f(s)dw Solution 1. Approximate f by piece-wise constant left-continuous non-random functions f n such that (f(s) f n (s)) 2 ds 0.

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1 Advanced Financial Models Example shee 3 - Michaelmas 217 Michael Tehranchi Problem 1. Le f : [, R be a coninuous (non-random funcion and W a Brownian moion, and le σ 2 = f(s 2 ds and assume σ 2 <. By considering he L 2 consrucion of he sochasic inegral, show ha f(sdw s is a normal random variable wih mean zero and variance σ 2. Soluion 1. Approximae f by piece-wise consan lef-coninuous non-random funcions f n such ha (f(s f n (s 2 ds. The exisence of such a sequence is a consequence of he densiy of sep funcions in L 2 (Leb. Bu for a concree consrucion, le MN f M,N = f( N i 11 ( N i 1, N i ] i=1 where N i = i/n. Noe (f(s f M,N (s 2 ds = M f(s 2 ds + M sup{ f(s f( : s, M, s 1 N }. For each n 1, by square-inegrabiliy of f we can find M n such ha M n f(s 2 ds < 1 n and by uniform coninuiy of f on he compac [, M n ] we can find N n such ha sup{ f(s f( : s, M n, s 1 N n } 1 nm n. Seing f n = f Mn,Nn does he job since Now f n (dw = i f( i 1 (W i W i 1 is he sum of independen mean-zero normal random variables, and hence is mean-zero normal wih variance f( i 1 2 ( i i 1 f( 2 d. i Since f n (dw f(dw in L 2 (P by he definiion of he sochasic inegral, we will be done once we appeal o he following sandard fac: Fac. Suppose X n N(µ n, σn 2 and µ n µ and σn 2 σ 2. If X n X in L 2 hen X N(µ, σ 2. 1

2 Proof. Since X n X in L 2, hen X n X in disribuion. Le F n and F be he disribuion funcions of X n and X, respecively. A he poins of coninuiy of coninuiy of F we have F (x = lim F n (x n ( x µn = lim Φ n σ n ( x µ = Φ σ by he coninuiy of he sandard normal disribuion funcion Φ. Problem 2. * (Ornsein Uhlenbeck process Le W be a Brownian moion, and le X = e a x + b e a( s dw s for some a, b, x R. (a Verify ha (X saisfies he following sochasic differenial equaion: dx = ax d + b dw, X = x. (b Show ha X N (e a x, b2 2a (e2a 1. (c Wha is he disribuion of he random variable X d? Soluion 2. (a Since we can apply Iô s formula (b Since X = e a (x + b dx = e a ( be a dw + (x + b = b dw + ax d e as dw s (e a( s 2 ds = e2a 1 2a e as dw s ae a d his par follows from Problem 1. (c Mehod 1: Noe ha by rearranging he sochasic differenial equaion we have X d = 1 a (X T x bw T 2

3 and hence X d is normally disribued wih mean (e at 1x/a. To compue he variance, firs noe ha ( Cov(X T, W T = Cov b e a(t, dw dw = b e a(t d = b a (eat 1 by Iô s isomery. Hence ( Var X d = 1 ( Var(XT 2b Cov(X a 2 T, W T + b 2 Var(W T Mehod 2: X d = = b2 2a 3 (e2at 4e at aT. = e at x d + e at x d + = eat 1 x + a s e a( s b dw s d e a( s b d dw s e a(t s 1 b dw s a Hence X d is normally disribued wih mean (e at 1x/a and variance b 2 a 2 (e a(t s 1 2 ds = b2 2a 3 (e2at 4e at aT This calculaion is useful in he sudy of he Vasicek ineres rae model. Problem 3. Le W be a Brownian moion. Show ha if Y = W 3 maringale (1 by hand, and (2 by Iô s formula. 3W hen Y is a Soluion 3. (1 By hand: Since Gaussian random variables have finie momens of all orders, Y is inegrable. Indeed, we have E( Y E( W 3 + 3E( W = C 3/2 < where C = 5 2/π. Therefore, using he independence of he incremens of W we have E(Y F s =E(W 3 3W F s =E[(W W s + W s 3 3(W W s + W s F s ] =E[(W W s 3 ] + 3E[(W W s 2 ]W s + 3E(W W s W 2 s + W 3 s 3E(W W s W s = + 3( sw s + + W 3 s + W s for s <. =Y s 3

