Solutions to Stochastic - Calculus for Finance Part 4

5.6 Exercise 5.14 1. Since  −rt  X(t) = −d ree−rtX( t)dt + e−rt dX(t) = − re−rtX(t)dt + e−rt ∆(t)dS(t) − e−rt a∆(t)dt + re−rt (X (t) − ∆(t)S (t)) dt h   i rS(t)dt + σS(t)d ˜ W (t) + adt − a∆(t)dt − r∆(t)S(t)dt =e−rt ∆(t) =σS(t)e−rt ∆(t)d ˜ W (t) contains no dt term, e−rt X(t) is a martingale. 2. (a) Let Z(t) = σ W̃ (t) + (r − 1 σ 2 )t, 2 then   dY (t) =dZ(t) eZ(t) 1 Z(t) e dZ (t)dZ (t) =e dZ(t) + 2 1 2  2 Y (t)σ (t)  W (t) + (r − 1 2 σ )dt + =Y (t) σd ˜ 2 =rY (t)dt + σd W̃ (t) (b) Since  −rt  = −d ree−rtYY(t(t) )dt + e−rt dY (t) = − re−rtY (t)dt + e−rt Y (t)dt + e−rt σY (t)d ˜ =σY (t)e−rt d ˜ W (t) W (t) contains no dt term, e−rt Y (t) is a martingale. (c) By Ito’s product rule, Z 0 t 1 a a dsdY (t) + Y (t) · dt + dY (t) · dt Y (t) Y (s) Y (t) dS(t) =S(0)dY (t) + 1  Z t  0 S(0) + = S(t) ds dY (t) + adt Y (s) = dY (t) + adt Y (t) S(t) = i h · rY (t)dt ˜ + σY (t)dW̃ (t) + adt =rS( W (t) + adt t)dt + σS(t)d Y (t) since dY (t)dt = 0 (5.8) which is (5.9.7) of Text. 3. First ˜ E[S(T )|F(t)] a Z T ds|F(t)] 0 Y (s) E[S(0)Y (T ) + Y (T ) a =S(0) Ẽ[Y (T )|F(t)] + ˜  Z =˜ a # "  Z T t ˜ ds|F(t) ds|F(t) + E Y (T ) E Y (T ) Y (s) Y (s) a =S(0) Ẽ[Y (T )|F(t)] + ˜  Z T ˜ Z t Y (T ) 0 t |F(t) ds ds + a E E [Y (T )|F (t)] Y (s) Y (s) by taking out the what is known. Secondly, since e−rt Y (t) is a martingale under P˜ , we have 0 (5.9) t Ẽ[Y (T )|F(t)] = ˜E[Y (T )e−rT |F (t)]erT = Y (t)e−rt erT = Y (t)er(T −t) And by the definition of Y (t), we have Y (T ) W (T ) − ˜     1 (5.10) σ ˜ W (s) + (r − 2 σ 2 )(T − s) )(T −s) Y (s) ˜ 2σ 2 h W (T ) − i 1 = exp σ ˜ W (s) e(r− ˜ (s) is normally distributed with mean 0 and variance T − s, by taking the Since W̃ (T ) − W expectation on both sides of the above equation, we have = exp W (T )− ˜ W (s)) (r− 1 2 σ 2 )(T −s) ]e  Y (T ) σ( ˜ =E[e E 1 2 2 1 σ (T −s) (r− Y (s) =e 2 2 σ )(T −s) ·e  (5.11) =er(T −s) Now by substituting (5.10) and (5.11) into (5.9), we have ˜ E [S(T )|F(t)] Z =S(0)e r(T −t) Y (t) + e Z −t) S(t) + aa =er(T =er(T −t) S(t) + 4. Now we differentiate (5.12) T t h r r(T −t) Y (t) er(T −s) ds er(T −t) − 1 a t 0 Y (s) Z ds + a t T er(T −s) ds (5.12) i d(E˜ [S(T )|F (t)])   a   =d er(T −t) S(t) + d er(T −t) − 1 r   =erT d e−rt S(t) − aer(T −t) dt (5.13) Since  −rt  S(t) = −d ree−rt S( t)dt + e−rt dS(t) = − re−rt S(t)dt + e−rt rS(t)dt + e−rt σS (t)d ˜ =e−rt σS(t)d ˜ W (t) + ae−rt dt W (t) + ae−rt dt (5.14) by substituting (5.14) into (5.13), we have d( E˜ [S(T )|F(t)])   =erT d e−rt) S(t) + −aer(T −t) dt =er(T −t) σS(t)d ˜ W (t) + aer(T −t) dt − aer(T −t) dt =er(T −t) σS(t)d ˜ W (t) ˜ There is no dt term in d( E [S(T )|F(t)]). Therefore ˜ E [S(T )|F(t)] is a martingale under ˜P . 5. Since ˜ h i e−r(T −t) (S(T ) − K)|F(t) E =˜ i h by linearility E e−r(T −t) S(T )|F(t) − e−r(T −t) K =e−r(T −t) ˜ E [S(T )|F(t)] − e−r(T −t) K by taking out what is known ˜ E  E [S(T )|F(t)], i.e.,  e−r(T −t) (S(T ) − K)|F(t) = 0 implies that K = ˜ F orS (t, T ) = K = E˜ [S(T )|F(t)] = F ut (t, T ) S 6. According to the hedgeing strategy, the value of the portfolio at time t is governed by (5.9.7) of Text with ∆(t) = 1, i.e., dX (t) = dS (t) − adt + r(X(t) − S(t))dt Then (5.8) implies that  −rt  σS(t)e−rt d ˜ By integrating both sides from d0 toe TX(t) , we =have Z e−rT X(T ) − X(0) = T Z = 0 T 0 W (t) σS(t)e−rt d ˜W (t)dt e−rt (dS (t) − rS (t)dt − adt) by (5.9.7) of Text Z T ae−rt dt Z T   0 0 = −rT d e−rt S(t) −a =e S(T ) − S(0) −   1 − e−rT Associated with the fact that X(0) = 0, we can rrewrite it as X(T ) = S(T ) − S(0)erT − a  r  erT − 1 By (5.12), S(0)erT + a  rT  e − 1 = F utS (0, T ) = F orS (0, T ) r X(T ) = S(T ) − F orS (0, T ) Therefore eventually, at time T , she delivers the asset at the price F orS (0, T ), which is exactly the amount that she needs to cover the debt in the money market X(T ) − S (T ) = S(0)erT + So a  r 6  erT − 1 . Chapter 6 6.1 Exercise 6.1 2 tR(b(v) − σ (v )/2)dv . Let Rt t 1. It is easy to see Z(t) = e0 = 1 sincet σ(v)dW (v) = 0 = Z u Z u(b(v) − σ 2 (v )/2)dv t t S(u) = σ(v)dW (v) + . Then Z(u) = eS(u) and   x dS (u) b(u) −then σ 2 (u)f/x2= fdv = σ(u)dW (u) +of text, Now set f (t, x) = ex in Ito’s Formula (4.4.23) xx = e and ft = 0. Then we have dZ (u) =df (t, S(u)) =eS(u) dS(u) + 1 2 eS(u) dS(u)dS(u)   =Z(u) · [σ1(u)dW (u) + b(u) − σ 2 (u)/2 dv] + Z(u)     · [σ(u)dW (u) + b(u) − σ 2 (u)/2 dv] · [σ(u)dW (u) + b(u) − σ 2 (u)/2 dv] 2    1 2 b(u) − σ 2 (u)/2 + =Z(u) · (since dW (u)dW (u) = du) σ (u) du 