Solutions to Stochastic - Calculus for Finance Part 3

4.4 Exercise 4.6 Define Ito’s process Z t (α(s) − X(t) = Then 1 Z 0 2 σ 2 )ds + 0 t σ(s)dW (s) 1 2 σ )dt + σdW (t) 2 dX(t) = (α − dX (t)dX (t) = σ 2 dt Now we use two different notations to compute d(S p (t)): 1. Under the notations in Theorem 1, G(x, t) = S p (0)epx , a = α −  p d(S (t)) = (α − 12 1 σ 2 ) · pS p (0)epX(t) + 0 + 12 pσ2 + 1 2  2 2 1 2σ 2 and b = σ. Then  σ 2 p2 · S p (0)epX(t) dt + σ · S p (0)pepX(t) dW (t) p p σ S (t)dt + σpS p (t)dW (t) 2 2. Under the notations in Theorem 4.4.6 of Text, S p (t) = f (X(t)), then f (x) = S p (0)epx , f ′ (x) = pS p (0)epx and f ′′(x) = p2 S p (0)epx . Now pα − = d(S p (t)) = df (X(t)) 1 = pS p (0)epX(t) dX (t) + 1 = pS p (t)dX (t) + 2 2 p2 S p (0)epX(t) dX(t)dX(t) p2 S p (t)dX(t)dX (t)  = pS p (t) · = 4.5 pα − 12 1 2 (α − σ )dt + σdW (t) 1 2 pσ + 2  2 Exercise 4.7  + 12 p2 S p (t) · σ 2 dt p2 σ 2 S p (t)dt + σpS p (t)dW (t) Here we use the following version (4.4.1)of Ito’s lemma: df (W (t)) = f ′ (W (t))dW (t) + 1 2 f ′′ (W (t))dt (4.2) 1. Set f (x) = x4 in (4.2), then f ′ (x) = 4x3 and f ′′ (x) = 12x2 . Then (4.2) becomes dW 4 (t) = 4W 3 (t)dW (t) + 1 2 · 12W 2 (t)dt = 4W 3 (t)dW (t) + 6W 2 (t)dt We may also write it in the integral form as: Z W 4 (T ) = W 4 (0)+4 0 T Z W 3 (t)dW (t)+6 T 0 Z W 2 (t)dt = 4 0 T Z W 3 (t)dW (t)+6 2. First we note that W (t) ∼ N (0, t), then by the symmetry of the integrand Z 3 E[W (t)] = ∞ −∞ 22 x3 √ 1 − x2 2t e dx = 0 2πt T 0 W 2 (t)dt (4.3) Now we take the expectation in (4.3), E  W 4 (T )  " Z =E 4 Z = T 0 0 4E Z T W 3 (t)dW (t) + 6   W 3 (t) dW (t) + 6 2 Z t dt Z T 0 = 2 40dW (t) + 6 =3T 0 # T 0 W 2 (t)dt Z 0 T E   W 2 (t) dt T 3. Set f (x) = x6 in (4.2), then f ′ (x) = 6x5 and f ′′ (x) = 30x4 . Then (4.2) becomes dW 6 (t) = 6W 5 (t)dW (t) + 1 · 30W 4 (t)dt = 6W 5 (t)dW (t) + 15W 4 (t)dt 2 We may also write it in the integral form as: Z 6 6 0 T Z 5 Z T 0 0 4 Z T 0 5 T W 4 (t)dt W (T ) = W (0) + 6 W (t)dW (t) + 15 W (t)dt = 6 W (t)dW (t) + 15 It is easy to see that E[W 5 (t)] = 0. Then by taking expectation in the above equation, we have E  W 6 (T )  " Z =E 6 Z = 0 0 T 6E T Z W 5 (t)dW (t) + 15   