof. Apply conditional Jensen’s inequality. 2. 4. (i) Proof.En [Mn+1 ] = Mn + En [Xn+1 ] = Mn + E[Xn+1 ] = Mn . (ii) 2 n+1 Proof. En [ SSn ] = En [e? Xn+1 e? +e ] = 2 ? Xn+1 ] e? +e E[e = 1. 2. 5. (i) 2 2 Proof. 2In = 2 j=0 Mj (Mj+1 ? Mj ) = 2 j=0 Mj Mj+1 ? j=1 Mj ? j=1 Mj = 2 j=0 Mj Mj+1 + n? 1 n? 1 n? 1 n? 1 2 2 2 2 2 2 2 2 Mn ? j=0 Mj+1 ? j=0 Mj = Mn ? j=0 (Mj+1 ? Mj ) = Mn ? j=0 Xj+1 = Mn ? n. n? 1 n? 1 n? 1 n? 1 n? 1 (ii) Proof. En [f (In+1 )] = En [f (In + Mn (Mn+1 ? Mn ))] = En [f (In + Mn Xn+1 )] = 1 [f (In + Mn ) + f (In ? Mn )] = 2 v v v g(In ), where g(x) = 1 [f (x + 2x + n) + f (x ? 2x + n)], since 2In + n = |Mn |. 2 2. 6. 4 Proof. En [In+1 ?In ] = En [? n (Mn+1 ? Mn )] = ? n En [Mn+1 ? Mn ] = 0. 2. 7. Proof. We denote by Xn the result of n-th coin toss, where Head is represented by X = 1 and Tail is 1 represented by X = ? 1. We also suppose P (X = 1) = P (X = ? 1) = 2 . De? ne S1 = X1 and Sn+1 = n Sn +bn (X1 ,  ·  ·  · , Xn )Xn+1 , where bn ( ·) is a bounded function on {? 1, 1} , to be determined later on. Clearly (Sn )n? 1 is an adapted stochastic process, and we can show it is a martingale. Indeed, En [Sn+1 ? Sn ] = bn (X1 ,  ·  ·  · , Xn )En [Xn+1 ] = 0. For any arbitrary function f , En [f (Sn+1 )] = 1 [f (Sn + bn (X1 ,  ·  ·  · , Xn )) + f (Sn ? n (X1 ,  ·  ·  · , Xn ))]. Then 2 intuitively, En [f (Sn+1 ] cannot be solely dependent upon Sn when bn ’s are properly chosen. Therefore in general, (Sn )n? 1 cannot be a Markov process. Remark: If Xn is regarded as the gain/loss of n-th bet in a gambling game, then Sn would be the wealth at time n. bn is therefore the wager for the (n+1)-th bet and is devised according to past gambling results. 2. 8. (i) Proof. Note Mn = En [MN ] and Mn = En [MN ]. (ii) Proof. In the proof of Theorem 1. 2. 2, we proved by induction that Xn = Vn where Xn is de? ned by (1. 2. 14) of Chapter 1. In other words, the sequence (Vn )0? n?N can be realized as the value process of a portfolio, Xn which consists of stock and money market accounts. Since ( (1+r)n )0? n? N is a martingale under P (Theorem Vn 2. 4. 5), ( (1+r)n )0? n? N is a martingale under P . (iii) Proof. (iv) Proof. Combine (ii) and (iii), then use (i). 2. 9. (i) (H) S1 (H) 1 = 2, d0 = S1S0 = 2 , S0 (T and d1 (T ) = S21 (TT)) = 1. S 1 1 0 ? d So p0 = 1+r? d0 0 = 2 , q0 = 2 , p1 (H) u0 5 q1 (T ) = 6 . Therefore P (HH) = p0 p1 (H) = 1 , 4 5 q0 q1 (T ) = 12 . Vn (1+r)n = En VN (1+r)N , so V0 , V1 1+r ,  ·Ã‚ ·Ã‚ ·, VN ? 1 , VN (1+r)N ? 1 (1+r)N is a martingale under P . Proof. u0 = u1 (H) = =S2 (HH) S1 (H) = 1. 5, d1 (H) = S2 (HT ) S1 (H) = 1, u1 (T ) = S2 (T H) S1 (T ) =4 1+r1 (H)? d1 (H) u1 (H)? d1 (H) 1 = 1 , q1 (H) = 2 , p1 (T ) = 2 1 4, 1+r1 (T )? d1 (T ) u1 (T )? d1 (T ) 1 12 1 = 6 , and P (HT ) = p0 q1 (H) = P (T H) = q0 p1 (T ) = and P (T T ) = The proofs of Theorem 2. 4. 4, Theorem 2. 4. 5 and Theorem 2. 4. 7 still work for the random interest rate m odel, with proper modi? cations (i. e. P would be constructed according to conditional probabilities P (? n+1 = H|? 1 ,  ·  ·  · , ? n ) := pn and P (? n+1 = T |? 1 ,  ·  ·  · , ? n ) := qn . Cf. notes on page 39. ). So the time-zero value of an option that pays o?V2 at time two is given by the risk-neutral pricing formula V0 = E (1+r0V2 1 ) . )(1+r (ii) Proof. V2 (HH) = 5, V2 (HT ) = 1, V2 (T H) = 1 and V2 (T T ) = 0. So V1 (H) = 2. 4, V1 (T ) = p1 (T )V2 (T H)+q1 (T )V2 (T T ) 1+r1 (T ) p1 (H)V2 (HH)+q1 (H)V2 (HT ) 1+r1 (H) = = 1 9, and V0 = p0 V1 (H)+q0 V1 (T ) 1+r0 ? 1. 5 (iii) Proof. ?0 = (iv) Proof. ?1 (H) = 2. 10. (i) Xn+1 Proof. En [ (1+r)n+1 ] = En [ ? n Yn+1 Sn + (1+r)n+1 (1+r)(Xn n Sn ) ] (1+r)n+1 Xn (1+r)n . V2 (HH)? V2 (HT ) S2 (HH)? S2 (HT ) V1 (H)? V1 (T ) S1 (H)? S1 (T ) = 1 2. 4? 9 8? 2 = 0. 4 ? 1 54 ? 0. 3815. = 5? 1 12? 8 = 1. = ?n Sn (1+r)n+1 En [Yn+1 ] + Xn Sn (1+r)n = ?n Sn (1+r)n+1 (up + dq) + Xn n Sn (1+r)n = ?n Sn +Xn n Sn (1+r)n = (ii) Proof . From (2. 8. 2), we have ? n uSn + (1 + r)(Xn ? ?n Sn ) = Xn+1 (H) ? n dSn + (1 + r)(Xn ? ?n Sn ) = Xn+1 (T ). So ? n = Xn+1 (H)? Xn+1 (T ) uSn ? dSn and Xn = En [ Xn+1 ]. To make the portfolio replicate the payo? at time N , we 1+r VN X must have XN = VN . So Xn = En [ (1+r)N ? n ] = En [ (1+r)N ? n ]. Since (Xn )0? n? N is the value process of the N unique replicating portfolio (uniqueness is guaranteed by the uniqueness of the solution to the above linear VN equations), the no-arbitrage price of VN at time n is Vn = Xn = En [ (1+r)N ? ]. (iii) Proof. En [ Sn+1 ] (1 + r)n+1 = = < = 1 En [(1 ? An+1 )Yn+1 Sn ] (1 + r)n+1 Sn [p(1 ? An+1 (H))u + q(1 ? An+1 (T ))d] (1 + r)n+1 Sn [pu + qd] (1 + r)n+1 Sn . (1 + r)n Sn (1+r)n+1 (1? a)(pu+qd) Sn+1 If An+1 is a constant a, then En [ (1+r)n+1 ] = Sn (1+r)n (1? a)n . = Sn (1+r)n (1? a). Sn+1 So En [ (1+r)n+1 (1? a)n+1 ] = 2. 11. (i) Proof. FN + PN = SN ? K + (K ? SN )+ = (SN ? K)+ = CN . (ii) CN FN PN Proof. Cn = En [ (1+r)N ? n ] = En [ (1+ r)N ? n ] + En [ (1+r)N ? n ] = Fn + Pn . (iii) FN Proof. F0 = E[ (1+r)N ] = 1 (1+r)N E[SN ? K] = S0 ? K (1+r)N . (iv) 6 Proof.At time zero, the trader has F0 = S0 in money market account and one share of stock. At time N , the trader has a wealth of (F0 ? S0 )(1 + r)N + SN = ? K + SN = FN . (v) Proof. By (ii), C0 = F0 + P0 . Since F0 = S0 ? (vi) SN ? K Proof. By (ii), Cn = Pn if and only if Fn = 0. Note Fn = En [ (1+r)N ?n ] = Sn ? So Fn is not necessarily zero and Cn = Pn is not necessarily true for n ? 1. (1+r)N S0 (1+r)N ? n (1+r)N S0 (1+r)N = 0, C0 = P0 . = Sn ? S0 (1 + r)n . 2. 12. Proof. First, the no-arbitrage price of the chooser option at time m must be max(C, P ), where C=E (SN ? K)+ (K ? SN )+ , and P = E . (1 + r)N ? m (1 + r)N ? That is, C is the no-arbitrage price of a call option at time m and P is the no-arbitrage price of a put option at time m. Both of them have maturity date N and strike price K. Suppose the market is liquid, then the chooser option is equivalent to receiving a payo? of max(C, P ) at time m. Therefore, its current no-arbitrage price should be E[ max(C,P ) ]. (1+r)m K K By the put-call parity, C = Sm ? (1+r)N ? m + P . So max(C, P ) = P + (Sm ? (1+r)N ? m )+ . Therefore, the time-zero price of a chooser option is E K (Sm ? (1+r)N ? m )+ P +E (1 + r)m (1 + r)m =E K (Sm ? (1+r)N ? m )+ (K ? SN )+ . +E (1 + r)N (1 + r)mThe ? rst term stands for the time-zero price of a put, expiring at time N and having strike price K, and the K second term stands for the time-zero price of a call, expiring at time m and having strike price (1+r)N ? m . If we feel unconvinced by the above argument that the chooser option’s no-arbitrage price is E[ max(C,P ) ], (1+r)m due to the economical argument involved (like â€Å"the chooser option is equivalent to receiving a payo? of max(C, P ) at time m†), then we have the following mathematically rigorous argument. First, we can construct a portfolio ? 0 ,  ·  ·  · , ? m? 1 , whose payo? at time m is max(C, P ).Fix ? , if C(? ) > P (? ), we can construct a portfolio ? m ,  ·  ·  · , ? N ? 1 whose payo? at time N is (SN ? K)+ ; if C(? ) < P (? ), we can construct a portfolio ? m ,  ·  ·  · , ? N ? 1 whose payo? at time N is (K ? SN )+ . By de? ning (m ? k ? N ? 1) ? k (? ) = ? k (? ) ? k (? ) if C(? ) > P (? ) if C(? ) < P (? ), we get a portfolio (? n )0? n? N ? 1 whose payo? is the same as that of the chooser option. So the no-arbitrage price process of the chooser option must be equal to the value process of the replicating portfolio. In Xm particular, V0 = X0 = E[ (1+r)m ] = E[ max(C,P ) ]. (1+r)m 2. 13. (i) Proof.Note under both actual probability P and risk-neutral probability P , coin tosses ? n ’s are i. i. d.. So n+1 without loss of generality, we work on P . For any function g, En [g(Sn+1 , Yn+1 )] = En [g( SSn Sn , Yn + = pg(uSn , Yn + uSn ) + qg(dSn , Yn + dSn ), which is a function of (Sn , Yn ). So (Sn , Yn )0? n? N is Markov un der P . (ii) 7 Sn+1 Sn Sn )] Proof. Set vN (s, y) = f ( Ny ). Then vN (SN , YN ) = f ( +1 Vn = where En [ Vn+1 ] 1+r = n+1 En [ vn+1 (S1+r ,Yn+1 ) ] N n=0 Sn N +1 ) = VN . Suppose vn+1 is given, then = 1 1+r [pvn+1 (uSn , Yn + uSn ) + qvn+1 (dSn , Yn + dSn )] = vn (Sn , Yn ), vn (s, y) = n+1 (us, y + us) + vn+1 (ds, y + ds) . 1+r 2. 14. (i) Proof. For n ? M , (Sn , Yn ) = (Sn , 0). Since coin tosses ? n ’s are i. i. d. under P , (Sn , Yn )0? n? M is Markov under P . More precisely, for any function h, En [h(Sn+1 )] = ph(uSn ) + h(dSn ), for n = 0, 1,  ·  ·  · , M ? 1. For any function g of two variables, we have EM [g(SM +1 , YM +1 )] = EM [g(SM +1 , SM +1 )] = pg(uSM , uSM )+ n+1 n+1 qg(dSM , dSM ). And for n ? M +1, En [g(Sn+1 , Yn+1 )] = En [g( SSn Sn , Yn + SSn Sn )] = pg(uSn , Yn +uSn )+ qg(dSn , Yn + dSn ), so (Sn , Yn )0? n? N is Markov under P . (ii) y Proof. Set vN (s, y) = f ( N ? M ).Then vN (SN , YN ) = f ( N K=M +1 Sk N ? M ) = VN . Suppose vn+1 is already given. a) If n > M , then En [vn+1 (Sn+1 , Yn+1 )] = pvn+1 (uSn , Yn + uSn ) + qvn+1 (dSn , Yn + dSn ). So vn (s, y) = pvn+1 (us, y + us) + qvn+1 (ds, y + ds). b) If n = M , then EM [vM +1 (SM +1 , YM +1 )] = pvM +1 (uSM , uSM ) + vn+1 (dSM , dSM ). So vM (s) = pvM +1 (us, us) + qvM +1 (ds, ds). c) If n < M , then En [vn+1 (Sn+1 )] = pvn+1 (uSn ) + qvn+1 (dSn ). So vn (s) = pvn+1 (us) + qvn+1 (ds). 