2 + Z(u) · [σ(u) + 0] du (since dW (u)du = 0 = dudu) =b(u)Z (u)du + σ(u)Z (u)dW (u) 2. It is clear that if X(u) = Y (u)Z (u), then X(t) = Y (t)Z(t) = x · 1 = x. Furthermore, by Ito’s product rule, dW (u)dW (u) = du and dW (u)du = 0 = dudu, we have dX(u) =d[Y (u)Z(u)] =Y (u)dZ (u) + Z(u)dY (u) + dZ (u)dY (u) γ +   aZ− σγ dW (u) · Z (u) = [bzdu + σzdW (u)] · Y (u) + γ Z +   aZ− σγ + [bzdu + σzdW (u)] · dW (u) Z   γσZ = [(a − σγ)du + γdW (u)] + [bZY du + σZY dW (u)] + du + 0 Z = [a − σγ + bZY + γσ] du + [γ + σZY ] dW (u) = (a(u) + b(u)X (u)) du + (γ(u) + σ(u)X(u)) dW (u) (since X = ZY ) i.e. X(u) = Z(u)Y (u) solve the stochastic differential equation (6.2.4). 6.2 Exercise 6.3 1. d h ds 2. e− Rs 0 b(v)dv i Rs C(s, T ) = −b(s)e− Z t e− RT 0 b(v )dv b(v)dv 0 C(s, T ) + e− d h T e − Rs 0 b(v )dv ds Rt C(T, T ) − e− Z i C(s, T ) ds = Rs 0 0 −e− T t Z b(v)dv b(v)dv C(t, T ) = − t T C ′ (s, T ) = −e− R s 0 b(v)dv e− Rs 0 b(v)dv ds R 0s b(v)dv ds By C(T, T ) = 0, R s b(v)dv 0 − Rt Te R t e− C(t, T ) = 0 3. ds Z = b(v)dv T t e− e − R s0 b(v)dv Z Rt 0 ds = b(v)dv Z T A(T, T ) − A(t, T ) = The result follows from A(T, T ) = 0. 6.3 −a(s)C(s, T ) + t T t e− R s b(v)dv t ds  σ 2 (s) 2 C (s, T ) ds 2 Exercise 6.4 1. d # φ(t)C(t, T ) Z T σ 22 σ2 t C(u, T )dt = − 2 dt σ 2 σ2 φ′′ (t) = − φ′ (t)C(t, T ) − φ(t)C ′ (t, T ) 2 2 φ′ (t) = φ(t) · " therefore, C ′ (t, T ) = −φ′′ (t) − σ2 σ2 2 2 φ′ (t)C(t, T ) φ(t) =− 2. Rearrange (6.5.14) into σ2 C′ − we have − σ2 2 φ′ (t) 2φ′′(t) 2φ′′(t) − C(t, T ) = − σ 2 φ(t) + 2 C (t, T ) 2 σ φ(t) φ(t) C2 = b · C − 1 2 2φ′ (t) 2φ′′(t) = −b · −1 σ 2 φ(t) σ 2 φ(t) which implies (6.9.10) 3. By the characteristic equation λ2 − bλ − 1 2σ 2 = 0, we have √ λ= Then b ± b2 + 2σ 2 2 φ(t) = a1 e( b +γ)t 2 = + a2 e( b b 2 ±γ 2 −γ)t 2 Indeed, by the facts φ(T ) = 1 and φ′ (T ) = −2σ φ(T )C(T, T ) = 0, we have a set of equations: ( 2 bb aa11e(( 2 which imples +γ)T + γ)e a = 1 a = 2  2 so c1 = c2 = σ /4γ . + a2 e( b ( b 2 +γ)T 2 −γ)T + a2 ( b γ− b 2 −( b 2 +γ)T 2γ e γ+ b 2 −( b −γ)T 2 2γ e = 1 − γ)e( 2 b 2 −γ)T = 0 σ2 −( b 2 +γ)T 4γ(r+ b ) e 2 2 σ −( b 2 −γ)T 4γ(r− b 2 ) e = = 2 4. (6.9.12) is straightforward from (6.9.11) by differentiatin w.r.t t. Again, since φ′ (T ) = − σ2 φ(T )C(T, T ) = 0, we have c1 · e0 − c2 · e0 = 0 i.e. c1 = c2 . √ 5. By γ = b2 + 2σ 2 /2, we have γ2 − b2 b b σ2 = (γ + )(γ − ) = 2 4 2 2 then φ(t) = c1 e− " b 2 (T −t) = c1 e− " b 2 (T −t) = c1 e− = c1 e− = c1 e− b 2 (T −t) b b 2 (T −t) 2 (T −t) b 12 + γ 1 e−γ(T −t) − # b 2 2 −γ e−γ(T −t) − 2 −γ 2 + b 4 eγ(T −t) −γ b +γ 2 b 2  2σ 2 2σ 2 σ2 −( b2  b· − γ)e−γ(T −t) + ( eγ(T −t) − 2e−γ(T −t) −γ 2 + b b2 4 # eγ(T −t)  + γ)eγ(T −t) 2 eγ(T −t) + e−γ(T −t) + 2γ · [b sinh(γ(T − t)) + 2γ cosh(γ(T − t))] 2 and φ′ (t) = c1 e− b −t) 2 (T b = −2c1 e− h e−γ(T −t) − eγ(T −t) sinh(γ(T − t)) i 2 (T −t) Therefore 2φ′ (t) C(t, T ) = − σ 2 φ(t) =− = b 2 −4c1 e− σ 2 c1 e− b 2 (T −t) b 2 (T −t) sinh(γ(T − t)) 2 [b sinh(γ (T − t)) + 2γ cosh(γ (T − t))] sinh(γ(T − t)) σ2 sinh(γ(T − t)) + γ cosh(γ(T − t))  6. By the fact φ(T ) = 1, we have c1 e− 2 b (T −T ) 2 σ2 [b sinh(γ(T − T )) + 2γ cosh(γ(T − T ))] = 1 c1 2 σ2 · 2γ = 1 c1 = σ2 4γ Eventually, Z T A′ (s, T )ds t A(t, T ) = A(T, T ) − ′ 2aφ (s) ds Z T σ 2 φ(s) t =0 − Z T a · C(s, T )ds t = log(φ(t)) =a· σ 2 /2 = 2a log 2 σ2a =− σ2 6.4  · 21 σ2c log " b sinh(γ(T − t)) + 2γ cosh(γ(T − t)) b  e (T −t) 2 2 (T −t) b γe b sinh(γ(T − t)) + 2γ cosh(γ(T − t)) # Exercise 6.5 1. Since g(t, x1 , x2 ) = E t,x1 ,x2 h(x1 (T ), x2 (T )) we have for any 0 ≤ s ≤ t ≤ T similarly as Theorem 6.3.1 of Text,  i h  E h(X1 (T ), X2 (T )) F(t) = g(t, X1 (t), X2 (t))  i h  E h(X1 (T ), X2 (T )) F(s) = g (s, X1 (s), X2 (s)) Then   i h i h   E g(t, X1 (t), X2 (t)) F(s) = E E [h(X1 (T ), X2 (T ))|F (t)] F(s)  i h  = T ),XX 2 (T )) F(t) =E g(s,h(XX11((s), 2 (s)) Similarly, Since   e−rt f (t, x1 , x2 ) = E t,x1 ,x2 e−rT h(x1 (T ), x2 (T )) = e−rT E t,x1 ,x2 [h(x1 (T ), x2 (T ))] e−rt f (t, X1 (t), X2 (t)) is also a martingale. 2. If W1 and W2 are independent, dW1 (t)dW1 (t) = dt = dW1 (t)dW1 (t) and dW1 (t)dW2 (t) = 0. By Ito’s formula, dg(t, X1 (t), X2 (t)) =gt dt + gx1 dX1 (t) + gx2 dX2 (t) + 1 2 gx1 x1 dX1 (t)dX1 (t) + gx1 x2 dX1 (t)dX2 (t) + 1 2 gx2 x2 dX2 (t)dX2 (t) =gt dt + gx1 · [β1 dt + γ11 dW1 (t) + γ12dW2 (t)] + gx2 · [β2 dt + γ21 dW1 (t) + γ22dW2 (t)] 1 + gx x [β1 dt + γ11 dW1 (t) + γ12dW2 (t)] · [β1 dt + γ11 dW1 (t) + γ12 dW2 (t)] 2 1 1 + gx1 x2 · [β1 dt + γ11 dW1 (t) + γ12dW2 (t)] · [β2 dt + γ21 dW1 (t) + γ22 dW2 (t)] 1 + g