W 5 (t) dW (t) + 15 2 Z 3t dt Z T 0 =6 3 0dW (t) + 15 =15T 4.6 0 0 T Z 0 # W 4 (t)dt T E   W 4 (t) dt T Exercise 4.8 1. In (4.4.23) of text, let f (t, x) = eβtx, then ft = eβt βx, fx = eβt and fxx = 0. Then (4.4.23) becomes   βt βR(t)dt + eβt βt dX(t) d eβt R(t) =eβt βR(t)dt [(α − βR(t))dt + σdW (t)] =αeβt dt + σeβtdW (t) 2. Integrating the both sides of the above equation, we have Z β·t β·0 e R(t) =e R(0) + α =R(0) +  β which implies R(t) = e−βt R(0) + t 0 e βt Z t 0 βs αe ds + −1  βs Z t e dW (s) +σ 0 α  β 1 − e−βt 23  σeβs dW (s) Z + σe−βt t 0 eβs dW (s) 4.7 Exercise 4.9 1. It is clear that 1 − y2 N ′ (y) = √ e 2π and 1 d± = d± (T − t, x) = √ Then N′  (d+ ) N ′ (d− ) +  = exp − d22 1 = exp = exp + 2σ 2 (T1 − t) − K x log σ T −t d2 2 + (r ± σ2  )(T − t) 2  2 " log  xK + (r − σ2 2 2 )(T − t) Kx log − + (r + σ2 2 2 # )(T − t)   σ2 x + r(T − t) · −4 · log (T − t) K 2 2σ 2 (T − t) h i x − log = exp + r(T − t) K −r(T = e −t) K x  2. First 1 ∂d± = √ ∂x σ T − tx By the definition c(t, x) = xN(d+ (T − t, x)) − Ke−r(T −t) N (d− (T − t, x)) cx = ∂c ∂x = N (d+ (T − t, x)) + xN ′ (d+ ) = N ( d+ ) + = N (d+ )  ∂d+ ∂d− − Ke−r(T −t) N ′ (d− ) ∂x ∂x xN ′ (d+ ) − Ke−r(T −t) N ′ (d− ) √ 1σ T − tx  3. ct = ∂c ∂t = xN ′ (d+ ) ∂d+ ∂t − Ke−r(T −t) rN (d− ) − Ke−r(T −t) N ′ (d− )  = −Ke−r(T −t) rN (d− ) + N ′ (d+ ) x = −Ke−r(T −t) rN (d− ) + xN ′ (d+ ) Now note that ∂d+ ∂t ∂d+ ∂t − Ke−r(T −t) − ∂d− ∂t √ d+ − d− = σ T − t ∂(d+ − d− ) ∂t 24 ∂d− σ =− √ 2 T −t  ∂t ∂d+  ∂t ∂xKe−r(T −t) so σx ct = −Ke−r(T −t) rN (d− ) − √ N ′ (d+ ) 2 T −t 4. 1 ct + rxcx + 2 σ 2 x2 cxx σx = − rKe−r(T −t) N (d− ) − σx = − rKe−r(T −t) N (d− ) − σx = − rKe−r(T −t) N (d− ) − √ 2 T √ 2 T 1 d [N (d+ (x))] dx 1 ∂d+ N ′ (d+ ) + rxN (d+ ) + σ 2 x2 N ′ (d+ ) ∂x 2 −t 1 1 N ′ (d+ ) + rxN (d+ ) + σ 2 x2 N ′ (d+ ) · √ 2 −t σ T − tx √ N ′ (d+ ) + rxN (d+ ) + 2 T −t 2 σ 2 x2 =rxN(d+ ) − rKe−r(T −t) N (d− ) =rc √ 5. As t → T −, T − t → 0. And ln > 0 if x > K, and lnK x< 0 if 0 < x < K. Therefore xK 1 lim d± = lim  + (r ± 1  √ 1 σ 2 (T − 1t)) log √ 2 (r ± = lim T − t h σ 2 T − t) i t→T − t→T − x 2 √ log + r + lim K T −t ( ∞,−∞, ififx0 >< Kx < k = t→T − t→T − K x Then ( lim t→T − So ( lim t→T − 6. For 0 ≤ t < T , ln c(t, x) = x K N (d± ) N (∞) = 1, = 0, N (−∞) if xif 0> 0, so X(t) = 0 for any t. 4.9 Exercise 4.13 Since B1 (t) and B2 (t) are Brownian motions, dB1 (t)dB1 (t) = dt and dB2 (t)dB2 (t) = dt. By substituting the definition dB1 (t) =dW1 (t) p dB2 (t) =ρ(t)dW1 (t) + 1 − ρ2 (t)dW2 (t) Note that these equations imply that W1 (t) is Brownian motion and W2 (t) is a continuous martingale p p 0 (since we with< ρW< −1 1).=Furthermore we can havewrite dW2 (t) = dB2 (t)/ 1 − ρ2 (t) − ρ(t)dW1 (t)/ 1 − ρ2 (t) with 2 (0) ( dW1 (t)dW1 (t) =p dt p (ρ(t)dW1 (t) + 1 − ρ2 (t)dW2 (t)) · (ρ(t)dW1 (t) + 1 − ρ2 (t)dW2 (t)) = dt Also by dB1 (t)dB2 (t) = ρ(t)dt, we have (4.4) p dW1 (t)(ρ(t)dW1 (t) + 1 − ρ2 (t)dW2 (t)) = ρ(t)dt p ρ(t)dt + 1 − ρ2 (t)dW1 (t)dW2 (t) = ρ(t)dt which implies dW1 (t)dW2 (t) = 0 (4.5) Then substitute (4.5) into the second equation of (4.4), p ρ2 dW t)dW ρ2 dt + (1 − ρ2 )dW2 (t)dW 2 (t)++(1ρ −1ρ−2 )dW 1 (dW 2 (t)+ =0 dt ρ2 dt = dt 2 (t) 2 (t) dW2 (t)dW2 (t) = dt (4.6) Based on the fact that W1 (t) and W2 (t) are continuous martingales with W1 (0) = 0 = W2 (0) , and (4.5)(4.4)(4.6), by applying Theorem 4.6.5, we conclude that W1 (t) and W2 (t) are independent Brownian motions. 4.10 Exercise 4.15 1. Note that ( and dWi (t)dWj (t) = d dBi (t) = X j=1 dt 0 if ifj =j 6= k k σij (t) dWj (t) σi (t) which implies Bi (t) is a continuous martingale with Bi (0) = 0 for any 1 ≤ i ≤ d. 27 (4.7) Then we compute the quadratic variation, or equivalently,  dBi (t)dBi (t) = d σij (t) X j=1 d σik (t) dWk (t) X σi (t)  k=1 · dWj (t)  σdi (t) 1 " d # 2 X σ i (t) X j=1 σ (t)dW (t) k=1 σ (t)dW (t) ij j ik k d 1 σij (t)dWj (t)σij (t)dWj (t) by (4.7) = 2 σ i (t) X = j=1 d 1 σij2 (t)dt = 2 σ i (t) X j=1 =1 2.  dBi (t)dBk (t) = = d X j=1 σij (t) d σkl (t) dWl (t) σk (t)  l=1 · dWj (t)  σi (t) d 1 σi (t)σk (t) X j=1 d X " σij (t)dWj (t) dX l=1 # σkl (t)dWl (t) = 1 σij (t)dWj (t)σkj (t)dWj (t) σi (t)σk (t) X j=1 = 1 σij (t)σkj (t)dt σi (t)σk (t) X j=1 d =ρik (t) 4.11 Exercise 4.18 1. In (4.4.13), let f (t, x) = e−θx−(r+θ 2 /2)t then ft (t, x) = −(r + θ2 /2)e−θx−(r+θ fx (t, x) = −θe 2 /2)t −θx−(r+θ 2 /2)t fxx (t, x) = θ2 e−θx−(r+θ 2 /2)t Then