3. State Prices 3. 1. Proof. Note Z(? ) := P (? ) P (? ) = 1 Z(? ) . Apply Theorem 3. 1. 1 with P , P , Z replaced by P , P , Z, we get the nalogous of properties (i)-(iii) of Theorem 3. 1. 1. 3. 2. (i) Proof. P (? ) = (ii) Proof. E[Y ] = (iii) ? Proof. P (A) = (iv) Proof. If P (A) = A Z(? )P (? ) = 0, by P (Z > 0) = 1, we conclude P (? ) = 0 for any ? ? A. So P (A) = A P (? ) = 0. (v) Proof. P (A) = 1 P (Ac ) = 0 P (Ac ) = 0 P (A) = 1. (vi) A P (? ) = Z(? )P (? ) = E[Z] = 1. Y (? )P (? ) = Y (? )Z(? )P (? ) = E[Y Z]. Z(? )P (? ). Since P (A) = 0, P (? ) = 0 for any ? ? A. So P (A) = 0. 8 Proof. Pick ? 0 such that P (? 0 ) > 0, de? ne Z(? ) = 1 P (? 0 ) 0, 1 P (? 0 ) , if ? = ? 0 Then P (Z ? 0) = 1 and E[Z] = if ? = ? 0 .  · P (? 0 ) = 1. =? 0 Clearly P (? {? 0 }) = E[Z1? {? 0 } ] = Z(? )P (? ) = 0. But P (? {? 0 }) = 1 ? P (? 0 ) > 0 if P (? 0 ) < 1. Hence in the case 0 < P (? 0 ) < 1, P and P are not equivalent. If P (? 0 ) = 1, then E[Z] = 1 if and only if Z(? 0 ) = 1. In this case P (? 0 ) = Z(? 0 )P (? 0 ) = 1. And P and P have to be equivalent. In summary, if we can ? nd ? 0 such that 0 < P (? 0 ) < 1, then Z as constructed above would induce a probability P that is not equivalent to P . 3. 5. (i) Proof. Z(HH) = (ii) Proof. Z1 (H) = E1 [Z2 ](H) = Z2 (HH)P (? 2 = H|? 1 = H) + Z2 (HT )P (? 2 = T |? 1 = H) = 3 E1 [Z2 ](T ) = Z2 (T H)P (? 2 = H|? = T ) + Z2 (T T )P (? 2 = T |? 1 = T ) = 2 . (iii) Proof. V1 (H) = [Z2 (HH)V2 (HH)P (? 2 = H|? 1 = H) + Z2 (HT )V2 (HT )P (? 2 = T |? 1 = T )] = 2. 4, Z1 (H)(1 + r1 (H)) [Z2 (T H)V2 (T H)P (? 2 = H|? 1 = T ) + Z2 (T T )V2 (T T )P (? 2 = T |? 1 = T )] 1 = , Z1 (T )(1 + r1 (T )) 9 3 4. 9 16 , Z(HT ) = 9 , Z(T H) = 8 3 8 and Z(T T ) = 15 4 . Z1 (T ) = V1 (T ) = and V0 = Z2 (HH)V2 (HH) Z2 (HT )V2 (HT ) Z2 (T H)V2 (T H) P (HH) + P (T H) + 0 ? 1. 1 1 1 1 P (HT ) + 1 (1 + 4 )(1 + 4 ) (1 + 4 )(1 + 4 ) (1 + 4 )(1 + 1 ) 2 3. 6. Proof. U (x) = have XN = 1 x, (1+r)N ? Z so I(x) = = 1 Z] 1 x. Z (3. 3. 26) gives E[ (1+r)N 1 X0 (1 + r)n Zn En [Z  ·X0 N Z (1 + r) . 0 = Xn , where ? Hence Xn = (1+r)N ? Z X En [ (1+r)N ? n ] N ] = X0 . So ? = = En [ X0 (1+r) Z n 1 X0 . By (3. 3. 25), we 1 ] = X0 (1 + r)n En [ Z ] = the second to last â€Å"=† comes from Lemma 3. 2. 6. 3. 7. Z ? Z Proof. U (x) = xp? 1 and so I(x) = x p? 1 . By (3. 3. 26), we have E[ (1+r)N ( (1+r)N ) p? 1 ] = X0 . Solve it for ? , we get ? ?p? 1 1 1 ? ? =? ? X0 p E 1 Z p? 1 Np ? ? ? = p? 1 X0 (1 + r)N p (E[Z p? 1 ])p? 1 1 p . (1+r) p? 1 ? Z So by (3. 3. 25), XN = ( (1+r)N ) p? 1 = 1 1 Np ? p? 1 Z p? 1 N (1+r) p? 1 = X0 (1+r) p? 1 E[Z p p? 1 Z p? 1 N (1+r) p? 1 = (1+r)N X0 Z p? 1 E[Z p p? 1 1 . ] ] 3. 8. (i) 9 d d Proof. x (U (x) ? yx) = U (x) ? y. So x = I(y) is an extreme point of U (x) ? yx. Because dx2 (U (x) ? yx) = U (x) ? 0 (U is concave), x = I(y) is a maximum point. Therefore U (x) ? y(x) ? U (I(y)) ? yI(y) for every x. 2 (ii) Proof. Following the hint of the problem, we have E[U (XN )] ? E[XN ? Z ? Z ? Z ? Z ] ? E[U (I( ))] ? E[ I( )], N N N (1 + r) (1 + r) (1 + r) (1 + r)N ? ? ? ? ? i. e. E[U (XN )] ? ?X0 ? E[U (XN )] ? E[ (1+r)N XN ] = E[U (XN )] ? ?X0 . So E[U (XN )] ? E[U (XN )]. 3. 9. (i) X Proof. Xn = En [ (1+r)N ? n ]. So if XN ? 0, then Xn ? 0 for all n. N (ii) 1 Proof. a) If 0 ? x < ? and 0 < y ? ? , then U (x) ? yx = ? yx ? and U (I(y)) ? yI(y) = U (? ) ? y? = 1 ? y? ? 0. So U (x) ? yx ? U (I(y)) ? yI(y). 1 b) If 0 ? x < ? and y > ? , then U (x) ? yx = ? yx ? 0 and U (I(y)) ? yI(y) = U (0) ? y  · 0 = 0. So U (x) ? yx ? U (I(y)) ? yI(y). 1 c) If x ? ? and 0 < y ? ? , then U (x) ? yx = 1 ? yx and U (I(y)) ? yI(y) = U (? ) ? y? = 1 ? y? ? 1 ? yx. So U (x) ? yx ? U (I(y)) ? yI(y). 1 d) If x ? ? and y > ? , then U (x) ? yx = 1 ? yx < 0 and U (I(y)) ? yI(y) = U (0) ? y  · 0 = 0. So U (x) ? yx ? U (I(y)) ? yI(y). (iii) XN ? Z Proof. Using (ii) and set x = XN , y = (1+r)N , where XN is a random variable satisfying E[ (1+r)N ] = X0 , we have ?Z ? Z ? E[U (XN )] ? E[ XN ] ? E[U (XN )] ? E[ X ? ]. (1 + r)N (1 + r)N N ? ? That is, E[U (XN )] ? ?X0 ? E[U (XN )] ? ?X0 . So E[U (XN )] ? E[U (XN )]. (iv) Proof. Plug pm and ? m into (3. 6. 4), we have 2N 2N X0 = m=1 pm ? m I( m ) = m=1 1 pm ? m ? 1{ m ? ? } . So X0 ? X0 ? {m : = we are looking for positive solution ? > 0). Conversely, suppose there exists some K so that ? K < ? K+1 and K X0 1 m=1 ? m pm = ? . Then we can ? nd ? > 0, such that ? K < < ? K+1 . For such ? , we have Z ? Z 1 E[ I( )] = pm ? m 1{ m ? ? } ? = pm ? m ? = X0 . N (1 + r) (1 + r)N m=1 m=1 Hence (3. 6. 4) has a solution. 0 2N K 2N X0 1 m=1 pm ? m 1{ m ? ? } . Suppose there is a solution ? to (3. 6. 4), note ? > 0, we then can conclude 1 1 1 m ? ? } = ?. Let K = max{m : m ? ? }, then K ? ? < K+1 . So ? K < ? K+1 and K N m=1 pm ? m (Note, however, that K could be 2 . In this case, ? K+1 is interpreted as ?. Also, note = (v) ? 1 Proof. XN (? m ) = I( m ) = ? 1{ m ? ? } = ?, if m ? K . 0, if m ? K + 1 4. American Derivative Securities Before proceeding to the exercise problems, we ? rst give a brief summary of pricing American derivative securities as presented in the textbook. We shall use the notation of the book.From the buyer’s perspective: At time n, if the derivative security has not been exercised, then the buyer can choose a policy ? with ? ? Sn . The valuation formula for cash ? ow (Theorem 2. 4. 8) gives a fair price for the derivative security exercised according to ? : N Vn (? ) = k=n En 1{? =k} 1 1 Gk = En 1{? ?N } G? . (1 + r)k? n (1 + r)? ?n The buyer wants to consider all the possible ? ’s, so that he c an ? nd the least upper bound of security value, which will be the maximum price of the derivative security acceptable to him. This is the price given by 1 De? nition 4. 4. 1: Vn = max? ?Sn En [1{? ?N } (1+r)? n G? ]. From the seller’s perspective: A price process (Vn )0? n? N is acceptable to him if and only if at time n, he can construct a portfolio at cost Vn so that (i) Vn ? Gn and (ii) he needs no further investing into the portfolio as time goes by. Formally, the seller can ? nd (? n )0? n? N and (Cn )0? n? N so that Cn ? 0 and Sn Vn+1 = ? n Sn+1 + (1 + r)(Vn ? Cn ? ?n Sn ). Since ( (1+r)n )0? n? N is a martingale under the risk-neutral measure P , we conclude En Cn Vn+1 Vn =? ? 0, ? n+1 n (1 + r) (1 + r) (1 + r)n Vn i. e. ( (1+r)n )0? n? N is a supermartingale. This inspired us to check if the converse is also true.This is exactly the content of Theorem 4. 4. 4. So (Vn )0? n? N is the value process of a portfolio that needs no further investing if and only if Vn (1+r)n Vn (1+r)n is a supermartingale under P (note this is independent of the requirement 0? n? N Vn ? Gn ). In summary, a price process (Vn )0? n? N is acceptable to the seller if and only if (i) Vn ? Gn ; (ii) is a supermartingale under P . 0? n? N Theorem 4. 4. 2 shows the buyer’s upper bound is the seller’s lower bound. So it gives the price acceptable to both. Theorem 4. 4. 3 gives a speci? c algorithm for calculating the price, Theorem 4. 4. establishes the one-to-one correspondence between super-replication and supermartingale property, and ? nally, Theorem 4. 4. 5 shows how to decide on the optimal exercise policy. 4. 1. (i) Proof. V2P (HH) = 0, V2P (HT ) = V2P (T H) = 0. 8, V2P (T T ) = 3, V1P (H) = 0. 32, V1P (T ) = 2, V0P = 9. 28. (ii) Proof. V0C = 5. (iii) Proof. gS (s) = |4 ? s|. We apply Theorem 4. 4. 3 and have V2S (HH) = 12. 8, V2S (HT ) = V2S (T H) = 2. 4, V2S (T T ) = 3, V1S (H) = 6. 08, V1S (T ) = 2. 16 and V0S = 3. 296. (iv) 11 Proof. First, we note the simple inequality max(a1 , b1 ) + max(a2 , b2 ) ? max(a1 + a2 , b1 + b2 ). >† holds if and only if b1 > a1 , b2 < a2 or b1 < a1 , b2 > a2 . By induction, we can show S Vn = max gS (Sn ), S S pVn+1 + Vn+1 1+r C P P pV C + Vn+1 pVn+1 + Vn+1 + n+1 1+r 1+r C C pVn+1 + Vn+1 1+r ? max gP (Sn ) + gC (Sn ), ? max gP (Sn ), P C = Vn + Vn . P P pVn+1 + Vn+1 1+r + max gC (Sn ), S P C As to when â€Å" C C pVn+1 +qVn+1 1+r or gP (Sn ) > P P pVn+1 +qVn+1 1+r and gC (Sn ) < C C pVn+1 +qVn+1 }. 1+r 4. 2. Proof. For this problem, we need Figure 4. 2. 1, Figure 4. 4. 1 and Figure 4. 4. 2. Then ? 1 (H) = and ? 0 = V2 (HH) ? V2 (HT ) 1 V2 (T H) ? V2 (T T ) = ? , ? 1 (T ) = = ? 1, S2 (HH) ? S2 (HT ) 12 S2 (T H) ?S2 (T T ) V1 (H) ? V1 (T ) ? ?0. 433. S1 (H) ? S1 (T ) The optimal exercise time is ? = inf{n : Vn = Gn }. So ? (HH) = ? , ? (HT ) = 2, ? (T H) = ? (T T ) = 1. Therefore, the agent borrows 1. 36 at time zero and buys the put. At the same time, to hedge the long position, he needs to borr ow again and buy 0. 433 shares of stock at time zero. At time one, if the result of coin toss is tail and the stock price goes down to 2, the value of the portfolio 1 is X1 (T ) = (1 + r)(? 1. 36 ? 0. 433S0 ) + 0. 433S1 (T ) = (1 + 4 )(? 1. 36 ? 0. 433 ? 4) + 0. 433 ? 2 = ? 3. The agent should exercise the put at time one and get 3 to pay o? is debt. At time one, if the result of coin toss is head and the stock price goes up to 8, the value of the portfolio 1 is X1 (H) = (1 + r)(? 1. 36 ? 0. 433S0 ) + 0. 433S1 (H) = ? 0. 4. The agent should borrow to buy 12 shares of stock. At time two, if the result of coin toss is head and the stock price goes up to 16, the value of the 1 1 portfolio is X2 (HH) = (1 + r)(X1 (H) ? 