x x · [β2 dt + γ21 dW1 (t) + γ22 dW2 (t)] 2 2 2 2 2   11 + γ 21 + γ  1    1 + gx1 x2 (γ11γ21 + γ12 γ22 ) + 12 = g t + β1 g x1 + β2 g x2 + dt g x1 x1 γ 2 g x2 x2 γ 2 22 + [γ11 gx1 + γ21gx2 ] dW1 (t) + [γ g + γ g ] dW (t) 2 2 12 x1 22 x2 2 Since g(t, X1 (t), X2 (t)) is a martingale, there is no dt term in dg(t, X1 (t), X2 (t)). Therefore by setting the dt term to zero in the above differential form, we will have (6.6.3) of Text immediately. Smilarly by Ito’s formula,   d e−rt f (t, X1 (t), X2 (t))  −rt  −rt 1 = +−re ft dt + e−rt fx1 dX−rt1 (t) + e−rt fx2 dX2 (t) 1 e−rt f +fxe1 x1 dX 1 (t)dX1 (t) + e fx1 x2 dX1 (t)dX2 (t) + 2  −rt 1 =e −rt−rf + ft + β1 fx1 + β2 fx2 + + e [γ11 fx1 + γ21 fx2 ] dW1 (t) + e−rt  11 2 +γ 2 2 e−rt fx2 x2 dX2 (t)dX2 (t)  + fx1 x2 (γ11γ21 + γ12 γ22 ) + 12 fx x γ 2 121fx11 + γ22 fx2 ] dW2 (t) [γ 1 2 fx2 x2  γ 21 2 + γ2 22 Since e−rt f (t, X1 (t), X2 (t)) is a martingale, there is no dt term in d [e−rt f (t, X1 (t), X2 (t))]. (6.6.4) of Text follows immediately by setting the dt term to zero in the above differential form. 3. If dW1 (t)dW2 (t) = ρdt, dW1 (t)dW1 (t) = dt = dW1 (t)dW1 (t), then by Ito’s formula, dg(t, X1 (t), X2 (t)) =gt dt + gx1 dX1 (t) + gx2 dX2 (t) + 1 2 gx1 x1 dX1 (t)dX1 (t) + gx1 x2 dX1 (t)dX2 (t) + 1 2 gx2 x2 dX2 (t)dX2 (t) =gt dt + gx1 · [β1 dt + γ11 dW1 (t) + γ12dW2 (t)] + gx2 · [β2 dt + γ21 dW1 (t) + γ22dW2 (t)] 1 + g x x [β1 dt + γ11 dW1 (t) + γ12 dW2 (t)] · [β1 dt + γ11 dW1 (t) + γ12 dW2 (t)] 2 1 1 + gx1 x2 · [β1 dt + γ11 dW1 (t) + γ12dW2 (t)] · [β2 dt + γ21 dW1 (t) + γ22 dW2 (t)] 1 + g x x · [β2 dt + γ21 dW1 (t) + γ22 dW2 (t)] 2 2 2 + γ2  + γ2   1  11  221  1 2 (γ γ + γ γ ) + + g x x 11 21 12 22 1 2 12 gx x γ gx x γ = g t + β1 g x1 + β2 g x2 + dt 22 + [gx1 x1 ργ11 γ12 + gx1 x2 (ργ 2 11γ1 221 + ργ12 γ21 ) + gx2 x2 ργ21γ22 ] dt 2 2 2 + [γ11 gx1 + γ21gx2 ] dW1 (t) + [γ12 gx1 + γ22 gx2 ] dW2 (t)   dt Since g(t, X1 (t), X2 (t)) is a martingale, there is no dt term in dg(t, X1 (t), X2 (t)). Therefore by setting the dt term to zero in the above differential form, we will have (6.9.13) of Text immediately. Smilarly by Ito’s formula,   d e−rt f (t, X1 (t), X2 (t))  −rt  −rt 1 = +−re ft (dt + e−rt fx dX−rt1 (t) + e−rt fx2 dX(2t)(t)+ 1 e−rt f +fxe1 x1 dX 1 t)dX1 (t) +1 