Ito’e Formula of (4.4.13) becomes 2 2 dζ (t) = − (r + θ2 /2)e−θW (t)−(r+θ /2)t dt − θe−θW (t)−(r+θ /2)t dW (t) 1 2 + θ2 e−θW (t)−(r+θ /2)t dW (t)dW (t) 2 1 θ2 ζ(t)dt = − (r + θ2 /2)ζ(t)dt − θζ(t)dW (t) + 2 = − θζ(t)dW (t) − rζ(t)dt 28 by (4.7) 2. By Ito’s Product rule, d(ζ(t)Z(t)) =ζ(t)d(Z(t)) + Z(t)d(ζ(t)) + d(ζ(t))d(Z(t)) =ζ · [rXdt + ∆(α − r)Sdt + ∆σSdW (t)] + X · [−θζdW (t) − rζdt] + [rXdt + ∆(α − r)Sdt + ∆σSdW (t)] · [−θζdW (t) − rζdt] = [ζrX + ζ∆(α − r)S − rζX − ∆σSθζ] dt + [ζ∆σS − Xθζ] dW (t) =∆ζS [(α − r) − σθ] dt + ζ [∆σS − Xθ]dW (t) =ζ[∆σS − Xθ]dW (t) (since θ = (θ − r)/σ) i.e.,d(ζ (t)Z (t)) has no dt term. Therefore ζ(t)Z (t) is a martingale. 3. If an investor begin with initial capital X(0) amd has portfolio value V (T ) at time T , which is X(T ) = V (T ), then ζ(T )X(T ) = ζ(T )V (T ) Since X(T ) and V (T ) are random, we take the expectation in the above formula E [ζ(T )X(T )] = E [ζ(T )V (T )] Now since ζ(t)X (t) is a martingale, E[ζ(T )X(T )] = ζ(0)X(0) = X(0) By equating the above two equations, we have X(0) = E[ζ(T )V (T )]. 4.12 Exercise 4.19 1. By the definition, dB(t) = sign(t)dW (t), B(t) is a martingale with continuous path. Furthermore, since dB(t)dB(t) = sign2 (t)dW (t)dW (t) = dt, B(t) is a Brownian motion. 2. By Ito’s product formula, d[B(t)W (t)] =d[B(t)]W (t) + d[W (t)]B(t) + d[B(t)]d[W (t)] =sign(W (t))W (t)dW (t) + B (t)dW (t) + sign(W (t))dW (t)dW (t) =[sign(W (t))W (t) + B(t)]dW (t) + sign(W (t))dt Integrate both sides, we will have Z B(t)W (t) =B(0)W (0) + t 0 Z t [sign(W (s))W (s) + B (s)]dW (s) 0 sign(W (s))ds + [sign(W ( s))W (s) + B (s)]dW (s) Z Z t t 0 0 = sign(W (s))ds + Note that the expectation of an Ito’s integral is 0 and W (s) ∼ N (0, s), which implies Z ∞ −∞ sign(x) 1 Z −x2 /2s √ e E[sign(W (s))] = dx = 2π sides, Then by taking expectation on both 1 ∞ 0 √ 2π e −x2 /2s Z dx − 0 √ −∞ 1 2π e−x 2 /2s dx = 0 Z t E[sign(W (s))]ds = 0 0 E[B(t)W (t)] = Note that E[W (t)] = 0, the above result also implies that B(t) and W (t) are uncorrelated. 