12 S1 (H)) + 12 S2 (HH) = 0, and the agent should let the put expire. If at time two, the result of coin toss is tail and the stock price goes down to 4, the value of the portfolio is 1 1 X2 (HT ) = (1 + r)(X1 (H) ? 12 S1 (H)) + 12 S2 (HT ) = ? 1.The agent should exercise the put to get 1. This will pay o? his debt. 4. 3. Proof. We need Figure 1. 2. 2 for this problem, and calculate the intrinsic value process and price process of the put as follows. 2 For the intrinsic value process, G0 = 0, G1 (T ) = 1, G2 (T H) = 3 , G2 (T T ) = 5 , G3 (T HT ) = 1, 3 G3 (T T H) = 1. 75, G3 (T T T ) = 2. 125. All the other outcomes of G is negative. 12 2 5 For the price process, V0 = 0. 4, V1 (T ) = 1, V1 (T H) = 3 , V1 (T T ) = 3 , V3 (T HT ) = 1, V3 (T T H) = 1. 75, V3 (T T T ) = 2. 125. All the other outcomes of V is zero. Therefore the time-zero price of the derivative security is 0. and the optimal exercise time satis? es ? (? ) = ? if ? 1 = H, 1 if ? 1 = T . 4. 4. Proof. 1. 36 is the cost of super-replicating the American derivative security. It enables us to construct a portfolio su? cient to pay o? the derivative security, no matter when the derivative security is exercised. So to hedge our short position after selling the put, there is no need to charge t he insider more than 1. 36. 4. 5. Proof. The stopping times in S0 are (1) ? ? 0; (2) ? ? 1; (3) ? (HT ) = ? (HH) = 1, ? (T H), ? (T T ) ? {2, ? } (4 di? erent ones); (4) ? (HT ), ? (HH) ? {2, ? }, ? (T H) = ? (T T ) = 1 (4 di? rent ones); (5) ? (HT ), ? (HH), ? (T H), ? (T T ) ? {2, ? } (16 di? erent ones). When the option is out of money, the following stopping times do not exercise (i) ? ? 0; (ii) ? (HT ) ? {2, ? }, ? (HH) = ? , ? (T H), ? (T T ) ? {2, ? } (8 di? erent ones); (iii) ? (HT ) ? {2, ? }, ? (HH) = ? , ? (T H) = ? (T T ) = 1 (2 di? erent ones). ? 4 For (i), E[1{? ?2} ( 4 )? G? ] = G0 = 1. For (ii), E[1{? ?2} ( 5 )? G? ] ? E[1{? ? ? 2} ( 4 )? G? ? ], where ? ? (HT ) = 5 5 1 4 4 ? 2, ? ? (HH) = ? , ? ? (T H) = ? ? (T T ) = 2. So E[1{? ? ? 2} ( 5 )? G? ? ] = 4 [( 4 )2  · 1 + ( 5 )2 (1 + 4)] = 0. 96. For 5 (iii), E[1{? ?2} ( 4 )? G? has the biggest value when ? satis? es ? (HT ) = 2, ? (HH) = ? , ? (T H) = ? (T T ) = 1. 5 This value is 1. 36. 4. 6. (i) Proof. The value of the put at time N , if it is not exercised at previous times, is K ? SN . Hence VN ? 1 = VN K max{K ? SN ? 1 , EN ? 1 [ 1+r ]} = max{K ? SN ? 1 , 1+r ? SN ? 1 } = K ? SN ? 1 . The second equality comes from the fact that discounted stock price process is a martingale under risk-neutral probability. By induction, we can show Vn = K ? Sn (0 ? n ? N ). So by Theorem 4. 4. 5, the optimal exercise policy is to sell the stock at time zero and the value of this derivative security is K ?S0 . Remark: We cheated a little bit by using American algorithm and Theorem 4. 4. 5, since they are developed for the case where ? is allowed to be ?. But intuitively, results in this chapter should still hold for the case ? ? N , provided we replace â€Å"max{Gn , 0}† with â€Å"Gn †. (ii) Proof. This is because at time N , if we have to exercise the put and K ? SN < 0, we can exercise the European call to set o? the negative payo?. In e? ect, throughout the portfolio’s lifetime, the portfolio has intrinsic values greater than that of an American put stuck at K with expiration time N . So, we must have V0AP ? V0 + V0EC ? K ?S0 + V0EC . (iii) 13 Proof. Let V0EP denote the time-zero value of a European put with strike K and expiration time N . Then V0AP ? V0EP = V0EC ? E[ K SN ? K ] = V0EC ? S0 + . (1 + r)N (1 + r)N 4. 7. VN K K Proof. VN = SN ? K, VN ? 1 = max{SN ? 1 ? K, EN ? 1 [ 1+r ]} = max{SN ? 1 ? K, SN ? 1 ? 1+r } = SN ? 1 ? 1+r . K By induction, we can prove Vn = Sn ? (1+r)N ? n (0 ? n ? N ) and Vn > Gn for 0 ? n ? N ? 1. So the K time-zero value is S0 ? (1+r)N and the optimal exercise time is N . 5. Random Walk 5. 1. (i) Proof. E[ 2 ] = E[? (? 2 1 )+? 1 ] = E[? (? 2 1 ) ]E[ 1 ] = E[ 1 ]2 . (ii) Proof. If we de? ne Mn = Mn+? ? M? m (m = 1, 2,  ·  ·  · ), then (M · )m as random functions are i. i. d. with (m) distributions the same as that of M . So ? m+1 ? ?m = inf{n : Mn = 1} are i. i. d. with distributions the same as that of ? 1 . Therefore E [ m ] = E[? (? m m? 1 )+(? m? 1 m? 2 )+ ·Ã‚ ·Ã‚ ·+? 1 ] = E[ 1 ]m . (m) (m) (iii) Proof. Yes, since the argument of (ii) still works for asymmetric random walk. 5. 2. (i) Proof. f (? ) = pe? ? qe , so f (? ) > 0 if and only if ? > f (? ) > f (0) = 1 for all ? > 0. (ii) 1 1 1 n+1 Proof. En [ SSn ] = En [e? Xn+1 f (? ) ] = pe? f (? ) + qe f (? ) = 1. 1 2 (ln q ? ln p). Since 1 2 (ln q ln p) < 0, (iii) 1 Proof. By optional stopping theorem, E[Sn 1 ] = E[S0 ] = 1. Note Sn 1 = e? Mn 1 ( f (? ) )n 1 ? e?  ·1 , by bounded convergence theorem, E[1{? 1 1 for all ? > ? 0 . v (ii) 1 1 Proof. As in Exercise 5. 2, Sn = e? Mn ( f (? ) )n is a martingale, and 1 = E[S0 ] = E[Sn 1 ] = E[e? Mn 1 ( f (? ) )? 1 ? n ]. Suppose ? > ? 0 , then by bounded convergence theorem, 1 = E[ lim e? Mn 1 ( n>? 1 n 1 1 ? 1 ) ] = E[1{? 1 K} ] = P (ST > K). Moreover, by Girsanov’s Theorem, Wt = Wt + in Theorem 5. 4. 1. ) (iii) Proof. ST = xe? WT +(r? 2 ? 1 2 1 2 t ( )du 0 = Wt ? ?t is a P -Brownian motion (set ? )T = xe? WT +(r+ 2 ? 1 2 1 2 )T . So WT v > ? d+ (T, x) T = N (d+ (T, x)). P (ST > K) = P (xe? WT +(r+ 2 ? )T > K) = P 46 5. 4. First, a few typos. In the SDE for S, â€Å"? (t)dW (t)† > â€Å"? (t)S(t)dW (t)†. In the ? rst equation for c(0, S(0)), E > E. In the second equation for c(0, S(0)), the variable for BSM should be ? ? 1 T 2 1 T r(t)dt, ? (t)dt? . BSM ? T, S(0); K, T 0 T 0 (i) Proof. d ln St = X = ? is a Gaussian with X ? N ( (ii) Proof. For the standard BSM model with constant volatility ? and interest rate R, under the risk-neutral measure, we have ST = S0 eY , where Y = (R? 1 ? 2 )T +? WT ? N ((R? 1 ? )T, ? 2 T ), and E[(S0 eY ? K)+ ] = 2 2 eRT BSM (T, S0 ; K, R, ? ). Note R = 1 T (rt 0 T T dSt 1 2 1 1 2 2 St ? 2St d S t = rt dt + ? t dWt ? 2 ? t dt. So ST = S0 exp{ 0 (rt ? 2 ? t )dt + 0 T 1 2 2 ? t )dt + 0 ? t dWt . The ? rst term in the expression of X is a number and the T 2 random variable N (0, 0 ? t dt), since both r and ? ar deterministic. Th erefore, T T 2 2 (rt ? 1 ? t )dt, 0 ? t dt),. 2 0 ?t dWt }. Let second term ST = S0 eX , 1 T (E[Y ] + 1 V ar(Y )) and ? = 2 T, S0 ; K, 1 T 1 T V ar(Y ), we can get 1 V ar(Y ) . T E[(S0 eY ? K)+ ] = eE[Y ]+ 2 V ar(Y ) BSM So for the model in this problem, c(0, S0 ) = = e? ? T 0 1 E[Y ] + V ar(Y ) , 2 rt dt E[(S0 eX ? K)+ ] e BSM T, S0 ; K, 1 T T 0 T 0 1 rt dt E[X]+ 2 V ar(X) 1 T ? 1 E[X] + V ar(X) , 2 1 V ar(X) T ? = 1 BSM ? T, S0 ; K, T 0 T rt dt, 2 ? t dt? . 5. 5. (i) 1 1 Proof. Let f (x) = x , then f (x) = ? x2 and f (x) = 2 x3 . Note dZt = ? Zt ? t dWt , so d 1 Zt 1 1 1 2 2 2 ? t ? 2 t = f (Zt )dZt + f (Zt )dZt dZt = ? 2 (? Zt )? t dWt + 3 Zt ? t dt = Z dWt + Z dt. 2 Zt 2 Zt t t (ii) Proof. By Lemma 5. 2. 2. , for s, t ? 0 with s < t, Ms = E[Mt |Fs ] = E Zs Ms . So M = Z M is a P -martingale. (iii) Zt Mt Zs |Fs . That is, E[Zt Mt |Fs ] = 47 Proof. dMt = d Mt  · 1 Zt = 1 1 1 ? M t ? t M t ? 2 ? t ? t t dMt + Mt d + dMt d = dWt + dWt + dt + dt. Zt Zt Zt Zt Zt Zt Zt (iv) Proof. In part (iii), we have dMt = Let ? t = 5. 6. Proof. By Theorem 4. 6. 5, it su? ces to show Wi (t) is an Ft -martingale under P and [Wi , Wj ](t) = t? ij (i, j = 1, 2). Indeed, for i = 1, 2, Wi (t) is an Ft -martingale under P if and only if Wi (t)Zt is an Ft -martingale under P , since Wi (t)Zt E[Wi (t)|Fs ] = E |Fs . Zs By It? ’s product formula, we have o d(Wi (t)Zt ) = Wi (t)dZt + Zt dWi (t) + dZt dWi (t) = Wi (t)(? Zt )? (t)  · dWt + Zt (dWi (t) + ? i (t)dt) + (? Zt ? t  · dWt )(dWi (t) + ? i (t)dt) d t M t ? t M t ? 2 ? t ? t ? t M t ? t t dWt + dWt + dt + dt = (dWt + ? t dt) + (dWt + ? t dt). Zt Zt Zt Zt Zt Zt then dMt = ? t dWt . This proves Corollary 5. 3. 2. ?t +Mt ? t , Zt = Wi (t)(? Zt ) j=1 d ?j (t)dWj (t) + Zt (dWi (t) + ? i (t)dt) ? Zt ? i (t)dt = Wi (t)(? Zt ) j=1 ?j (t)dWj (t) + Zt dWi (t) This shows Wi (t)Zt is an Ft -martingale under P . So Wi (t) is an Ft -martingale under P . Moreover,  ·  · [Wi , Wj ](t) = Wi + 0 ?i (s)ds, Wj + 0 ?j (s)ds (t) = [Wi , Wj ](t) = t? ij . Combined, this proves the two-dimensional Girsanov’s Theorem. 5. 7. (i) Proof. Let a be any strictly positive number. We de? e X2 (t) = (a + X1 (t))D(t)? 1 . Then P X2 (T ) ? X2 (0) D(T ) = P (a + X1 (T ) ? a) = P (X1 (T ) ? 0) = 1, and P X2 (T ) > X2 (0) = P (X1 (T ) > 0) > 0, since a is arbitrary, we have proved the claim of this problem. D(T ) Remark: The intuition is that we invest the positive starting fund a into the money market account, and construct portfolio X1 from zero cost. Their sum should be able to beat the return of money market account. (ii) 48 Proof. We de? ne X1 (t) = X2 (t)D(t) ? X2 (0). Then X1 (0) = 0, P (X1 (T ) ? 0) = P X2 (T ) ? X2 (0) D(T ) = 1, P (X1 (T ) > 0) = P X2 (T ) > X2 (0) D(T ) > 0. 5. 