e fx1 x2 dX1 (t)dX 2 2  1 −rt =e −rt−rf + ft + β1 fx1 + β2 fx2 + + e [gx1 x1 ργ11 γ12 + gx1 x2 (ργ11 γ22 +e −rt [γ11 fx1 + γ21 fx2 ] dW1 (t) + e  11 2 + γ2 2 e−rt fx2 x2 dX2 (t)dX2 (t)  + fx1 x2 (γ11γ21 + γ12 γ22 ) + 12 fx1 x1 γ +2ργ12 γ21) + gx2 x2 ργ21 γ22 ] dt −rt 1 2 fx2 x2  γ 21 2 + γ2 22 [γ12 fx1 + γ22 fx2 ] dW2 (t) −rt Since e f (t, X1 (t), X2 (t)) is a martingale, there is no dt term in d [e−rt f (t, X1 (t), X2 (t))]. (6.9.14) of Text follows immediately by setting the dt term to zero in the above differential form. 6.5 Exercise 6.7 1. By the iterated conditioning, for 0 ≤ s ≤ t ≤ T , we have ˜  i i  h h i h    E e−rt c(t, S(t), V (t)) F(s) =˜Ẽ e−rt ˜E e−r(T −t) (S(T ) − K)+ F(t) F(s) = h  i  E e−rt e−r(T −t) (S(T ) − K)+ F(s) =˜ h  i  E e−rT (S(T ) − K)+ F(s) =e−rs ˜ h  i −s) +  e−r(T =e−rs c(E s, S(s ), V(S(T (s)) ) − K) F(s) Then since it is a martingale, the dt term in the differential d[e−rt c(t, S(t), V (t))] should be 0, d[e−rt c(t, S(t), V (t))]  =e −rt −rcdt + ct dt + cs dS(t) + cv dV (t) + 12 cssdS(t)dS (t) + csv dS(t)dV (t) + 1 1 2 cssV S 2 + csv V Sσρ +  1 2 −rc + ct + cs rS + cv (a − bV ) + cvv σ 2 V dt 2 h √ i h √ i W2 (t) + e−rt cs V S d ˜W1 (t) + e−rt cv σ V d ˜ which is 1 1 cssvs2 + csv vsσρ + cvv σ 2 v = 0 −rc + ct + cs rs + cv (a − bv) + 2 2 This is the same as (6.9.26) of text.   cvv dV (t)dV (t) =e−rt 2. To simplify the notations, we denote f = f (t, log s, v), g = g(t, log s, v) and their paritial deriva-   dt tives as well. Then c(t, s, v) =sf (t, log s, v) − e−r(T −t) Kg(t, log s, v) ct =sft − e−r(T −t) Kgt − re−r(T −t) Kg 1 1 1 cs =f + sfx · − e−r(T −t) Kgx · = f + fx − e−r(T −t) Kgx · s s cv =sfv − e−r(T −t) Kgv 1 1 css = fx + fxx − e−r(T −t) Kgxx s s csv =fv + fxv − e−r(T −t) Kgxv s 1 s2 1 s cvv =sfvv − e−r(T −t) Kgvv Then the left handside of (6.9.26) of Text becomes ct + rscs + (a − bv)cv + 1 2 s2 vcss + ρσsv +  2 v fx + v 2 1 2 σ 2 vcvv fxx + ρσvfv + ρσvfxv + =s ft + rf + rfx + (a − bv)fv +  v + e−r(T −t) K gt + rg − rgx + (a − bv)gv + −r(T −t) =rsf − re Kg =rc 2 gxx + ρσvgxv + σ2 v σ2 v 2  2 fvv  gvv which verifies (6.9.26) of Text. 3. One method is to observe first that f (t, X(t), V (t)) is a martingale, and take differentiation and then set dt = 0 to get the