29 3. By Ito’s product formula, d[W (t)W (t)] =d[W (t)]W (t) + d[W (t)]W (t) + d[W (t)]d[W (t)] =W (t)dW (t) + W (t)dW (t) + dW (t)dW (t) =2W (t)dW (t) + dt 4. By Ito’s product formula, d[B(t)W 2 (t)] =d[B(t)]W 2 (t) + d[W 2 (t)]B(t) + d[B(t)]d[W 2 (t)] =sign(W (t))W 2 (t)dW (t) + [2W (t)dW (t) + dt] B(t) + sign(W (t))dW (t) [2W (t)dW (t) + dt] =[sign(W (t))W 2 (t) + 2B(t)W (t)]dW (t) + [B(t) + 2sign(W (t))W (t)]dt By taking the integral on both sides, Z B(t)W 2 (t) = t 0 Z [sign(W (s))W 2 (s) + 2B (s)W (s)]dW (s) + t 0 [B(s) + 2sign(W (s))W (s)]ds Note that again the expectation of an Ito’s integral is 0 and W (s) ∼ N (0, s). Then by taking the expectation, we have Z E[B(t)W 2 (t)] = E [[B(s) + 2sign(W (s))W (s)] ds t 0 Z t E [sign(W (s))W (s)] ds =2 Now 1 2 sign(x) √ e−x /2s dx 2π x x 2 √ e−x /2s dx = 0 Z ∞ Z 0 2π 2 0 −∞ √ e−x /2s dx − = 2πx −x2 /2s dx Z ∞ √ e 2π 0 =2 >0 Z E[sign(W (s))] = 0 ∞ −∞ Thererfore E[B(t)W 2 (t)] > 0 6= 0 = E[B(t)]E[W 2 (t)]. If B(t) and W (t) are independent, then B(t) and W 2 (t) will also be independent, which implies E [B(t)W 2 (t)] = E[B(t)]E [W 2 (t)]. Contradiction. Therefore B(t) and W (t) are uncorrelated but not independent. 5 Chapter 5 5.1 Exercise 5.1 Recall that dtdt = 0, dtdW = 0 and dW (t)dW (t) = dt. 1. First note that dX(t) = σ(t)dW (t) + (α(t) − R(t) − 30 1 2 σ 2 (t))dt For f (x) = S(0)ex , we have f ′ (x) = f ′′(x) = f (x) and f (X(t)) = S(0)eX(t) = D(t)S(t), therefore by Ito’s Lemma, d(D(t)S(t)) =df (X(t)) =f ′ (X(t))dX(t) + 1 =S(0)eX(t) dX(t) + 1 2 f ′ (X(t))dX(t)dX(t) 2 S(0)eX(t) dX (t)dX (t)  2 1 =D(t)S(t) σ (t)dW (t) + (α(t) − R(t) − =D(t)S(t) [σ (t)dW (t) + (α(t) − R(t))dt] =D(t)S(t)σ (t) [dW (t) + Θ(t)dt] σ 2 (t))dt + 1  2 σ 2 dt + 0 + 0   with Θ(t) = (α(t) − R(t))/σ(t). 2. The other way is by Ito’s Product Rule, d(D(t)S(t)) =S(t)dD(t) + D(t)dS (t) + dD(t)dS (t) =S(t) [−R(t)D(t)dt] + D(t) [α(t)S (t)dt + σ (t)S(t)dW (t)] + [−R(t)D(t)dt] · [α(t)S(t)dt + σ(t)S(t)dW (t)] = [α(t)S(t)D(t) − S (t)R(t)D(t)] dt + D(t)σ(t)S(t)dW (t) =D(t)S(t) [σ (t)dW (t) + (α(t) − R(t))dt] =D(t)S(t)σ (t) [dW (t) + Θ(t)dt] 5.2 Exercise 5.4 1. Let us consider d(ln S(t)), by Ito’s Lemma d(ln S(t)) = 1 1 S(t) + S(t)  1 · − 2 2 dS (t)dS (t) σ 2 S (t) d =r(t)dt + σ (t)d ˜ W (t) − W̃ (t)d W̃ (t) 2 W (t)  σ 2 (t) = r(t) − dt + σ(t)d ˜ 2 By writing it in the