8.The basic idea is that for any positive P -martingale M , dMt = Mt  · sentation Theorem, dMt = ? t dWt for some adapted process ? t . So martingale must be the exponential of an integral w. r. t. Brownian motion. Taking into account d iscounting factor and apply It? ’s product rule, we can show every strictly positive asset is a generalized geometric o Brownian motion. (i) Proof. Vt Dt = E[e? 0 Ru du VT |Ft ] = E[DT VT |Ft ]. So (Dt Vt )t? 0 is a P -martingale. By Martingale Represent tation Theorem, there exists an adapted process ? t , 0 ? t ? T , such that Dt Vt = 0 ? s dWs , or equivalently, ? 1 t ? 1 t ? 1 Vt = Dt 0 ? dWs . Di? erentiate both sides of the equation, we get dVt = Rt Dt 0 ? s dWs dt + Dt ? t dWt , i. e. dVt = Rt Vt dt + (ii) Proof. We prove the following more general lemma. Lemma 1. Let X be an almost surely positive random variable (i. e. X > 0 a. s. ) de? ned on the probability space (? , G, P ). Let F be a sub ? -algebra of G, then Y = E[X|F] > 0 a. s. Proof. By the property of conditional expectation Yt ? 0 a. s. Let A = {Y = 0}, we shall show P (A) = 0. In? 1 1 deed, note A ? F, 0 = E[Y IA ] = E[E[X|F]IA ] = E[XIA ] = E[X1A? {X? 1} ] + n=1 E[X1A? { n >X? n+1 } ] ? 1 1 1 1 1 P (A? {X ? 1})+ n=1 n+1 P (A? n > X ? n+1 }). So P (A? {X ? 1}) = 0 and P (A? { n > X ? n+1 }) = 0, ? 1 1 ? n ? 1. This in turn implies P (A) = P (A ? {X > 0}) = P (A ? {X ? 1}) + n=1 P (A ? { n > X ? n+1 }) = 0. ? ? t Dt dWt . T 1 Mt dMt . By Martingale Repre? dMt = Mt ( Mtt )dWt , i. e. any positive By the above lemma, it is clear that for each t ? [0, T ], Vt = E[e? t Ru du VT |Ft ] > 0 a. s.. Moreover, by a classical result of martingale theory (Revuz and Yor [4], Chapter II, Proposition (3. 4)), we have the following stronger result: for a. s. ?, Vt (? ) > 0 for any t ? [0, T ]. (iii) 1 1 Proof. By (ii), V > 0 a. s. so dVt = Vt Vt dVt = Vt Vt Rt Vt dt + ? t Dt dWt ? t = Vt Rt dt + Vt Vt Dt dWt = Rt Vt dt + T ?t Vt dWt , where ? t = 5. 9. ?t Vt Dt . This shows V follows a generalized geometric Brownian motion. Proof. c(0, T, x, K) = xN (d+ ) ? Ke? rT N (d? ) with d ± = then f (y) = ? yf (y), cK (0, T, x, K) = xf (d+ ) 1 v ? T x (ln K + (r  ± 1 ? 2 )T ). Let f (y) = 2 y v1 e? 2 2? 2 , ?d+ ? d? ? e? rT N (d? ) ? Ke? rT f (d? ) ? y ? y ? 1 1 = xf (d+ ) v ? e? rT N (d? ) + e? rT f (d? ) v , ? TK ? T 49 and cKK (0, T, x, K) x ? d? e? rT 1 ? d+ d? ? v ? e? rT f (d? ) + v (? d? )f (d? ) xf (d+ ) v f (d+ )(? d+ ) 2 ? y ? y ? y ? TK ? TK ?T x xd+ ? 1 ? 1 e? rT d? ?1 v v ? e? rT f (d? ) v ? v f (d? ) v f (d+ ) + v f (d+ ) ? T K2 ? TK K? T K? T ? T K? T x d+ e? rT f (d? ) d? v [1 ? v ] + v f (d+ ) [1 + v ] 2? T K ? T K? T ? T e? rT x f (d? )d+ ? 2 2 f (d+ )d? . K? 2 T K ? T = = = = 5. 10. (i) Proof. At time t0 , the value of the chooser option is V (t0 ) = max{C(t0 ), P (t0 )} = max{C(t0 ), C(t0 ) ? F (t0 )} = C(t0 ) + max{0, ? F (t0 )} = C(t0 ) + (e? r(T ? t0 ) K ? S(t0 ))+ . (ii) Proof. By the risk-neutral pricing formula, V (0) = E[e? rt0 V (t0 )] = E[e? rt0 C(t0 )+(e? rT K ? e? rt0 S(t0 )+ ] = C(0) + E[e? rt0 (e? r(T ? t0 ) K ? S(t0 ))+ ]. The ? st term is the value of a call expiring at time T with strike price K and the second term is the value of a put expiring at time t0 with strike price e? r(T ? t0 ) K. 5. 11. Proof. We ? rst make an analysis which leads to the hint, then we give a formal proof. (Analysis) If we want to construct a portfolio X that exactly replicates the cash ? ow, we must ? nd a solution to the backward SDE dXt = ? t dSt + Rt (Xt ? ?t St )dt ? Ct dt XT = 0. Multiply Dt on both sides of the ? rst equation and apply It? ’s product rule, we get d(Dt Xt ) = ? t d(Dt St ) ? o T T Ct Dt dt. Integrate from 0 to T , we have DT XT ? D0 X0 = 0 ? d(Dt St ) ? 0 Ct Dt dt. By the terminal T T ? 1 condition, we get X0 = D0 ( 0 Ct Dt dt ? 0 ? t d(Dt St )). X0 is the theoretical, no-arbitrage price of the cash ? ow, provided we can ? nd a trading strategy ? that solves the BSDE. Note the SDE for S ? R gives d(Dt St ) = (Dt St )? t (? t dt + dWt ), where ? t = ? t? t t . Take the proper change of measure so that Wt = t ? ds 0 s + Wt is a Brownian motion under the new measure P , we get T T T Ct Dt dt = D0 X0 + 0 T 0 ?t d(Dt St ) = D0 X0 + 0 ?t (Dt St )? t dWt . T This says the random variable 0 Ct Dt dt has a stochastic integral representation D0 X0 + 0 ? t Dt St ? dWt . T This inspires us to consider the martingale generated by 0 Ct Dt dt, so that we can apply Martingale Representation Theorem and get a formula for ? by comparison of the integrands. 50 (Formal proof) Let MT = Xt = ?1 Dt (D0 X0 T 0 Ct Dt dt, and Mt = E[MT |Ft ]. Then by Martingale Representation Theot 0 rem, we can ? nd an adapted process ? t , so that Mt = M0 + + t 0 ?t dWt . If we set ? t = T 0 ?u d(Du Su ) ? t 0 ?t Dt St ? t , we can check Cu Du du), with X0 = M0 = E[ Ct Dt dt] solves the SDE dXt = ? t dSt + Rt (Xt ? ?t St )dt ? Ct dt XT = 0. Indeed, it is easy to see that X satis? es the ? rst equation.To check the terminal condition, we note T T T XT DT = D0 X0 + 0 ? t Dt St ? t dWt ? 0 Ct Dt dt = M0 + 0 ? t dWt ? MT = 0. So XT = 0. Thus, we have found a trading strategy ? , so that the corresponding portfolio X replicates the cash ? ow and has zero T terminal value. So X0 = E[ 0 Ct Dt dt] is the no-arbitrage price of the cash ? ow at time zero. Remark: As shown in the analysis, d(Dt Xt ) = ? t d(Dt St ) ? Ct Dt dt. Integrate from t to T , we get T T 0 ? Dt Xt = t ? u d(Du Su ) ? t Cu Du du. Take conditional expectation w. r. t. Ft on both sides, we get T T ? 1 ? Dt Xt = ? E[ t Cu Du du|Ft ]. So Xt = Dt E[ t Cu Du du|Ft ].This is the no-arbitrage price of the cash ? ow at time t, and we have justi? ed formula (5. 6. 10) in the textbook. 5. 12. (i) Proof. dBi (t) = dBi (t) + ? i (t)dt = martingale. Since dBi (t)dBi (t) = P. (ii) Proof. dSi (t) = = = R(t)Si (t)dt + ? i (t)Si (t)dBi (t) + (? i (t) ? R(t))Si (t)dt ? ?i (t)Si (t)? i (t)dt d d ? ij (t) ? ij (t) d d j=1 ? i (t) ? j (t)dt = j=1 ? i (t) dWj (t) + ? ij (t)2 d e j=1 ? i (t)2 dt = dt, by L? vy’s Theorem, Bi ? ij (t) d j=1 ? i (t) dWj (t). So Bi is a is a Brownian motion under R(t)Si (t)dt + ? i (t)Si (t)dBi (t) + j=1 ?ij (t)? j (t)Si (t)dt ? Si (t) j=1 ?ij (t)? j (t)dt R(t)Si (t)dt + ? (t)Si (t)dBi (t). (iii) Proof. dBi (t)dBk (t) = (dBi (t) + ? i (t)dt)(dBj (t) + ? j (t)dt) = dBi (t)dBj (t) = ? ik (t)dt. (iv) Proof. By It? ’s product rule and martingale property, o t t t E[Bi (t)Bk (t)] = E[ 0 t Bi (s)dBk (s)] + E[ 0 t Bk (s)dBi (s)] + E[ 0 dBi (s)dBk (s)] = E[ 0 ?ik (s)ds] = 0 ?ik (s)ds. t 0 Similarly, by part (iii), we can show E[Bi (t)Bk (t)] = (v) ?ik (s)ds. 51 Proof. By It? ’s product formula, o t t E[B1 (t)B2 (t)] = E[ 0 sign(W1 (u))du] = 0 [P (W1 (u) ? 0) ? P (W1 (u) < 0)]du = 0. Meanwhile, t E[B1 (t)B2 (t)] = E[ 0 t sign(W1 (u))du [P (W1 (u) ? 0) ? P (W1 (u) < 0)]du = 0 t = 0 t [P (W1 (u) ? ) ? P (W1 (u) < u)]du 2 0 = < 0, 1 ? P (W1 (u) < u) du 2 for any t > 0. So E[B1 (t)B2 (t)] = E[B1 (t)B2 (t)] for all t > 0. 5. 13. (i) Proof. E[W1 (t)] = E[W1 (t)] = 0 and E[W2 (t)] = E[W2 (t) ? (ii) Proof. Cov[W1 (T ), W2 (T )] = E[W1 (T )W2 (T )] T T t 0 W1 (u)du] = 0, for all t ? [0, T ]. = E 0 T W1 (t)dW2 (t) + 0 W2 ( t)dW1 (t) T = E 0 W1 (t)(dW2 (t) ? W1 (t)dt) + E 0 T W2 (t)dW1 (t) = ? E 0 T W1 (t)2 dt tdt = ? 0 1 = ? T 2. 2 5. 14. Equation (5. 9. 6) can be transformed into d(e? rt Xt ) = ? t [d(e? rt St ) ? ae? rt dt] = ? t e? rt [dSt ? rSt dt ? adt]. So, to make the discounted portfolio value e? t Xt a martingale, we are motivated to change the measure t in such a way that St ? r 0 Su du? at is a martingale under the new measure. To do this, we note the SDE for S is dSt = ? t St dt+? St dWt . Hence dSt ? rSt dt? adt = [(? t ? r)St ? a]dt+? St dWt = ? St Set ? t = (? t ? r)St ? a ? St (? t ? r)St ? a dt ? St + dWt . and Wt = t ? ds 0 s + Wt , we can ? nd an equivalent probability measure P , under which S satis? es the SDE dSt = rSt dt + ? St dWt + adt and Wt is a BM. This is the rational for formula (5. 9. 7). This is a good place to pause and think about the meaning of â€Å"martingale measure. † What is to be a martingale?The new measure P should be such that the discounted value pro cess of the replicating 52 portfolio is a martingale, not the discounted price process of the underlying. First, we want Dt Xt to be a martingale under P because we suppose that X is able to replicate the derivative payo? at terminal time, XT = VT . In order to avoid arbitrage, we must have Xt = Vt for any t ? [0, T ]. The di? culty is how to calculate Xt and the magic is brought by the martingale measure in the following line of reasoning: ? 1 ? 1 Vt = Xt = Dt E[DT XT |Ft ] = Dt E[DT VT |Ft ]. You can think of martingale measure as a calculational convenience.That is all about martingale measure! Risk neutral is a just perception, referring to the actual e? ect of constructing a hedging portfolio! Second, we note when the portfolio is self-? nancing, the discounted price process of the underlying is a martingale under P , as in the classical Black-Scholes-Merton model without dividends or cost of carry. This is not a coincidence. Indeed, we have in this case the relation d(Dt Xt ) = ? t d(Dt St ). So Dt Xt being a martingale under P is more or less equivalent to Dt St being a martingale under P . However, when the underlying pays dividends, or there is cost of carry, d(Dt Xt ) = ? d(Dt St ) no longer holds, as shown in formula (5. 9. 