4 (2 By Iô s rule: dy = d(w 3 3W = (3W 2 dw + 3W d 3( dw + W d = 3(W 2 dw ( and hence Y is a local maringale. Recall ha if E α2 s ds < for all hen ( he process α s dw s is a maringale. Again, i s clear ha he inegrand is square inegrable in his case since Guassian random variables have finie momens of all orders. Bu, jus o be explici, E [3(W 2 s s] 2 ds = 9 and hence (Y is a maringale. E(W 4 s 2W 2 s s + s 2 ds = 9 2s 2 ds = 6 3 < Problem 4. (Hea equaion Le W be a scalar Brownian moion, and le g : [, T ] R R be a smooh funcion ha saisfy he parial differenial equaion wih erminal condiion g g 2 x = 2 g(t, x = G(x. (a Show ha (g(, W [,T ] is a local maringale. (b If he funcion g is bounded, deduce he formula g(, x = G(x + T z e z2 /2 dz. (c Use Problem 2 o find explicily he unique bounded soluion o he PDE wih erminal condiion h + x h x h x = 2 h(t, x = cos x. Useful fac: If Z N(µ, σ 2 hen E(cos Z = e σ2 /2 cos µ. Soluion 4. (a By Iô s formula: ( g dg(, W = g d + g 2 x 2 x dw = g x dw and hence (g(, W [,T ] is a local maringale. 4

5 (b Recall a bounded local maringale is a rue maringale. In paricular, by he independence of he incremens of Brownian moion, we have g(, W = E[g(T, W T F ] = E[G(W T F ] = E[G(W + W T W F ] = G(W + T z e z2 /2 dz since W T W N(, T. Since he above formula holds idenically, we have he desired inegral represenaion of he soluion of he hea equaion. (c Le X be he Ornsein Uhlenbeck process dx = X d + dw. By Iô s formula, he process (h(, X [,T ] is a local maringale if and only if h is soluion o he PDE. Now if h is bounded, hen (h(, X [,T ] is a rue maringale. Using problem 2, we have h(, X = E[cos X T F ] [ ( = E cos e T X + e T s dw s F ] = cos(e T X exp( (e 2(T 1/4 since he condiional disribuion of X T given F is normal wih mean e T X and variance (e 2(T 1/2. In paricular, he unique soluion is h(, x = cos(e T x exp( 1 4 (e2(t 1 Problem 5. (Sricly local maringale This is a echnical exercise o exhibi a local maringale ha is no a rue maringale. Le W = (W 1, W 2, W 3 be a hree-dimensional Brownian moion and le u = (1,,. I is a fac ha P(W u for all = 1. (a Le X = W u 1. Use Iô s formula and Lévy s characerisaion of Brownian moion o show ha dx = X 2 dz, X = 1 where Z is a Brownian moion. In paricular, show ha X is a posiive local maringale. (b By direcly evaluaing he inegral or oherwise, show ha E(X = 2Φ( 1/2 1 for all >, where Φ is he disribuion funcion of a sandard normal random variable. Why does his imply ha X is a sricly local maringale? Soluion 5. (a Le f(x 1, x 2, x 3 = ((x x x 2 3 1/2 so ha ( f, f, f = [f(x 1, x 2, x 3 ] 3 (x 1 1, x 2, x 3 x 1 x 2 x 3 5