integral form, we have Z ln S(T ) = ln S(0) + "Z0 S(T ) =S(0) exp Under the form S(T ) = S(0)eX , 0 r(t) − σ 2 (t) 2 r(t) − σ 2 (t) 2 T 0  Z dt +  0 Z dt + 0 T T ˜ (t) σ(t)d W σ(t)d ˜ # W (t)  Z T σ(t)d W̃ (t) σ 2 (t) 0 r(t) − dt + 0 ˜W (t) is a2 normal random variable with mean 0 and variance σ(t)d Z X= T T RT By Theorem 4.4.9 of Text, R T RT 2 2 σ ((tt))dt. dt, which implies X is normal distributed with mean 0 r(t) − 00 σ RT 31 σ 2 (t) 2  dt and variance 2. To simplify the notations, denote R= 1 Z 0 Σ = T s 1 T r(t)dt Z 0 T σ 2 (t)dt T Then X is normal with mean RT − T Σ2 /2 and variance T Σ2 . c(0, S(0)) =˜  e−RT (S(T ) − K)+ 1 " # 2 Z T Σ2 √ √ (S(0)ex − K) exp − x − RT + /2T Σ2 dx =e−RT 2 2πΣ T x:S(T )≥K e−RT " # 2 Z 2 TΣ √ √ ex exp − x − RT + /2T Σ2 dx =S(0) 2 2πΣ T 1 x:S(T )≥K " # 2 Z T Σ2 √ √ exp − x − RT + /2T Σ2 dx − Ke−RT 2 2πΣ T x:S(T )≥K E Now  1 Z " √ √ exp − x − RT + x:S(T )≥K =P (S(T ) ≥ K) 2πΣ T =P =P =P =N S(0) K √ X − (RT Σ− 1T X ≥ ln √ X − (RT Σ− 1T 1 √ Σ T  ln # 2 T Σ2 (5.1) /2T Σ2 dx 2  2TΣ 2 2TΣ 2 K S(0) ) ) ≥ ≤ ln ln + (RT − K S(0) S(0) K 1 2 32 − (RT − √ Σ T + (RT − √ Σ T  T Σ2 ) 1 2TΣ 2 ) ! 1 T Σ2 ) 2 ! (5.2) and e−RT " # 2 T Σ2 √ ex exp − x − RT + /2T Σ2 dx 2 2πΣ T x:S(T )≥K 1 # " 2 Z T Σ2 √ exp − x − RT − /2T Σ2 dx = 2 2π x:S(T 1 )≥K " # = 2 Z ∞ K 2 ln S(0) TΣ √ exp − x − RT −K /2T Σ2 dx 2 2 2π T Σ2 ) ln S(0) − (RT + 1 2 T Σ ) ! √ 2 ≥ 1 X − (RT Σ+ T =P √ S(0) Σ T + 1 T Σ2 ) T Σ2 ) ln K + (RT 2 √ 2 ! ≤ 1T Σ X − (RT + =P √ 1 Σ T + (RT +   K S(0) 1 2 √ ln TΣ ) =N Σ T (5.3) into (5.1), we 2have By substituting (5.2) and Z √ c(0, S (0)) = BSM (T, S (0); K, R, Σ) 5.3 Exercise 5.5 1. By the definition 1 Z Z(t) Let = exp t 0 Θ(u)dW (u) +  d 1 Z(t)  2 Z 0 t t  Θ2 (u)du Θ2 (u)du 2 d h X(t) i = e 1 dt =eX(t) dX(t) + 1 1 0 X(t) = Z 0 Z t Θ(u)dW (u) + then 1 2 eX(t) dX(t)dX(t) 2  2Z(t) Θ (t)dt 1 2 1 Θ(t)dW (t) + Θ (t)dt + 1 2 = Z(t)   Θ(t)dW (t) + Θ2 (t)dt Z(t) 2. By Lemma 5.2.2 of text, sinceM ˜(t) is a martingale, for 0 ≤ s ≤ t, =  Z(t) ˜ M (s)    i   i h h Z(t) 1    M (t) F(s) = Z (s)E M˜(t) F(s) = Z (s) ˜ E Z(t) M̃ (t) F(s) = Z (s)E which means M̃ (t)Z(t) is a martingale. 