6). The portfolio is no longer self-? nancing, but self-? nancing with consumption. What we still want to retain is the martingale property of Dt Xt , not that of Dt St . This is how we choose martingale measure in the above paragraph. Let VT be a payo? at time T , then for the martingale Mt = E[e? rT VT |Ft ], by Martingale Representation rt t Theorem, we can ? nd an adapted process ? t , so that Mt = M0 + 0 ? s dWs . If we let ? t = ? t e t , then the ? S value of the corresponding portfolio X satis? es d(e? rt Xt ) = ? t dWt . So by setting X0 = M0 = E[e? T VT ], we must have e? rt Xt = Mt , for all t ? [0, T ]. In particular, XT = VT . Thus the portfolio perfectly hedges VT . This justi? es the risk-neutral pricing of Europea n-type contingent claims in the model where cost of carry exists. Also note the risk-neutral measure is di? erent from the one in case of no cost of carry. Another perspective for perfect replication is the following. We need to solve the backward SDE dXt = ? t dSt ? a? t dt + r(Xt ? ?t St )dt XT = VT for two unknowns, X and ?. To do so, we ? nd a probability measure P , under which e? rt Xt is a martingale, t then e? rt Xt = E[e? T VT |Ft ] := Mt . Martingale Representation Theorem gives Mt = M0 + 0 ? u dWu for some adapted process ?. This would give us a theoretical representation of ? by comparison of integrands, hence a perfect replication of VT . (i) Proof. As indicated in the above analysis, if we have (5. 9. 7) under P , then d(e? rt Xt ) = ? t [d(e? rt St ) ? ae? rt dt] = ? t e? rt ? St dWt . So (e? rt Xt )t? 0 , where X is given by (5. 9. 6), is a P -martingale. (ii) 1 1 Proof. By It? ’s formula, dYt = Yt [? dWt + (r ? 2 ? 2 )dt] + 2 Yt ? 2 dt = Yt (? dWt + rdt). So d(e? rt Yt ) = o t a ? e? rt Yt dWt and e? rt Yt is a P -martingale.Moreover, if St = S0 Yt + Yt 0 Ys ds, then t dSt = S0 dYt + 0 a dsdYt + adt = Ys t S0 + 0 a ds Yt (? dWt + rdt) + adt = St (? dWt + rdt) + adt. Ys This shows S satis? es (5. 9. 7). Remark: To obtain this formula for S, we ? rst set Ut = e? rt St to remove the rSt dt term. The SDE for U is dUt = ? Ut dWt + ae? rt dt. Just like solving linear ODE, to remove U in the dWt term, we consider Vt = Ut e Wt . It? ’s product formula yields o dVt = = e Wt dUt + Ut e Wt 1 ( )dWt + ? 2 dt + dUt  · e Wt 2 1 ( )dWt + ? 2 dt 2 1 e Wt ae? rt dt ? ? 2 Vt dt. 2 53 Note V appears only in the dt term, so multiply the integration factor e 2 ? e get 1 2 1 2 d(e 2 ? t Vt ) = ae? rt Wt + 2 ? t dt. Set Yt = e? Wt +(r? 2 ? (iii) Proof. t 1 2 1 2 t on both sides of the equation, )t , we have d(St /Yt ) = adt/Yt . So St = Yt (S0 + t ads ). 0 Ys E[ST |Ft ] = S0 E[YT |Ft ] + E YT 0 t a ds + YT Ys T t T a ds|Ft Ys E YT |Ft ds Ys E[YT ? s ]ds t = S0 E[YT |Ft ] + 0 a dsE[YT |Ft ] + a Ys t t T = S0 Yt E[YT ? t ] + 0 t a dsYt E[YT ? t ] + a Ys T t = = S0 + 0 t a ds Yt er(T ? t) + a Ys ads Ys er(T ? s) ds S0 + 0 a Yt er(T ? t) ? (1 ? er(T ? t) ). r In particular, E[ST ] = S0 erT ? a (1 ? erT ). r (iv) Proof. t dE[ST |Ft ] = aer(T ? t) dt + S0 + 0 t ads Ys a (er(T ? ) dYt ? rYt er(T ? t) dt) + er(T ? t) (? r)dt r = S0 + 0 ads Ys er(T ? t) ? Yt dWt . So E[ST |Ft ] is a P -martingale. As we have argued at the beginning of the solution, risk-neutral pricing is valid even in the presence of cost of carry. So by an argument similar to that of  §5. 6. 2, the process E[ST |Ft ] is the futures price process for the commodity. (v) Proof. We solve the equation E[e? r(T ? t) (ST ? K)|Ft ] = 0 for K, and get K = E[ST |Ft ]. So F orS (t, T ) = F utS (t, T ). (vi) Proof. We follow the hint. First, we solve the SDE dXt = dSt ? adt + r(Xt ? St )dt X0 = 0. By our analysis in part (i), d(e? t Xt ) = d(e? rt St ) ? ae? rt dt. Integrate fr om 0 to t on both sides, we get Xt = St ? S0 ert + a (1 ? ert ) = St ? S0 ert ? a (ert ? 1). In particular, XT = ST ? S0 erT ? a (erT ? 1). r r r Meanwhile, F orS (t, T ) = F uts (t, T ) = E[ST |Ft ] = S0 + t ads 0 Ys Yt er(T ? t) ? a (1? er(T ? t) ). So F orS (0, T ) = r S0 erT ? a (1 ? erT ) and hence XT = ST ? F orS (0, T ). After the agent delivers the commodity, whose value r is ST , and receives the forward price F orS (0, T ), the portfolio has exactly zero value. 54 6. Connections with Partial Di? erential Equations 6. 1. (i) Proof. Zt = 1 is obvious.Note the form of Z is similar to that of a geometric Brownian motion. So by It? ’s o formula, it is easy to obtain dZu = bu Zu du + ? u Zu dWu , u ? t. (ii) Proof. If Xu = Yu Zu (u ? t), then Xt = Yt Zt = x  · 1 = x and dXu = = = = Yu dZu + Zu dYu + dYu Zu au ? ?u ? u ? u du + dWu Zu Zu [Yu bu Zu + (au ? ?u ? u ) + ? u ? u ]du + (? u Zu Yu + ? u )dWu Yu (bu Zu du + ? u Zu dWu ) + Zu (bu Xu + au )du + (? u Xu + ? u )dWu . + ? u Z u ? u du Zu Remark: To see how to ? nd the above solution, we manipulate the equation (6. 2. 4) as follows. First, to u remove the term bu Xu du, we multiply on both sides of (6. 2. 4) the integrating factor e? bv dv . Then d(Xu e? ? Let Xu = e? u t u t bv dv ) = e? u t bv dv (au du + (? u + ? u Xu )dWu ). u t bv dv Xu , au = e? ? u t bv dv au and ? u = e? ? bv dv ? ? u , then X satis? es the SDE ? ? ? dXu = au du + (? u + ? u Xu )dWu = (? u du + ? u dWu ) + ? u Xu dWu . ? ? a ? ? ? ? To deal with the term ? u Xu dWu , we consider Xu = Xu e? ? dXu = e? u t u t ?v dWv . Then ?v dWv ?v dWv ? ? [(? u du + ? u dWu ) + ? u Xu dWu ] + Xu e? a ? u t u t 1 ( u )dWu + e? 2 u t ?v dWv 2 ? u du ? +(? u + ? u Xu )( u )e? ? ?v dWv du 1 ? 2 ? ? ? = au du + ? u dWu + ? u Xu dWu ? ?u Xu dWu + Xu ? u du ? ?u (? u + ? u Xu )du ? ? ? 1 ? 2 = (? u ? ?u ? u ? Xu ? u )du + ? u dWu , a ? ? 2 where au = au e? ? ? ? 1 d Xu e 2 u t ?v dWv 2 ? v dv and ? u = ? u e? ? ? = e2 1 u t 2 ? v dv u t ?v dWv . Finally, use the integrating factor e u t 2 ? v dv u 1 2 ? dv t 2 v , we have u t 1 ? ? 1 2 (dXu + Xu  · ? u du) = e 2 2 [(? u ? ?u ? u )du + ? u dWu ]. a ? ? Write everything back into the original X, a and ? , we get d Xu e? i. e. d u t bv dv? u t 1 ? v dWv + 2 u t 2 ? v dv = e2 1 u t 2 ? v dv? u t ?v dWv ? u t bv dv [(au ? ?u ? u )du + ? u dWu ], Xu Zu = 1 [(au ? ?u ? u )du + ? u dWu ] = dYu . Zu This inspired us to try Xu = Yu Zu . 6. 2. (i) 55 Proof.The portfolio is self-? nancing, so for any t ? T1 , we have dXt = ? 1 (t)df (t, Rt , T1 ) + ? 2 (t)df (t, Rt , T2 ) + Rt (Xt ? ?1 (t)f (t, Rt , T1 ) ? ?2 (t)f (t, Rt , T2 ))dt, and d(Dt Xt ) = ? Rt Dt Xt dt + Dt dXt = Dt [? 1 (t)df (t, Rt , T1 ) + ? 2 (t)df (t, Rt , T2 ) ? Rt (? 1 (t)f (t, Rt , T1 ) + ? 2 (t)f (t, Rt , T2 ))dt] 1 = Dt [? 1 (t) ft (t, Rt , T1 )dt + fr (t, Rt , T1 )dRt + frr (t, Rt , T1 )? 2 (t, Rt )dt 2 1 +? 2 (t) ft (t, Rt , T2 )dt + fr (t, Rt , T2 )dRt + frr (t, Rt , T2 )? 2 (t, Rt )dt 2 ? Rt (? 1 (t)f (t, Rt , T1 ) + ? 2 (t)f (t, Rt , T2 ))dt] 1 = ? 1 (t)Dt [? Rt f (t, Rt , T1 ) + ft (t, Rt , T1 ) + ? t, Rt )fr (t, Rt , T1 ) + ? 2 (t, Rt )frr (t, Rt , T1 )]dt 2 1 +? 2 (t)Dt [? Rt f (t, Rt , T2 ) + ft (t, Rt , T2 ) + ? (t, Rt )fr (t, Rt , T2 ) + ? 2 (t, Rt )frr (t, Rt , T2 )]dt 2 +Dt ? (t, Rt )[Dt ? (t, Rt )[? 1 (t)fr (t, Rt , T1 ) + ? 2 (t)fr (t, Rt , T2 )]]dWt = ? 1 (t)Dt [? (t, Rt ) ? ?(t, Rt , T1 )]fr (t, Rt , T1 )dt + ? 2 (t)Dt [? (t, Rt ) ? ?(t, Rt , T2 )]fr (t, Rt , T2 )dt +Dt ? (t, Rt )[? 1 (t)fr (t, Rt , T1 ) + ? 2 (t)fr (t, Rt , T2 )]dWt . (ii) Proof. Let ? 1 (t) = St fr (t, Rt , T2 ) and ? 2 (t) = ? St fr (t, Rt , T1 ), then d(Dt Xt ) = Dt St [? (t, Rt , T2 ) ? ?(t, Rt , T1 )]fr (t, Rt , T1 )fr (t, Rt , T2 )dt = Dt |[? t, Rt , T1 ) ? ?(t, Rt , T2 )]fr (t, Rt , T1 )fr (t, Rt , T2 )|dt. Integrate from 0 to T on both sides of the above equation, we get T DT XT ? D0 X0 = 0 Dt |[? (t, Rt , T1 ) ? ?(t, Rt , T2 )]fr (t, Rt , T1 )fr (t, Rt , T2 )|dt. If ? (t, Rt , T1 ) = ? (t, Rt , T 2 ) for some t ? [0, T ], under the assumption that fr (t, r, T ) = 0 for all values of r and 0 ? t ? T , DT XT ? D0 X0 > 0. To avoid arbitrage (see, for example, Exercise 5. 7), we must have for a. s. ?, ? (t, Rt , T1 ) = ? (t, Rt , T2 ), ? t ? [0, T ]. This implies ? (t, r, T ) does not depend on T . (iii) Proof. In (6. 9. 4), let ? 1 (t) = ? (t), T1 = T and ? (t) = 0, we get d(Dt Xt ) = 1 ? (t)Dt ? Rt f (t, Rt , T ) + ft (t, Rt , T ) + ? (t, Rt )fr (t, Rt , T ) + ? 2 (t, Rt )frr (t, Rt , T ) dt 2 +Dt ? (t, Rt )? (t)fr (t, Rt , T )dWt . This is formula (6. 9. 5). 1 If fr (t, r, T ) = 0, then d(Dt Xt ) = ? (t)Dt ? Rt f (t, Rt , T ) + ft (t, Rt , T ) + 2 ? 2 (t, Rt )frr (t, Rt , T ) dt. We 1 2 choose ? (t) = sign ? Rt f (t, Rt , T ) + ft (t, Rt , T ) + 2 ? (t, Rt )frr (t, Rt , T ) . To avoid arbitrage in this case, we must have ft (t, Rt , T ) + 1 ? 2 (t, Rt )frr (t, Rt , T ) = Rt f (t, Rt , T ), or equivalently, for any r in the 2 range of Rt , ft (t, r, T ) + 1 ? (t, r)frr (t, r, T ) = rf (t, r, T ). 2 56 6. 3. Proof. We note d ? e ds s 0 bv dv C(s, T ) = e? s 0 bv dv [C(s, T )(? bs ) + bs C(s, T ) ? 1] = ? e? s 0 bv dv . So integrate on both sides of the equation from t to T, we obtain e? T 0 bv dv C(T, T ) ? e? t 0 t 0 T bv dv C(t, T ) = ? t s 0 e? T t s 0 bv dv ds. Since C(T, T ) = 0, we have C(t, T ) = e 1 ? a(s)C(s, T ) + 2 ? 2 (s)C 2 (s, T ), we get A(T, T ) ? A(t, T ) = ? bv dv T t e? bv dv ds = T e t s bv dv ds. Finally, by A (s, T ) = T a(s)C(s, T )ds + t 1 2 ? 2 (s)C 2 (s, T )ds. t

