Solutions to Stochastic - Calculus for Finance Part 1

Solutions to Stochastic Calculus for Finance Introduction This solution manual will be updated anytime, and is NOT intended for any business use. The author suggests this manual as a reference book to the metioned book by Steven Shreve. Also anyone involved in any mathematical finance courses shall not copy the solutions in this book directly. This is ideally for self-check purpose. If anyone find any mistakes or misprints in this book, please inform me. Thanks. 1 Chapter 1 1.1 Exercise 1.4 (i) On the infinite coin-toss space Ω∞ , for n = 1, 2, . . . we define ( Yn (ω) = 0, 1, if ωn=TH Yn 2n n=1 P let We set X = ∞ Then by Example 1.2.5, X is uniformly distributed on [0, 1]. Furthermore, Z z −∞ where φ(ζ)dζ φ(ζ) = √ N (z) = Then for Z = N −1 (X), by Example 1.2.6, 1 2π e−ζ 2 /2 µZ [a, b] = P (ω : a ≤ Z(ω) ≤ b) = P (ω : a ≤ N −1 (X(ω)) ≤ b) = P (ω : N (a) ≤ X ≤ N (b)) = N (b)−N (a) for any −∞ < a ≤ b < ∞. So we conculde that Z = N −1 (X) is a standard normal distributionon on (Ω∞ , F∞ , P ). (ii) Let n Zn (ω) = N −1 (Xn (ω )) where Xn ( ω ) = X k=1 Yk (ω) 2k It is clear that lim Zn (ω) = lim N −1 (Xn (ω)) = N −1 ( lim n→∞ n→∞ n→∞ Xn (ω)) = N −1 (X (ω)) = Z(ω) for every ω ∈ Ω∞ Since Yk (ω) only depends on the k−th coin toss, Xn will only depend on the first n coin tosses, which means Zn will also only depend on the first n coin tosses. 1.2 Exercise 1.5 First, since ( I(0,X(ω))(x) = we can compute the double integral as: 0, 1, (1.1) ∈ ,(0, X(ω)) if ifx ∈x/ (0 X(ω )) Z [X(ω) − 0]dP (ω) = E[X] (1.2) I(0,X(ω))(x)dxdP (ω) = Ω Ω In the other hand, we may also compute the same double integral by changing the order f the integration Z Z ∞ Z Z ∞ 0 as: Z ∞ Z I(0,X(ω))(x)dP (ω)dx I(0,X(ω))(x)dxdP (ω) = Now let us consider the innner integral with respect 0to PΩ(ω), in which we shall consider the function 0 Ω I(0,X(ω))(x) as a function of ω instead of x. Now note the condition x ∈ (0, X(ω)) in (1.1) is equivalent to X(ω) ≥ x. So Z Z Ω I(0,X(ω))(x)dP (ω) = Then Z Z Iω:X(ω)≥x (ω )dP (ω) = P (ω : X(ω ) ≥ x) = P (X ≥ x) = 1−P (X ≤ x) = 1−F (x) Ω ∞ Z ∞ Z Z I(0,X(ω))(x)dxdP (ω) = I(0,X(ω))(x)dP (ω)dx = By equating (1.1) and (1.3), we conculde 0that Ω 0 Ω Z ∞[1 − F (x)]dx E[X] = 0 ∞ 0 [1 − F (x)]dx (1.3) 1.3 Exercise 1.6 1. Z E[euX ] = = = = = = = ux 1e f (x)dx Z √ exp 1 2πσ Z √ exp 1 2πσ Z √ exp 1 2πσ Z √ exp 1 2πσ Z √ exp 1 2πσ Z √ exp 2πσ = 1 · exp 2. 