33 (5.3) 3. Note that dM(t) = Γ(t)dW (t), then h i d M̃ (t)   1 =d M (t) · Z(t)     1 1 1 =M (t)d + d [M 2 (t)] + d [M (t)] d Θ2 (t) Z(t) Θ (t) Z(t) Z(t) dW (t) +    Z(t) dW (t) +  Z(t) Θ( t) 1 Θ(t) =M1(t) · dt + 1 · Γ(t)dW (t) + dt · Γ(t)dW (t) Z(t) (t) + Z(t) Z(t) [Γ(t) + M (t)Θ(t)] dW = Z(t)   Γ(t)Θ(t) + M (t)Θ2 (t)M (t) dt Z(t) 4. Now since dW˜(t) = dW (t) + θ(t)dt, if we define Γ̃(t) = Γ(t) + M (t)Θ(t) Z(t) then h i d 1M̃ (t) Θ(t) [Γ(t) + M (t)Θ(t)M (t)] dt = [Γ(t) + M (t)Θ(t)] dW (t) + Z(t) Z(t) ˜ ˜ = Γ(t)dW (t) + Γ(t)Θ(t)dt ˜ (t) = ˜Γ(t)d W Integrate both sides, we will have M̃ (t) = ˜ Z t Γ̃(u)d W̃ (u) M (0) + 5.4 0 Exercise 5.12 1. First each ˜ P dBi (t)dt = 0Biand furthermore we have (t) is a continuous martingale. Secondly, since dBi (t) = j σij (t)/σi (t)dWt (t), d B̃i (t)d ˜B (t) = [dB (t) + γ (t)] · [dB (t) + γ (t)] = dB (t)dB (t) = dt i i i i i i i B̃i (t) must be a brownian motion for every i. 2. By dB̃i (t) = dBi (t) + γi (t) we have dSi (t) =αi (t)Si (t)dt + σi (t)Si (t)dBi (t) =αi (t)Si (t)dt + σi (t)Si (t)[d ˜ Bi (t) − γi (t)] = [αi (t)Si (t) − σi (t)γi (t)Si (t)] dt + σi (t)Si (t)d ˜ =R(t)Si (t)dt + σi (t)Si (t)d ˜ Bi (t) 34 Bi (t) 3. Since dBi (t)dBk (t) = ρik (t)dt, dB̃i (t)d ˜Bk (t) = [dBi (t) + γi (t)dt] · [dBk (t) + γk (t)dt] =dBi (t)dBk (t) + γi (t)dtdBk (t) + γk (t)dtdBi (t) + γi (t)γk (t)dtdt =dBi (t)dBk (t) =ρik (t)dt 4. By Ito’s product rule, d[Bi (t)Bk (t)] = Bi (t)dBk (t) + Bk (t)dBi (t) + dBi (t)dBk (t) then since dBi (t)dBk (t) = ρik (t)dt, we have Z t Z t Z t ρik (u)du 0 0 0 Bi (t)Bk (t) = Bi (u)dBk (u) + Bk (u)dBi (u) + Now note the fact that the expectation of an Ito’s integral is 0, if we take the expectation on the above formula, then Z t  Z t ρik (u)du 0 0 E[Bi (t)Bk (t)] = E ρik (u)du = ρ ( u ) du. Similarly we have E[B˜i (t) ˜ ik R0 t Bk (t)] = 5. For E[B1 (t)B2 (t)], since dB 1 (t) = dW2 (t) and dB2 (t) = sign(W1 (t))dW2 (t) by Ito’s product rule, d[B1 (t)B2 (t)] =B1 (t)dB2 (t) + B2 (t)dB1 (t) + dB1 (t)dB2 (t) =W2 (t)dB2 (t) + B2 (t)dW2 (t) + sign(W1 (t))dW2 (t)dW2 (t) By dW2 (t)dW2 (t) = dt, we may write it in the integral version as: Z t Z t Z t sign(W1 (u))du 0 0 0 B1 (t)B2 (t) = B1 (0)B2 (0) + W2 (u)dB2 (u) + B2 (u)dW2 (u) + Note