A supermarket in california

Poetry Poetry is a form of literary art that uses aesthetic and rhythmic qualities of language to evoke meaning to an audience. In â€Å"A supermarket in California† by Allen Ginsberg, he uses symbolism and literary allusions to convey a man going through a crisis between the modern American consumerism, an individual's detachment with nature; following the ways of his idol Walt Whitman by living a spiritual natural lifestyle and also tell a story about his search for sexual acceptance among the society for homosexuals.Walt Whitman's work deals with nature and how modern ocieties have effects on the natural world. Also the use of Whitman in this poem is a device which Allen Ginsberg used to contract his idols version of reality and also sexuality. Allen Ginsbergs first literary device in the poem, â€Å"A supermarket in California† is the use of symbolism. Symbolism is the use of symbols to represent ideas or qualities. In (line 4) â€Å"l went into the neon fruit mark et†.The supermarket is symbolic of mass production in modern America while the neon is symbolic of how artificial the society has become and also opposes Allen's naturalistic way of living because here is nothing natural about a neon light. However he continues to seek some kind of approval from Walt by asking â€Å"Where are we going â€Å"Walt Whitman†, Overall the supermarket symbolizes how artificial the supermarket has become as a venue for food sales.These symbols relates Allen's his perspective of the supermarket compared to natural food/ street stores/ farmers market and he describes it as a symbol of the man-made consumerist nature of a supermarket. He then went on to talk about the peaches and the penumbras. â€Å"What peaches and what penumbras! Whole families shopping at night! Aisles full of husbands! Wives in the avocados, babies in the tomatoes! And you Garcia Lorca, what were you doing by the watermelon? (Lines 6-8).The fruit symbolizes his yearn for t he American family to be close to nature and got things in little quantities rather than mass produced goods. He also wished the society was still close; his description of the family showed the family all over the place not making unilateral decision and not doing things as a whole rather as individuals. The families is symbolic of how everyone is going about their lives based on the â€Å"society's idea† of what a mainstream family should be hence eading to people lacking uniqueness, authenticity and a sense of self also becoming indistinguishable from the produce (not unique) .In the final part of his poem Allen goes on to talk about how less optimistic he was about the world he lives in. He then questions Whitman; â€Å"where are we going the doors close in an hour† (18) this symbolizes his realization that everything is going to change no matter what. His glorified idea of the natural world seems to be falling apart around him and he realizes it might not last due to the new consumerist nature of modern America. He admits that he feels absurd for having such high opes of seeing the beauty of supermarket commodities in lines 20-21.He realizes there is no place that ne and Walt can to tind W t's ideal place and realizes that their journey through the â€Å"solitary streets† past symbols of a â€Å"lost America† such as cars would only lead them to a lonely life (line 22-25). Allen Ginsbergs second literary device in the poem is the use of literary allusion. Literary allusion is a reference to a work of art, music, history, religion, or another work of literature within a piece of literature.It is pretty obvious that Walt Whitman is the driving force behind the oem; he is mentioned several times and also portrayed as a lonely man and also as a lifestyle along with other poets; however he had several other allusions such as Garcia Lorca in (line 8) who was an influential Spanish poet and an inspiration to Walt, Charn in (line 28) w ho was from the Greek mythology. Allen uses these individuals as a point of reference for his audience to refer to see how much this gotten â€Å"worse† and changed since their era so his reader could get a feel of then and now.Later on it became quiet known that Allen was gay and parts of the poem symbolized and represented a person who is being unsure of his sexuality and omeone who is going through a Journey of self-discover, this can be seen to some clues of Allen referencing Walt and Lorca who were both gay right activists and also homosexual individuals. It is quite clear when Allen states â€Å"l saw you Walt Whitman, childless, lonely grubber, poking among the meats in the refrigerator and eyeing the grocery boys†.Allen is making Walt Whitman's sexuality obvious by calling him a homosexual. Allen isolating Walt, stating that he is childless, lonely and eyeing boys are all clearly accusations at the society since during that time the society was not ccepting of homosexual of the idea of homosexuality and the ideology behind the â€Å"American dream† was a man, a woman, and their children in a house not a man or a woman and a man or a woman and their children.He is makes these comparisons in order to made his readers see how similar he is to Walt. And how he feels they do not belong since the idea of the American dream/modern America didn't accept them for who they are and the only way they can survive is to pretend about who they are. In Allen's poem he uses these literary devices to show the reader that America's onsumerism has caused a detachment between humans and natures.The use of symbolism allows the reader the chance to see Allen's point of view and to also show how much individuals have conformed to the societys new perspective of mainstream and modern America and also show how the society was not accepting of being homosexual and if one tries to defy the societies opinion of â€Å"relationships and marriage† he/she w ould only end up alone. With Literary allusion it helped the readers connect to him; to see where he is coming from literary and also have a historic background of what he is trying to relay.

Compare and Contrast Psychodynamic Theory and Person Centerd

Title – Critically compare and contrast two counselling theories Module code- Cg2030 Module tutor – Phill Goss Word count-2500 Counselling takes place when a counsellor sees a client in a private and confidential setting to explore a difficulty a client is having, distress they may be experiencing or perhaps their dissatisfaction with life, or loss of a sense of direction and purpose. It is always at the request of the client as no one can properly be ‘sent' for counselling. (Counselling central) By listening the counsellor can begin to perceive the difficulties from the client's point of view and can help them to see things more clearly, possibly from a different perspective. Counselling is a way of enabling choice or change or of reducing confusion. It does not involve giving advice or directing a client to take a particular course of action. Counsellors do not judge or exploit their clients in any way. ’(Counselling central). The two main therpaies withi n counselling that I will focus on comparing on contrasting are person centerd counselling.PCT is a form of talk-psychotherapy  developed by  psychologist  Carl Rogers  in the 1940s and 1950s. The goal of PCT is to provide clients with an opportunity to develop a sense of self wherein they can realize how their attitudes, feelings and behavior are being negatively affected and make an effort to find their true positive potential  In this technique, therapists create a comfortable, non-judgmental  environment  by demonstrating  congruence  (genuineness),  empathy, and unconditional positive regard toward their clients while using a non-directive approach.This aids clients in finding their own solutions to their problems. Secondly Psychoanalysis  is a  psychological  and  psychotherapeutic  theory conceived in the late 19th and early 20th centuries by Austrian neurologist Sigmund Freud. Psychoanalysis has expanded, been criticized and developed in differe nt directions, mostly by some of Freud's colleagues and students, such as  Alfred Adler,  Carl Gustav Jung  and  Wilhelm Reich, and later by neo-Freudians such as  Erich Fromm,  Karen Horney,  Harry Stack Sullivan  and  Jacques Lacan.The basic tenets of psychoanalytic therapy include the following beside the inherited constitution of personality, a person's development is determined by events in early childhood, human behaviour, experience, and cognition are largely determined by irrational drives, those drives are said to be largely  unconscious. This main aim of this assignment will be taking both person centred and psychoanalytic therapy and critically comparing them I will do this by giving an overview of the two therapies how they were developed, the principles they lie upon, what their relationships are based on.We will then look at how they are similar and also how they differ by looking at the skills used and how they work as therapies to come to both do the same thing which is to help a client overcome an issue in their life. Person centred therapy is based on the idea that humans have a drive to grow towards their potential and will act with the best interest to themselves, if they are provided the right atmosphere (Mcleod,2008). The direction of therapy is guided by the client with the support of the therapist. The client is always the expert on themselves if they are provided with the right atmosphere for self-actualisation to occur.The central component being the client knows best the client knows where it hurts the most. The counsellors main motive is to relate to the client in such a way that he or she can find there sense of self direction. Carl Rodgers was the founder of person centred therapy. The development stemmed from Rodgers experiences of being a client and working as a counsellor (casemore) Rodgers didn’t like the view of behaviourist that humans were organisms that react to stimuli and developed habits from learned experiences.Therefor in his work as a counsellor became increasingly uncomfortable with being â€Å"the expert† so began to develop a different view of what clients needed to experience in counselling, empathy congruence and unconditional positive regard. Rodgers proposed that human beings were always in the process of becoming rather than being in a fixed state. As humans we have the captivity to develop in a basically positive direction given the right conditions.That’s not to say that he thought people are not sometimes cruel or hurtful but every person has the tendencies towords strong positive direction. Rodgers proposed that when the conditions were growth promoting an individual could develop into a fully functioning person. Rodgers described his approach as a basic philosophy rather than a simple technique which empowers the individual and leads to personal and social transformation, grounded in empathic understanding being non-judgemental and congruent .For this reason the person centred approach is often seen as touchy feel relation and seen as soft skilled that lacks structure, sometimes said as a way of preparing the road for real therapy. However it does have a clear theory of self, the creation of distrees and the tehraputic process. It’s aims are transformational and asks the counsellor to be a human and transparent â€Å"be real†(langridge). Freud is regarded as the founder of modern psychology, developing psychoanalysis. The therapy is based on the idea that a great deal of the individual behavior are not within conscious control.Therefor the main emphasis is to help the client get to the deep root of the problem often thought, to stem from childhood. Based on the principle that childhood experiences effect our behavior as adults and effect out thinking processe, Freud belived that these thoughts and feelings can become repressed and may manifest themselves as depression or other negative symptoms. The client is able to reveal unconscious thought by talking freely aboput thought that enter their mind the analysis will attempt to interpratate and make sence of the clients experiances.Deeply burtied experiances are expressed and the opportunity to share tehase thoughts and feelings can help the client work through thease problems. Clients are asked to try and transfer thougts and feelings they have towords people in their life on the analyst this process is called transfernace the success depends on how the analysats and client work together. Psychoanalysis can be life changing if successful howver around 7 years of therpay is needed to discover the full unconcious mind . regular sessions provide a setting to explore there thoughts and make sence of them.Psychoanalytic therapy is based on freuds work of pychoanalysis but less intensive it is found to be bennaficial for clients who want to understand more about tehmselves and useful to people who feel tehir problems have affcted them for a long period of time and need reliving of emotional disstress. Through deep exploration client and therapist try to understand the inner life of the client. Uncovering the uncocncious needs and thoughts may help the client understand how their past experiances affect their life today.It can also help them to work out how they can live a more fulfilling life. Person centred counselling and psychoanalytic therapy are both off springs of two great minded people Sigmund Freud and carl Rodgers the originators of these two approaches. Freud based his framework on his medical background , Rodgers was influenced by excistential phillosphy were the person is there central role in their growth and change. Some similarities can be drawn from a comparison betwewen the two models they both want to widen the concioussness but this is done in different means.Psychonalsis aims to make the unconscious conscious and by doing this helps the person gain controle over their thought and feelings. And the person cneterd approach helps the client to overcome a state of incongruence whilst psychoanalytic objective is two seek the repressed childhood experiances. The person centerd works through the concioussness by focusing on the here and now. Kahn (1985) compares the term incongruence with psychoanalytic defence mechanism repression. To him both are the same thing but different versions.Both prevent the person from being aware of his their own feelings the onluy diffrnece being that Rodgers belives by imputing the core conditions of Two persons are in Psychological contact, The first, whom we shall term the client, is in a state of incongruence, being vulnerable or anxious,The second person, whom we shall term the therapist is congruence or integrated in the relationship,The therapist experiences unconditional positive regard for the client. The therapist experiences an empathic understanding of the clients internal frame of reference and endeavours to communicate this experience to the client.The communication to the client of the therapist’s empathic understanding and unconditional positive regard is to a minimal degree achieved. Then if thease are imputed incongruence will be shifted and psychoanalytic belive through interpretation of childhoon events repression can be acchived. Both psychoanalytic and person centred involve empathy that is applied to client and therapist, enabling the client to gain new understanding and move away from distress and towards harmony with self and others. Therefor potentially both try to increase insight and strength towards the self.Other areas that overlap are also evident simple areas such as setting bounderies, along with assessment for therapy. From the first minuite of contact both persons become aware of their own and others aims, values and lifestyle by empathic attending. Both types of therapist are warm and open minded and accepting. In both therapies the therapist is bidden to provide a non judgemental respe ctful attitude towords the client. Both also share a commitment to the use of silence beliving it is an effective tool for therapy commiting to listening without impediment.The similarities are strongly guided by personal growth and development both have the interest of promoting self-reflection of the client. Their interest is to promote self-reflection of the clirnt. The relationship is very important in both and the main reason for this is to gain a beter insight and clearer understanding of the client. The aim is to build a relationship built on trust honesty and reliance on one another. The relationship is crucial in both therpays as without a tight relationship there is no ground work in place for the client to feel safe to explore.Both models encourage the client to relase emotions and it is through tehase that empathy can be experienced. On the other hand psychoanalytic theory stresses the importance of unconscious procedures and sexuality as the key terms for a deep underst anding of the human pychopathology. Freud thought that dreams were the best way to explore the unconscious since they are disquised as the id whishes repressed by the ego in order to escape from awareness. The goal of counselling to Rodgers is the congruence of personality acchived when the self gains access to a variety of experiances. he need for self actualization can only be atteneded to once counsellor creates an atmosphere of unconditional positive regard and empathic understanding. There are considerable diffrences between the two approaches ialthough stated that they both create a strong knitted relationship between client ancd counsellor . person centers counselling the counsellor takes on a non directice role more like a companion rather than a leader and the client takes lead of the session were as psychodynamic counsellors take more of an authority figure in the relationship.Person centred firmly believes that the 6 core condition are necessary for change. Hoewever psych odynamic use of transference is to make interpretation to the client for effective change. Person centred would see using transference would not create a genuine relationship between counsellor and client. One of the central themes in person centred is the emphasis on the present behaviours of the client. Psychodynamic focuses on the past and how it determines the present behaviours.Pychodyanmic counsellors choose to remain neutral during a session as this encourages transference a major tool in psychodynamic therapy. McLeod (2004) points out that in person centered counselling, questions are only asked to clients when necessary and may also answer questions if asked by clients, as this was supports to create the quality of the relationship. This again differs to the way Psychodynamic counsellors' work; as asking questions are pivotal during therapy as this elps to explore and build up relevant material, furthermore, it would be unlikely for a psychodynamic counsellor to answer any personal questions by the client and instead try to figure out why the question is important. Although both see the relationship as crucial each therapy maintains a diferent way to responding to the client attitudes and values. for instance defences and transferance excist in both forms but handled in different ways. For person centerdit is a requirement that the core conditions are stimuiltaneous for the therapist thease core conditions repersent an openess to self experience and to the experience of another.Rodgers belived counsellors should be egalitarian in their meetings with clinets and a major diffrence concerns pychodynamic interpretations. Appearing as all knowing and going beyond what is un-concious based mostly on theory rather than a clients spacific experiances. Thorne (1996) states that â€Å"pychodynamic therpay may go to early in interpratation to make sence to clients† Pychodynamic interpratation specifically concerns the naming of the unconcious for causeing problems that the clinet may be having.Only in the hermanutic meaning does the word interpratation make sence of things Hermeneutics applies to all persons who make sense of all situations, whereas psychodynamic interpretation in the narrow sense is the most specific ingredient of psychodynamic therapy’s efforts to make positive changes for clients. All in all psychodynamic approaches are based on freuds work based on the unconscious of the ID ego and superego which emphasis on sexual aggression.Person centerd is based on the belief that humans have unique qualities for freedom and growth beliving that we are beyond being controlled by ID ego and superego not controlled by sexual urges. The similarities between both approaches promote and guide the idea of progress and development. There main interest to promote self reflection and awarenss for the client in order to do this they effectively use communication. Both approaches are the result of hypothesis the main diffrences lie at the foundations and what is belived to work best however sometimes they meet for what may be the best too to use at the time of hearapy. prehaps what makes any thrapy work is the belief that they can work if the client wants it to work there emotional needs and expectataions I belive that theory used does play a a very important role but it is the client who has the ability to change which ever root of therapy they take they can change no matter how they recive therapy weather it be in a humanistic approach or psychodynamic because if they expect it to work it will so maybe I am more for the person cneterd view as I belive that self actualisation and awarenss is very important to be able to facilitate growth.