1.4 (x − µ)2 2σ 2  dx  x2 − 2µx + µ2 dx ux − 2 2σ   2  µ2  2 1 −2σ 2 x − (2µ + 2uσ )x − dx µ2 2σ 2 (µ + uσ2 )2 −   2   1 x − (µ + uσ2 ) 2 + 12σ −2σ 2 dx2 2 µ 2σ 2 2   2 (µ + uσ ) −   2σ1 2 1  2 dx · exp − 2 x − (µ + uσ ) µ2 σ 4 2σ 2σ 2   2σ 2 2+  1 2uµσ − 2 y 2 dy · exp 2σ 2σ 2  1 uµ + u2 σ 2 2 ux − E[φ(X)] = E[euX ] = euµ+ 1 2u 2 2 σ ≥ euµ = φ(µ) = φ(EX) Exercise 1.7 1. f (x) = 0 for any x; R R 2. limn→∞ fn (x)fdxis =not 1 6= 0 = fRefer (x)dx;to the following diagram for the graphs of the functions: 3. The sequence monotone. i Figure 1: Graphs of Exercise 1.7 with n=1,2,3,4 from the top respectively 1.5 Exercise 1.8 (i) We consider the sequence etX(ω) − esn X(ω) t − sn Yn (ω) = First note that lim Yn (ω) = sn →t d [etX(ω) ] = X(ω)etX(ω) dt Also by the Mean value Theorem, there exists θn ∈ [sn , t] (or [t, sn ]) such that Yn (ω) = etX(ω) − esn X(ω) = X(ω)eθn (ω)X(ω) t − sn Note that for n sufficiently large, θn ≤ max{t, sn } ≤ 2|t| since sn → t, so |Yn (ω)| = X(ω )eθn (ω)X(ω) ≤ X(ω )e2|t|X(ω) Also note that for any n and E[XetX ] < ∞ for any t, so by Dominated Convergence Theorem, we have lim EYn = lim Z Z n→∞ n→∞ tXω∈Ω ] =E[Xe Yn (ω )dP (ω ) = i.e., lim ω∈Ω n→∞ X(ω)etX(ω) dP (ω) Z Yn (ω )dP (ω ) = ω∈Ω lim EYn = E[ lim Yn ] = E[XetX ] n→∞ n→∞ (ii) If X can take both positive and negative values, then write X = X + − X − , where X + = max{X, 0} and X − = max{−X, 0}. Then by the Mean value Theorem, there exists θn ∈ [sn , t] (or [t, sn ]) such that Yn (ω ) = X(ω )eθn (ω)X(ω) Then     |Yn (ω)| = X(ω)eθn (ω)X (ω)  ≤ |X| eθn (ω)X(ω) ≤ |X| e2|t||X| (1.4) Now we need to show that E[|X| et|X| ] < ∞ for any t to apply Dominated Convergence Theorem. For any t, Z E[|X| et|X| ] = |X| et|X| dP (ω) Zω∈Ω 1X(ω)≤0 · |X| et|X| dP (ω) Z = Zω∈Ω 1X(ω)≥0 · |X| et|X| dP (ω) + + tX + Z ω∈Ω 1X(ω)≤0 · X − etX X+ e dP (ω) + = 1X(ω)≥0+· tX − tX − ] + E [1X(ω)≤0 · Xω∈Ω e =E[1 ω∈Ω X(ω)≥0 · X e − dP (ω) ] Since E[|X| etX ] < ∞ for any t, so Z E[|X| etX ] = |X| etX dP (ω) Z ω∈Ω = ω∈Ω 1X(ω)≤0 · |X| etX dP (ω) Z 1X(ω)≥0 · |X| etX dP (ω) + ω∈Ω Z Z 1X(ω)≤0 · X − e−tX + X++ etX dP (ω) + = 1X(ω)≥0+ · tX − − −tX ] + E [1X(ω)≤0 · Xω∈Ω e =E[1 ω∈Ω X(ω)≥0 · X e ]<∞ − dP (ω) which implies for any t + E[1X(ω)≥0 · X + etX ] < ∞ and E[1X(ω)≤0 · X − e−tX and E[1X(ω)≤0 · X − etX − ]<∞ Or equivalently for any t, we have + E[1X(ω)≥0 · X + etX ] < ∞ Hence − + E [|X| et|X| ] = E [1X(ω)≥0 · X + etX ] + E[1X(ω)≤0 · X − etX ]<∞ − ]<∞ for any t. Then E[|X| e|t||X| ] < ∞ for any t. Then by (1.4) and Dominated Convergence Theorem, we have lim EYn = E[ lim Yn ] = E[XetX ] n→∞ n→∞ which is φ′ (t) = E[X etX ]. 