equation. Here we apply the Two Dimensional Feynman-Kac Theorem (Exercise 6.5 of Text)directly2 Now by the definition of dX(t) and dV (t), 1 β1 = r + 2 β2 = a − bv + ρσv v √ , γ11 = v, , γ21 = 0, γ12 = 0 √ γ12 = σ v Then (6.9.13) yields ft + (r + 1 2 v)fx + (a − bv + ρσv)fv + v 2 fxx + ρσvfxv + 1 2 σ 2 vfvv = 0 Also for the boundary condition, in (6.6.1)of Text h(x, v) = Ix≥log K , then f (T, x, v) = Ix≥log K 4. Here we still apply the Two Dimensional Feynman-Kac Theorem directly, by the definition of dX(t) and dV (t), β1 = r − 1 2 β2 = a − bv v √ , γ11 = v, , γ21 = 0, γ12 = 0 √ γ12 = σ v 2 Here we cannot apply (6.6.3) or (6.6.4) of Text since they are for independent Brownian motions. Then (6.9.13) yields v 1 1 v)gx + (a − bv)gv + gxx + ρσvgxv + σ 2 vgvv = 0 2 2 2 Also for the boundary condition, in (6.6.1)of Text h(x, v) = Ix≥log K , then gt + (r − gT, x, v) = Ix≥log K 5. Apparently the function c(t, x, v) defined by (6.9.34) satisfies the equation (6.9.26). Now by the precise definition c(T, s, v) =sf(T, log s, v) − e−r(T −T ) Kg(T, log s, v) ( s− ≥ log K s ·K, 0 − K · if 0, log sotherwise = =(s − K)+ which implies that such c satisfies the boundary condition. 7 Chapter 7 8 Chapter 8 9 Chapter 9 10 Chapter 10 11 Chapter 11 11.1 Exercise 11.1 Since M 2 (t) − M 2 (s)   = N 2 (t) − N 2 (s) + λ2 (t2 − s2 ) − 2λtN (t) + 2λsN (s)   2 2 2 2 2 + 2N (t)N (s) − 2N 2 (s) 2+ λ = N((t) N (t) − (s N)](s)) ) − 2λt[N N (s)] + 2λN(s)( =[ −N + 2N (s)[N (t) − N (s)] + λ (t2 −(ts2−) −s 2λt[N (t) −(t)N−(s)] + 2λN (s)(s − ts) − t) we have   E[M 2 (t) − M 2 (s) =E[[N (t) − N (s)]2 + F(s)] 2N (s)[N (t) − N (s)] + λ2 (t2 − s2 )   − 2λt[N (t) − N (s)] + 2λN(s)(s − t) =λ2 (t − s)2 + λ(t − s) + 2N (s) · λ(t − s) F(s)] + λ2 (t2 − s2 ) − 2λt · λ(t − s) + 2λN(s)(s − t) =λ2 (t2 + s2 − 2ts + t2 − s2 − 2t2 + 2ts) + λ(t − s) =λ(t − s) Therefore, for any 0 ≤ s ≤ t, 1.   E[M 2 (t) F(s)]   =E[M 2 (t) − M 2 (s) + M 2 (s) F(s)]   =E[M 2 (t) − M 22 (s) F(s)] + M 2 (s) =λ(t − s) + M (s) ≥M 2 (s) i.e., M 2 (t) is submartingale. 2.   E[M 2 (t) − λt F(s)]   =E[M 2 (t) − M 2 (s) − λ(t − s) + M 2 (s) − λs F(s)]   =E[M 2 (t) − M 2 (s) F(2s)] − λ(t − s) + M 2 (s) − λs =λ(t − s) − λ(t − s) + M (s) − λs =M 2 (s) − λs i.e., M 2 (t) − t is martingale.