that the expectation of an Ito’s integral is 0 and W1 (u) ∼ N (0, u), which implies Z E[sign(W1 (u))] = Then we have ∞ −∞ 1 sign(x) √ 2πu e−x 2 Z /2u dx = 1 ∞ 0 √ 2πu e−x 2 Z /2u dx− 0 √ −∞ 1 2π e−x 2 /2u E[B1 (t)B2 (t)] = 0 dx = 0 (5.4) For E[B̃1 (t) ˜B2 (t)], since and dB̃1 (t) = dB1 (t) = dW2 (t) = d W̃2 (t) (5.5) dB̃2 (t) = dB2 (t) = sign(W1 (t))dW2 (t) = sign( ˜ W1 (t) − t)d ˜W2 (t) (5.6) by Ito’s product rule, d[ B̃1 (t) ˜B2 (t)] = B̃1 (t)d ˜B2 (t) + B̃2 (t)d ˜B1 (t) + d B̃1 (t)d ˜B2 (t) =W2 (t)sign( ˜W1 (t) − t)d ˜W2 (t) + B2 (t)d ˜W2 (t) + sign(W1 (t))dt =W2 (t)sign( ˜W1 (t) − t)d ˜W2 (t) + B2 (t)d ˜W2 (t) + sign( ˜W (t) − t)dt 1 35 we may write it in the integral version as: B̃1 (t) ˜B2 (t) = B̃1 (0) ˜ ˜ Z t [W2 (u)sign( ˜W1 (u) − u) + B2 (u)]d ˜ Z t sign( W 1 (u) − u)du 0 0 B2 (0) + W2 (u) + ˜ (u)−u ∼ Note that the expectation of an Ito’s integral is 0 andW1˜(u) ∼ N (0, u), which implies W 1 N (−u, u) and Z sign(x) √ ∞ −∞ E[sign(W1 (u) − u)] = Z = Z 1 2πu e−(x−u) 1 0 ∞ √ 1 2πu e −(x−u)2 /2u 2 /2u Z dx 0 dx − 1 −(x−u)2 /2u e dx 2πu 1 −(x+u)2 /2u √ e dx 2πu √ −∞ Z ∞ 0 √ e−(x−u) /2u dx − 2πu Z ∞ h i 2 2 √ e−(x−u) /2u − e−(x+u) /2u du >0 2πu 0 since u > 0 = = 0 ∞ 2 1 Then we have E[B̃1 (t) ˜ ˜ Z t E[sign( W1 (u) − u)du] > 0 0 B2 (t)] = (5.4) and (5.7) are the exmaples that E[B˜1 (t) ˜B (t)] > 0 6= 0 = E[B (t)B (t)]. 2 1 2 5.5 (5.7) Exercise 5.13 1. Ẽ[W1 (t)] = ˜E[W1˜(t)] = 0 Ẽ[W2 (t)] = ˜E[ ˜ W2 (t) − Z W1 (u)du] = ˜E[ ˜ t 0 Z W2 (t)] − t 0 Ẽ[W1 (u)]du = 0 2. Note that the covariance ˜ ˜ Cov[W 1 (T ), W2 (T )] = Ẽ[W1 (T )W2 (T )] − Ẽ[W1 (T )] E[W2 (T )] = Ẽ[W1 (T )W2 (T )] ˜ 1 (T )] = 0 = E[W ˜ 2 (T )]. because of E[W Now by Ito’s product rule, we have Z Here since W1 (T )W2 (T ) = W1 (0)W2 (0) + 0 Z T W1 (t)dW2 (t) + T 0 W2 (t)dW1 (t) d W̃1 (t) =dW1 (t) d W˜2 (t) =dW2 (t) + W1 (t)dt therefore, dW1 (t) =d W̃1 (t) dW2 (t) =d W̃2 − W̃1 (t)dt then W1 (T )W2 (T ) = W̃1 (0) ˜ Z TW̃1 (t)d ˜ Z T W̃1 (t) ˜ Z T W̃2 (t)d W̃1 (t) 0 0 0 W2 (0) + W2 (t) − W1 (t)dt + Now since the expectation of any Ito’s integration is 0, and W1 (0) = 0 = W2 (0), ˜ Z T 0 Ẽ[ W̃1 2 (t)]dt = − E[W1 (T )W2 (T 2)] = − [W̃ (t)] = Cov[W˜1 ] = t. Z T 0 tdt = − T2 2