1.6 Exercise 1.10 1. Since Z is a nonnegative random variable and 1 1 Z Z(ω)dω = 0 ·2 + 2 ·2 = 1 E[Z] = Ω by Theorem 1.6.1, P̃ is a probability measure. 2. ˜ Z Z Z P (A) = Z(ω)dP (ω) = Z(ω )dP (ω ) + } 2 · P (A ∩ {ω : 1/2 A∩{≤ω:1/2≤ ≤1}) ω ≤ω:0≤ω≤1/2 1/2}) + ω ≤ω1} = 0 A· P (A ∩ {ω : 0 ≤ A∩{ Z(ω)dP (ω) ≤ 0 + 2 · P (A) = 0 since P˜(A) ≥ 0, ˜P (A) = 0. 0dω = 0. 3. Let A = {ω : 0 ≤ ω ≤ 1/2}, then P (A) = 1/2 > 0, ˜ R 1/2 R0 1/2 P (A) = 0 Z(ω )dω = 1.7 Exercise 1.11 By (1.6.4) of Text, we have for any u ∈ R ˜ E  euY i h   2 1 22 θ 1 =E euY Z = E e2u(X+θ) e−θX− 1 θ2 = e 2 (u−θ) · euθ− h i 1 (u−θ)X uθ− 2 1 2 2 1 1 =E =e 2 ue −uθ+ 2·θe+uθ− 2 θ  =eu 2 /2 therefore Y is standard normal under P̃ . 2θ 2 1.8 Exercise 1.14 Since for a ≥ 0 Z P (X ≤ a) = 1 − e−λa = where a f (x)dx −∞ ( x 0< 0 λe0,−λx , x ≥ f (x) = f (x) is the density function for such expoential r.v. 1. ˜ Z Z(ω )dP (ω ) P (Ω) = Z = Z = Z = = 1 λ̃ −( ˜ λ−λ)x e f (x)dx λ Ω ∞ −∞ λ̃ −( ˜ λ−λ)x e · (λe−λx )dx λ ∞ 0 λ̃e− ∞ 0 ˜ λx dx 2. ˜ χω:X(ω)≤a Z(ω)dP (ω) Z P (X ≤ a) = Ω a 0 Z = Z a 0 λ̃ −( ˜ λ−λ)x e · (λe−λx )dx λ λ̃e− ˜ λx dx = ˜ λa = 1 − e− 2 Chapter 2 2.1 Exercise 2.2 1. By the definition, σ(X) = {{X ∈ B}; B ranges over the subsets of R}. when B = {1}, {X ∈ B} = {X = 1} = {ω ∈ Ω2 : S2 (ω) = 4} = {HT , T H} when B = {0}, {X ∈ B} = {X = 0} = {ω ∈ Ω2 : S2 (ω) 6= 4} = {HH, T T } So all the sets in σ(X) are φ, Ω2 , {HT, T H} and {HH, T T }. 2. By the definition, σ(S1 ) = {{S1 ∈ B}; B ranges over the subsets of R}. when B = {8}, {S1 ∈ B} = {S1 = 8} = {ω ∈ Ω2 : S1 (ω) = 4} = {HT, HH } when B = {2}, {S1 ∈ B} = {S1 = 2} = {ω ∈ Ω2 : S1 (ω) 6= 4} = {T H, T T } So all the sets in σ(X) are φ, Ω2 , {HT, HH} and {T H, T T }. 3. Here we only investigate the non-trivial cases: ˜ P (A) P̃ (B) B ∈ σ(S1 ) A ∩ B 1 1 {HT, HH} {HT} ( 41 + 4 ) · ( 14 + 4 ) 1 1 1 1 + + {T H, T T } {TH} 4 ( 4 ) · ( 4 4 ) 1 1 {HH, T T } {HT, HH} {HH} (4 1+ 4 ) · ( 14 + 4 ) 1 1 ) · (1 + 1 ) + {HH, T T } {T H, T T } {TT} 4 ( 4 4 4 A ∈ σ(X) {HT, T H} {HT, T H} = = = = 1 4 1 4 1 4 1 4 ˜ P (A ∩ B) 1 4 1 4 1 4 1 4 therefore for any A ∈ σ(X) and B ∈ σ(S1 ), we have P˜(A ∩ B) = P̃ (A) P̃ (B). So σ (X) and σ (S1 ) are independent under ˜P . 4. Again we only investigate the non-trivial cases: A ∈ σ (X) {HT, T H} {HT, T H} {HH, T T } {HH, T T } B ∈ σ (S1 ) A ∩ B {HT, HH} {HT} P2 (A)P (B)4 ( 92 + 9 ) · ( 92 + 9 ) = 2 1 {T H, T T } {TH} 9 (+2 9 ) · ( 92 + 9 ) = 1 4 {HT, HH} {HH} (9 4+ 9 ) · ( 92 + 9 ) = 1 1 {T H, T T } {TT} 9 (+4 9 ) · ( 92 + 9 ) = 8 27 4 27 10 27 5 27 P (A ∩ B) 2 9 2 9 4 9 1 9 therefore for some A ∈ σ(X) and B ∈ σ(S1 ) (indeed for any non-trivial elements as shown), we have P (A ∩ B) 6= P (A)P (B). So σ(X) and σ(S1 ) are not independent under P . 