Use of RFID technologies for identification, tracking Term Paper

Use of RFID technologies for identification, tracking - Term Paper Example There are advanced technologies associated with this method of keeping watch applied by the US while in the battle field. Evidences in support of this are the depth and range of research on this topic carried out by various scholars and interested parties. Their recent operations and raids have utilized space surveillance as a method of staying alert in the battle field. Technology and military theory, as well as the manner, in which it is applied during war and the impact of the aforementioned on the society, will be looked into keenly with the analysis of the works of various researchers on the topic. Another aspect will be that of information warfare. There are other researches on related topics, which support this view from different angles but with similar findings as will be realized subsequently. Revolution in military affairs has put technology in the right perspective when it comes to battlefield awareness. The US military as history has it, made heavy investments on military technology with space craft's as a priority . Use of devices that collect data before putting up a plan of attack while at war takes center stage in the US military operation layout. The socioeconomic paradigm of each era in the US reflects on the investments made in ensuring victory over the wars. In so doing, poor return on technological investments scales escalate. Errors in judgments are associated with such technologies resulting to retarded economic progress. ... Introduction Many organizations are fast adopting the use of RFID systems for easy tracking of counterfeit goods and all sorts of other criminal activities that people might engage in to the detriment of a company. Despite the growing popularity of the technology, the opposition has slammed RFID for its potential threats to privacy of individuals. It is imperative to acknowledge the opposition could be right since every person has a constitutional right to their own privacy. The deployment of RFID systems in the contemporary society is widespread. Philips Semiconductors that manufactures RFID chips has sold more than a billon chips across the world [9]. This may mean that many companies across the world have already installed RFID into their surveillance and tracking systems. Industrialized countries have adopted the technology more than in any other parts of the world [9]. While people may think that RFID systems are some of the latest technology breakthroughs in tracking systems, R FID have been in existent since the World War II. They were used in the war to detect Friend or Foe systems in military aircrafts [9]. In the contemporary world, there are several applications of the RFID systems including; automobile immobilization, inventory management, payments systems, tracking of animals, and in automated traffic toll collection. These widespread applications of RFID have improved efficiencies in accomplishing tasks for companies but on the other hand, have invigorated the debates on their infringement on people’s privacy. How RFID Works RFID tag has two components that aid in its collection, processing and transmission of information. Its part that is integrated into

Zappos as an Online Shoe and Clothing Trading Company Essay

Zappos as an Online Shoe and Clothing Trading Company - Essay Example The company utilizes the internet to undertake most of the business operations, including marketing and customer service. Relationship marketing and loyalty business models have been the marketing approaches adopted by the company since inception. An organizational culture is the values and behaviors that enable the customers to define the organizational environment. The organizational culture at Zappos remains focused on ensuring satisfaction of the customers from the services received. One fundamental value at the company remains the value for customers, commonly achieved through providing customer service beyond the traditional customer service. The company ensures relationships become established and nurtured between the organization and the customers. The employees within the organization have been empowered to undertake any action deemed relevant, towards attending to a customer needs (Schein, 2010). Within the organization, customer service representatives do not need to cons ult their supervisors when presented with complaints. Empowerment of employees has enabled the company to pursue the goal of excellent customer service, for a long period. The activities and actions undertaken by the customer representative employees remain fundamental in determining the organizational culture at Zappos. One instance when a customer called seeking to return a pair of shoes she’d bought for the husband proves the level of empowerment among the employees. The customer wanted to return the shoes since the husband had been deceased before he could receive the boots. The customer service employees offered to inform the customer about the procedure. The customer service employees instead, sent flowers to console the customer for the demise of her husband. This presented a human touch to the virtual company which undertakes to trade online. Many employees of the company have created  social media accounts through which to communicate with customers.  

CONSTITUTION & ADMINISTRATIVE LAW Coursework Example | Topics and Well Written Essays - 1000 words

CONSTITUTION & ADMINISTRATIVE LAW - Coursework Example He argues that they are indeed acts of necessity, which the government performs in sudden and extreme emergency, especially when it is in the public interest to do so. This essentially contradicts Dicey’s understanding of the royal prerogative. This is so considering that the power of war, which is generally, one of the significant powers of the prerogative, would anyway be an exception. The above differences in argument, therefore, call for reexamination into other theories in order to understand the significance of the royal prerogative2. In Attorney General v DE Keyser’s Royal Hotel Ltd , Lord Parmoor said that a right may be common to the rulers and the subjects; nevertheless, that does not qualifies it to be a prerogative right. Instead, he argued that Royal prerogative means a privilege in the executive that may be of an exclusive and a special character. Similarly, Blackstone concurs that the prerogative covers those actions that no person or institution other than the executive may undertake3. Among prerogatives contemplated in this explanation include the making of treaties and the deployment of armed forces. However, it is observable that the Blackstone’s argument contradicts the judicial reasoning as to what the prerogative is. To shed more light on this, an examination of the case of R v Criminal Injuries Compensation Board, Ex parte Lain will be of help. In this case, the board that was to investigate and make recommendation on the compensation to victims of violence through ex gratia payments, was appointed by the executive4. The key issue here is that the board was set up by the executive in disregarded the agreement between the Judiciary and the applicant. Clearly the prerogative act was used in doing this, arguably though, giving out money to the victims of violence is not such unique to deserve direct action of the executive. Moreover, Wade an administrative theorist agrees with the arguments of Dicey and Blackstone albeit at

Database and Enterprise Application Security Essay - 1

Database and Enterprise Application Security - Essay Example Through the identification of problems and issues near the beginning of the projects initialization phase the operating system, environment, system architecture, and database can be designed and integrated with security included features. In addition, it also ensures that system development process followed the rules and regulations, legislation and standards application. This paper presents a detailed analysis of web-security issues which need to be considered by the developers of enterprise web-applications. This paper also outlines fundamental security features offered by database management systems and use of these features in securing the database from security breaches. Â  Websites and web applications normally interact and communicate with other back-office applications, remote services, and distributed systems those are competent to be placed with the range of local premises, locations, and facilities at some other location. In this scenario, the difficult to manage and complicated nature of web-based system presents the need for better communication among the systems and this aspect leads to a greater likelihood of experiencing security vulnerabilities or weaknesses. This condition initiates elevated chances of the security infringement.

Student will be required to go to zoo observe two types of non human Research Paper

Student will be required to go to zoo observe two types of non human primates and write two pages on each primate observed - Research Paper Example His movement on the ground is fast and fluid, and he leaps along, using hands and feet, and somersaulting as he goes. He looks left and right all the time, as if to scan for predators, and as soon as he reaches a tree trunk he swings up in an arm over arm gesture. His movement from branch to branch is even faster than his ground movement, showing an adaptation to life in the forest canopy. On the ground slightly apart from the tree equipment there is an older female and a younger orang utan who are engaging in what looks like affectionate teasing. This is probably a mother and child, but the younger individual is not a baby. This may be an adolescent. The pair roll about on the grass, and hug each other from time to time as they do so. Communication appears to be by touch, since they do not have much eye contact, and their interaction is mostly silent. The bond between the two is obviously close. At one point the solitary younger male approaches the two and reaches out as if to touch them, but the two ignore him and he distances himself again from them. None of the orang utans take any notice of the human observers on the perimeter of the enclosure. High up on a shelf there is another adult individual sitting in a crouched position. The full face is hidden, and so it is not obvious whether this is a male or female. It is easy to overlook this one because there was little movement and no sound. The orang utan looked down on the pair playing on the grass, but appeared not to be interested in what was going on. It may be that this orang utan was depressed, or simply bored with the same routine. There were several spells when the young male positioned himself at the edge of the enclosure, looking out and turning his back to the other individuals. This appeared to be a deliberate statement of independence from the rest, as if he were imitating the senior sitting up on the high shelf. He did not sustain this pose for long, however, and soon resumed his hyper-active s winging, grasping the suspended toys, and running along the ground. The key activities viewed were therefore play (both solo and in a mother/child pair) and observation of each other and the surrounding area. There was plenty of independent activity, but only the mother and child had any real close interaction with each other. This suggests that orang utans are fairly solitary creatures outside the mother/child unit. There is evidence of group awareness but it appears not to be the main concern of the orang utans observed. The chimpanzee group is much more vocal and there is a lot of interaction between individuals, with groups forming and dissolving all the time. There is also a larger number present, with at least 12 individuals moving around in a steady walk on their hands and feet. From time to time there is some screeching from one or two individuals, and the rest appear to be uneasy when they hear this. The screeches are made with bared teeth and agitated movements. The other chimpanzees look at the screeching chimpanzee and then look away again, sometimes making lip movements and raising eyebrows. Some chimpanzeess get up and move out of the way when an agitated individual approaches them. Many individuals sit for a time on the grass, picking items up and looking at them which suggests a foraging instinct. They do not appear to be eating what they find. It is not always evident which are males and which are females, especially in the younger individuals. There is one

