Sử dụng cùng một hạt giống, các kết quả tương ứng là 5,4751, 0,00820, và 14. Các . bây giờ tốt hơn như Wi bây giờ là tuyến tính trong Ái và Si-1 hơn trong (a). Tại sao? 4. (b) (i) t = 2 8386. 5. Hàm mật độ xác suất có điều kiện là | 196 Solutions restart with stats anova describe fit importdata random statevalf statplots transform randomize 96341641 n 10000 m 10 n 10000 m 10 for i from 1 to n do for j from 1 to m do a j evalf rand 1Cf12 end do b seq a j j c transform statsort b d 0 for j from 1 to m do d d m - j 1 A2 c j end do end do f seq e i i g1 describe mean f m g2 describe standarddeviation 1 f m Ínterval evalf g2 sqrt n g1 1 96 g2 sqrt n g1 .00003042677684 g2 .9879697859e-5 interval .00003023313476 .00003062041892 Solutions 6 1. Put call parity gives c t Ke r -T t p t x t e rf T t . Using the result derived for the delta of the corresponding call option dp t dc t - e-rf T-t dx t dx t e-rf T drf - e-rf T-t -e-rẠT-o 1 - 0 drf -e-rf T - -drỳ 2. Use put call parity with rf 0. This gives a p b p and c p . The seller of a put will initially hedge his or her position by having Solutions 6 197 a portfolio consisting of -1 put and A blocks of shares where A -0 -d see Problem 1 and d r a2 2 T -t ln x t K a T t . The number of shares shorted initially is a 328 b 440 and c 553. 3. a Let P X T t x t denote the payoff for the bond holder at time T given that the current FTSE is at x t . Then P X T t x t In a risk-neutral world 1 XT 1 Ấ X 0 1 I 1 r X T _ 1 1 X T 1 6 1 2 X 0 X 1 X 0 1 6 1 XS- X T x t exp r - T - t a T - tZ ỵ where Z N 0 1 . Let Q Z t x t denote the payoff as a function of Z rather than of X T and let d1 ln x 0 x t - r - T - t Then and so d2 ơ T -1 ln 0 x t - r - T - t Q Z i x i V x t t e-r T-t a T -1 1 Z dp 1 1 r XiT X 0I1 d Z d 1 2 X 0 d1 Z d2 1 d2 Z X -d2 b The following Maple procedure computes this for given t and xt where xt x t is expressed in terms of x0. capped_bond proc xt t local r sigma T b d1 d2 P price r sigma T 4 b xt x0 d1 ln 1 b - sigma 2 T-t sqrt T-t sigma