5. If we are told that X = 1, then we have S2 = 4, i.e., the event must be HT or T H. But since P (HT ) = P (T H) = 2/9, we canot use the un-conditioned probability for S1 any longer. In other words, based on the current information, we shall estimate P (S1 = 8|X = 1) = 1/2 = P (S1 = 2|X = 1). 2.2 Exercise 2.3 Since V is normal random variable and E[V ] = E[X cos θ + Y sin θ] = cos θE [X] + sin θE [Y ] = cos θ · 0 + sin θ · 0 = 0 V ar[V ] = V ar[X cos θ + Y sin θ] = cos2 θ · V ar[X] + sin2 θ · V ar[Y ] = cos2 θ · 1 + sin2 θ · 1 = 1 so V is standard normal. Similarly W is also standard normal. To show the independence, we quote the fact that if Z is a standard normal random variable, then for any u, E[euZ ] = exp(u2 /2) (Exercise 1.6). Then for any real numbers a and b, E[eaV +bW ] = E[ea(X cos = E[e θ+Y sin θ )+b(−X sin θ +Y cos θ) X(a cos θ−b sin θ)+Y (a cos θ+b sin θ) = E[eX(a cos θ−b sin θ) ] ] ] · E[eY (a cos θ+b sin θ) ] by the independence of X and Y (a cos θ + b sin θ)2 +   (a cos θ 2− b sin θ)2 = exp 2 a cos2 θ + b2 sin2 θ + 2ab cos θ sin θ 2 +  2  22 2 2 a cos θ + b sin θ − 2ab cos θ sin θ = exp 2 b 2  22 + a = expaV = E[e ]E [ebW2 ] So V and W are independent. 2.3 Exercise 2.4 1. By the definition, E  euX+vY  ux+vxz dµZ (z)dµX (x) Z ∞Z ∞ e −∞ = 1 ux+vx·1 −∞ +  Z ∞  2e 1 = eux+vx·(−1) dµX (x) 1 −∞ 2 1 (u+v )x dµX (x) + = Z ∞ e Z ∞ e(u−v)x dµX (x) −∞ 2 e(u+v)X ] + E [e(u−v)X ] =E[ 1 (u+v)2 /2 1 2 + = e e(u−v) /2 2 2 2 2 euv + e−uv =e(u +v )/2 · 2 2 −∞ for any u and v . 2. Let u = 0 in the last result, then E[evY ] = e(0 2 +v 2 )/2 · e0·v + e−0·v 2 = ev 2 /2 So Y must be a standard normal random variable. 3. If X and Y are independent, then by Theorem 2.2.7 of text, the joint moment generating function must factor as: EeuX+vY = EeuX EevY But apparently EeuX+vY = e(u 2 +v 2 )/2 · euv + e−uv 2 6= eu 2 /2 · ev 2 /2 = EeuX EevY So they are not independent. 