Art History Essay Example | Topics and Well Written Essays - 1250 words - 6

Art History - Essay Example Long time ago, pottery vessels would be used mainly for four functions. These functions include; eating, drinking, cooking and storage purposes. With respect to the artwork under analysis, the Red Figure Column Krater is one form of pottery that has an outstanding history rooted in the culture and lifestyles of the Greeks (Museum of Fine Arts). The Column Krater is made out of ceramic clay, and was mainly used by the Greeks to mix and drink wine. The Column Krater was valued as a special vessel, thus it was used majorly in households to serve wine to esteemed guests. The vessel is estimated to have come into existence around 470BC (Museum of Fine Arts). This paper will contextualize the Red Figure Column Krater within the parent culture. The red figure column krater originated for Greece. It should be understood that Greeks started engaging in pottery as early as the 7th Millennium BC (Museum of Fine Arts). Original use of pots specifically happened at the eastern peninsula of the Me diterranean Sea, in the Neolithic era. There have been various pieces of evidence which suggest that Greek culture might have been the starting point of all form of pottery. Pots made in the era range from the clay-made vessels to bronze-aluminum vessels. Most signatures of the artists behind ancient Greek artistic works have been found either on the vessels themselves are where they were found. Currently, signatures can be seen in ancient pots kept in most of the archives and museums in the world. Art History Essay Example | Topics and Well Written Essays - 500 words Art History - Essay Example The video considers a variety of early 20th century artists that implemented African art techniques, and argues that it was these early Western artists that in great part shaped the way later Western artists would use and implement African sculpture. The video goes on to demonstrate through side-by-side comparisons ways that Western artists appropriated African sculpture within their own work. It argues that in this appropriation Western artists oftentimes misinterpreted the African art. One such example comes in terms of a sculpture of an African face, and its appropriation in a Western painting. One of the weaknesses of the video is that it takes a somewhat pedantic view of influence in criticizing Western appropriation of these African sculptures. For instance, the video never gives an in-depth explanation of how the artists misread the African art. It also neglects to note that it may not have been the intention of the Western artists to accurately interpret the African art, but instead to implement its structural or artistic dimensions as a means of influence. Perhaps some of the most engaging elements are the video are the biographical footage it contains into the lives and apartments of early 20th century artists. These photos provide the viewer with direct insight into the lives and habitats of these Western artists.

The glass menagerie Research Paper Example | Topics and Well Written Essays - 1000 words

The glass menagerie - Research Paper Example In the production notes , Tennessee Williams says that ‘The Glass menagerie’ is a memory play This statement refers to a major theme of the play , namely , all the characters getting stuck up in memories of their past. He himself confesses that the play is string of memories ‘of his own youth. It follows the events and people in William’s true life experience in St. Louise between the years of 1934 and 1936. His mother, his sister, his job in shoe factory, and the glass menagerie were all part of his’ earlier life†. Besides Tom Amanda also live in constant pursuit of her bygone youth. She was extremely well-liked and cute young lady but she lost her chances. Now the realities in front of her are consciously ignored y her. The way she treats Tom and Laura is another evidence of her fanciful life. She prevents these two from becoming responsible young people by treating them as children. Without accepting the reality, she insists Laura to imitate her youth and hopes to recreate what she missed in life through her daughter. Thus Laura is bound to her mother always. The glass menagerie Laura lulls the infantile world with the glass menagerie. Thus she is also not raised to the level of a woman. Another character hounded by his memory is the father, who had left the family to travel long distance. The whole family lacks a father who supports them. The absence of such a father figure adds something more to their memory, the love and care of a father is also a memory to the children . Thus memory prevents all the characters to live in present and also to lead a happy life. Through Tom Winfield, Tennessee Williams was portraying himself as a young man. According to Presley â€Å"No one has even reviewed the bare details of his biography can overlook the obvious similarities between the record of his early life and the events described in The Glass Menagerie†. There are many similarities between his life and Tom’s lif e. Tom says â€Å" I am the narrator of the play , and also a character in it. The other characters are my mother, Amanda, my sister, Laura (1147). Thus the first resemblance between Tom and Williams is, he stands for the writer, who tells the story. Tennessee dropped out of the school according to his father’s instruction. After that he went to work in a shoe factory. The reaction of Tom in the play is relevant here... Tom says, â€Å" Listen ! You think I am crazy about the ware house! You think I am in love with the Continental Shoemakers? You think I want to spend fifty-five years down there in that Celotex interior!! With Fluorescent tubes...†. Like Tennessee, To also liked to write poems and plays during leisure times. â€Å"He called me Shakespeare (1168). The mental stress that Tennessee suffered is shown by his narrator character Tom also, and that lead both of them to write poems and plays. Tennessee William’s father was a salesman who always tried t o keep away from home. The father character presented here was also a telephone man who loved to travel. The children were brought up by their mother. The place where Tennessee lives is similar to what Tom explains in the play.† The apartment faces an alley and is entered by a fire escape.’ (1146). Tennessee was unwilling to remain in St. Louise school. In the play we can see Tom’s mother advises him to attend night –school course in accounting at Washington –â€Å" U Just think what a wonderful thing that would be for you son(1162). Another similarity we can see is the relation between Tennessee

Critical Analysis of a Childs Reading Essay Example for Free

Critical Analysis of a Childs Reading Essay Reading is an essential skill in modern society. Not only does it enable people to access information, it provides people with a great deal of pleasure. It is vital that primary schools equip children with effective strategies for reading as well as foster a desire to read that will stay with them throughout their lives. This analysis of reading will firstly give a brief outline of the context of my school placement. It will analyse two pupils as readers and their strategies. The school’s policy indicates that the context of teaching reading is very important suggesting a variety of text styles. English Williamson (2005) inform us that the introduction of the National Literacy Strategy (DfES 2001) broadened the range of texts children are introduced to at primary level. The school is superbly resourced, with thousands of books available to all pupils. Silent reading is also practised daily. X Primary is a larger than average three-form entry primary school with 472 pupils. It’s in an area of average to high socio-economic status and the majority of pupils are from White British backgrounds with few pupils who speak English as an additional language. The number of pupils with learning difficulties is below average. (Ofsted 2010). Below is an analysis of a childs reading. I will focus on analysing the childs mistakes in reading, called miscues (Hall, 2003) to gain information of the child as a reader. See more: how to write a critical analysis outline Pupil A was chosen for assessment as he enjoys reading and is a strong reader. He has had several school moves due to family issues, and has received intervention and support throughout his time at Primary X due to his level of absence. He is eager to learn, and was keen to read for me. The assessment involved analysing his word recognition and comprehension skills. This provides an opportunity to understand how Pupil A as a fairly fluent reader may process a text. The text which was read by Pupil A was chosen as it was unknown to him. It was also chosen as a text that was suitable for his level of reading. Many of Pupil As miscues take place in the form of substitution. These miscues often relate to his syntactic knowledge. He reads a instead of one (line 3) and but instead of and (line 8). He also produces the miscue even (line 9) as an insertion. These miscues suggest that he is making predictions about a text using his syntactic knowledge. This suggests that Pupil A brings his own knowledge to a text which causes him to make predictions (Smith cited in Hall, 2003), resulting in a miscue. This suggests that Pupil A uses his syntactic knowledge to obtain meaning in what he reads (Hall, 2003). This miscue can also alter the meaning of the text which may affect his understanding. Pupil A makes the same miscue when he substitutes for for from (Page 2 line 1). He self corrects and asks for reassurance in his correction. Pupil A also corrects himself on the word quickly (Pg 3 line 3). This self-correction reveals that he uses syntactic knowledge to process the text, but also that the text Pupil A sees is different from the text on the page (Goodman cited in Hall, 2003). Goodman suggests that there are two texts in question when reading takes place, being the actual text, and the perceived text. Pupil A demonstrated a comprehensive understanding of the text. He demonstrated an understanding for the organisation of the text and recalled events of the story. Pupil A demonstrated an ability to infer from the text and evaluate it. When questioned, he was able to express that he liked the text and expressed why. He demonstrated emotional or psychological response to the text and characters feelings. This suggests that Pupil A engaged with the text and was interested in the plot. In conclusion Pupil A demonstrates a clear ability to read fluently and uses different strategies for decoding words. His ability to decode unknown words could be extended by knowledge of consonant digraphs. Pupil A shows an ability to understand a text on a literal level, as well as engaging in a text by making inferences and evaluating a text. We could improve this motivation to read by encouraging Pupil A to read regularly for pleasure. Pupil B was chosen as although she receives literacy support, she does not enjoy reading. She has accessed literacy support since starting at Primary school in Year 3 and has made considerable progress and is able to read certain texts independently however she also shows little interest in reading for pleasure. She is willing to read with me, as she is used to reading with adults, in a 1-1 situation, and is comfortable with me as I have been in her class for several weeks. She is reluctant to pick a book she in unfamiliar with and cannot think of a favourite author/book when questioned. Pupil B begins well, with her decoding strategies revealing her processing of a text but also her phonic knowledge. She reads the word spider lings (line 8) correctly, by segmenting the word in her head first. She then blends ‘ling’ quietly, to herself, and then asks for reassurance to put both words together. This is because this is an unusual, unknown word, and Pupil B is unfamiliar with the term. She stumbles over the word ‘different’ (line9). She did not segment the word out loud and so it is difficult to determine which strategy she used to decode the word. Nevertheless, it is possible that Pupil B may have used one of two strategies. For the first strategy, it’s possible that she segmented and blended the word in silently. This suggests that Pupil B is confident in segmenting and blending. For the second strategy, Pupil B may have used her graphophonic knowledge to decode the word. Therefore it is possible that she recognised the word from previous reading exercises. She demonstrates her grapheme-phoneme correspondence knowledge in her unsuccessful attempt to decode the word notice (line 10). She fell silent which suggests she attempted to segment the word in her head. However, Pupil B finds this strategy unsuccessful and then chooses to segment the word out loud Pupil B often falls silent throughout the exercise, and waits for a prompt. I feel this is due to her lack of confidence rather than lack of knowledge. Pupil B demonstrates her grapheme and phonemic knowledge (Hall, 2003) by successfully sounding out the first syllable of the word ‘children’ (line11). She was unable to sound out the second syllable. This suggests that she struggled to sound out a particular grapheme. It’s possible that Pupil B was unfamiliar with the consonant digraph il. However, Pupil B demonstrates a fluency in reading which may suggest that she uses sight reading as a strategy (Ehri cited in Hall, 2003) to process a text. Erhi (cited in Hall, 2003) suggests that readers find new ways of identifying words. Finding new methods to identify a word can help a reader to become a more fluent in reading. My reading assessment can provide an insight to how a reader may process a text (Ellis Lewis, 2006 but it’s only an insight. I cannot be certain that the suggested reading strategy is the method used. The childs responses is dependent on the text. Another influence could be the texts difficulty. Too difficult a text may cause them to make miscues and create an unfair representation of the reader (Campbell, 2011). A readers inability to engage in the text may be because the reader is not interested in the text. To remedy this, it would be useful to find out what books the reader prefers. Another strategy for developing reading is shared reading which provides opportunities for children to peer assess. Iversen Reeder (1998) suggest that this allows children to actively participate when they feel comfortable. This is useful when children haven’t developed full confidence in their own reading ability, it provides a ‘safe’ structure encouraging contribution. This would be beneficial if both pupils could work together as Pupil A may help Pupil B become more engaged with the text. After analysing both Pupil A and Pupil B, I was surprised at how both pupils used similar techniques, however they were different when reading aloud. I felt there was a gap in understanding and intonation from both pupils, despite being close in age, and both receiving support. I believe another difference was the pupils was desire to read, with Pupil A keen to read books, demonstrating a clear opinion on authors or genre, however Pupil B was reluctant to name a book she’d read, and didn’t have a favourite author/style. I believe this lack of enthusiasm for reading will hinder her development, regardless of support put in. In conclusion, both pupils show an understanding and varying strategies to break down a text, however the major difference seems to be their attitude towards reading itself. References EDP 4120 Assessing Reading Riley, J Reedy, D. (2000) Developing writing for different purposes: teaching about genre in the early years. Paul Chapman Publishing, London. Iversen, S. Reeder, T. (1998) Organising for a Literacy Hour, London: Kingscourt Publishing. Hall, K 2003 Listening to Stephen Read: Multiple perspectives on Literacy Buckingham: Open University English, E. and Williamson, J. (2005) Meeting the Standards in Primary English. Routledge Falmer. DfES. (2006) The Primary Framework for literacy and mathematics, London: Department for Education and Skills. DfES. (2006) The Primary Framework for literacy and mathematics: Core position papers underpinning the renewal of guidance for teaching literacy and mathematics, London: Department for Education and Skills. DfEE. (1999) The National Curriculum: Handbook for primary teachers in England, London: Department for Education and Employment. Ofsted 2010 Campbell, R 2011 Miscue Analysis in the Classroom Leicester: UKLA