2.4 Exercise 2.5 Z ∞ −∞ fX (x) = fX,Y (x, y)dy 2|x| + y   (2|x| + y)2 dy 2  2  Z ∞ z |x| dz (z = y + 2|x|) √ exp − = 2 1 2π z 2 z=∞ = − √ exp[−  2π z=|x|  ]  1 x2 2 = √ exp[− ] 2 2π Z = Z fY (y) = ∞ −|x| ∞ −∞ √ z 2π exp − fX,Y (x, y)dx 2|x| + y Z √ = |x|≥−y 2π  exp − (2|x| + y)2 2  dx if y ≥ 0, then 2|x| + y   (2|x| + y)2 dx 2   Z ∞ 0 (2x + y)2 dx √ exp − =2 z2π 2 (z = 2x + y)  2  Z ∞ 2dz z y √ exp − =2 2 1 2 2π y = √ exp[− ] 2π 2 Z ∞ −∞ fY (y) = √ exp 2x2π +y − if y ≤ 0, then 2|x| + y    Z y (2|x| + y)2 (2|x| + y)2 −∞ √ exp − dx + dx 2 2 −2x +2π y   Z y Z ∞ 2  2  (2x + y) (−2x + y) −y √ √ exp − exp − dx + −∞ dx = 2 2 2x 2π 2π+ y   Z ∞ 2 (2x + y) −y dx √ exp − =2 2 z2π (z = 2x + y)  2  Z ∞ 2dz z − = 2 −y √ exp y2 1 2 2π ] = √ exp[− 2 2π Z ∞ −y fY (y) = 2|x| + y  √ exp 2x 2π +y − Therefore both X and Y are standard random variable. It is clear that fX,Y (x, y) 6= fX (x) · fY (y), so they are not independent. And it is also clear that E[X] = E[Y ] = 0, and E[XY ] Z ∞ Z Z −∞ ∞ Z ∞ −∞ = = Z −∞ ∞ Z 0 ∞ Z = Z ∞ −|x| = ∞ −x ∞ −x 0 Z = =0 ∞ 0 Z ∞ −x xyfX,Y (x, y)dydx 2|x| + y  (2|x| + y)2 dydx 2 −2x + y xy ·     Z 0 Z ∞ xy · 2 (2x + y) (−2x + y)2 x √ √ exp − exp − dydx + dydx 2t + y 2x2π +y 2 2 2π −∞ xy ·     Z 0Z ∞ (−t)y · 2 2 (2x + y) (2t + y) −t √ √ exp − exp − dydx + dy(−dt) 2 2 2π ∞ (t = −x) 2π xy · √ 2x +2π y xy ·  exp 2x + y √ 2π  exp − (2x + y)2 − 2  Z dydx − 0 ∞ Z ∞ −x xy · 2x + y √ 2π so E[XY ] = E[X ]E[Y ], which implies X and Y are uncoorelated.  exp − (2x + y)2 2  dydx 2.5 Exercise 2.7 Let µ = E[Y − X], then we consider   V ar(Y − X) =E (Y − X − µ)2 h i 2 =E ((Y − E[Y |G]) + (E[Y |G] − X − µ)) h i 2 =E (Y − E[Y |G])2 + (E[Y |G] − X − µ) + 2 (Y − E [Y |G]) (E[Y |G] − X − µ)   2 ≥E ar(Err) (Y − E[Y |G ]) 0 +|G]) 2E [((E[Y Y − |G] E[Y−|G =V + 2E [(Y −+ E[Y X])− (E[Y µ)] |G ] − X − µ)] (2.1)   2 bothE(Err) X and =E[Y |G]V are G−measurable TheNow last note step that is since E[Y |G])variables, . 0, so ar(Err) = E (Y −random E [(Y − E[Y |G]) (E[Y |G] − X − µ)] =E [Y E[Y |G] − XY − µY − E[Y |G ]E [Y |G] + XE [Y |G] + µE[Y |G]] =E [Y E[Y |G] − XY − µY − E[Y E [Y |G ]|G] + E[XY |G ] + µE[Y |G]] by G−measurable =E[Y E[Y |G]] − E [XY ] − µE [y] − E [E [Y E[Y |G ]|G]] + E [E[XY |G]] + µE [E [Y |G]] =E[Y E[Y |G]] − E[XY ] − µE [y] − E [Y E[Y |G]] + E [XY ] + µE[Y ] =0 by iterated conditioning E[E [Z|G]] = E[Z ] By substituting this result into (2.1), we will have V ar(Y − X) ≥ V ar(Err) 2.6 Exercise 2.10 By the notations, for any A ∈ σ(X), Z g(X)dP Z g(X(ω ))dP (ω ) A = {ω ∈Ω:X(ω )∈B} Z g(X(ω)) · (X ∈ B )dP (ω) = Z = Z Ω ∞ −∞ g(x)fX (x) · Z ∞ ∞ −∞ = −∞ (x ∈ B)dx yfX,Y (x, y)  (x ∈ B)dx · fX (x) · fX (x) y · (x ∈ B)fX,Y (x, y)dxdy dy Z ∞ Z ∞ −∞ = −∞ · (X ∈ B)] =E[Y by the notation of (2.6.3) =E[Y · by the notation of (2.6.2) and A = {X ∈ B } (A)] Y dP Z = A whic verifies the partial averaging property, hence proves E[Y |X] = g(X).