6 2 f x f x f x 2 3 = f 3 + 3f 5 (x f 3 + 3f 5 x 2 2 f 3 + 3f 5 x 2 3 =. In paricular, Iô s formula yields dx = X 3 [(W 1 1dW 1 + W 2 dw 2 + W 3 dw 3 ]. Since X can be wrien as a sochasic inegral of a hree dimensional Brownian moion, i is a local maringale. Now le Z be he local maringale such ha Z = and Since dz = X [(W 1 1dW 1 + W 2 dw 2 + W 3 dw 3 ]. d Z = X 2 [(W (W (W 3 2 ]d = d by consrucion, he process Z is a Brownian moion by Lévy s characerisaion heorem. (b Swich o spherical coordinaes: E(X = ( 3/2 e x2 1 /2 x2 2 /2 x2 3 /2 ( dx 1 dx 2 dx 3 x x x 2 3 = ( 3/2 = ( 1/2 π r= θ= π r= = ( 1/2 θ= φ= re r2 /2 r= r 2 sin θe r2 /2 r2 2 dφ dθ dr cos θ + 1 r 2 sin θe r2 /2 r2 2 dθ dr cos θ + 1 π r 2 2 cos θ + 1 θ= = ( 1/2 2(r1 {r> 1/2 } + r 2 1 {r }e 1/2 r2 /2 dr r= 1/2 e r2 /2 = 2 dr Noe ha E(X < X for all >, so X is a sricly local maringale. Problem 6. (sricly local maringales again (a Suppose ha X is posiive maringale wih X = 1. Fix T > and le dq dp = X T. Le Y = 1/X for all. Use Girsanov s heorem o show ha (Y T is a posiive maringale under Q. (b Coninuing from par (a, now suppose ha X has dynamics dx = X σ dw where W is a Brownian moion under P. Show ha here exiss a Q-Brownian moion Ŵ such ha dy = Y σ dŵ 6 dr

7 (c Le X be a posiive local maringale wih X = 1 and dynamics dx = X 2 dw. Our goal is o show ha X is a sricly local maringale. For he sake of finding a conradicion, suppose X is a rue maringale. In he noaion of pars (a and (b, show ha P(Y > = 1 bu Q(Y > = Φ( 1/2. Why does his conradic he assumpion ha X is a rue maringale? Soluion 6. (a Since P and Q are equivalen and P(X > for all = 1 hen Q(Y > for all = 1. Now o show ha Y is a Q-maringale, noe ha (b By Iô s formula, E Q (Y T F = EP (X T Y T F E Q (X T F = 1 X dy = dx 1 = X 2 dx + X 3 d X = Y σ (dw σ d Now by Girsanov s heorem, he process d ˇW = dw σ d defines a Q Brownian moion. And of course Ŵ = ˇW is a Brownian moion also. (c Now assuming X is a rue maringale, hen Girsanov s heorem applies and hence dy = Y σ dŵ = dŵ since σ = X. Hence Q(Y > = Q(Ŵ > 1 = Φ( 1/2 < 1. Bu since P(Y > = P(X > = 1. Therefore P and Q are no equivalen afer all. Problem 7. Consider a hree asse marke wih prices given by db B = 2 d ds (1 S (1 ds (2 S (2 Consruc an absolue arbirage. = 3 d + dw (1 2 dw (2 = 5 d 2 dw (1 + 4 dw (2. Soluion 7. If he pure invesmen sraegy is decomposed as (φ, π a good choice for he holding is sock is given by ( 2 1 π =, S (1 7 S (2

8 bu, of course, i is no unique. I remains o find he holding in he bank accoun φ. Noe ha he wealh X evolves as so he unique soluion wih X = is dx = r(x π S d + π ds = (2X + 5d X = 5 2 (e2 1. Since X > a.s. for >, his is an arbirage wih he holding in he bank accoun given by φ = X π S B = 5 11e 2 2B. Problem 8. * Consider a Black Scholes marke wih wo asses wih dynamics given by db = B r d ds = S (µ d + σdw Find a replicaing sraegy H and he associaed wealh process for a claim wih payou (1 ξ T = S p T for some p R (2 ξ T = (log S T 2 (3 ξ T = S sds Show ha your answer o par (3 is unchanged if we only assume ha S is a posiive Iô process. Soluion 8. Noe ha S T = S e (r σ2 /2(T +σ(ŵt Ŵ where Ŵ = W + (µ r/σ defines a Brownian moion for he equivalen maringale measure Q. (1 We could solve he Black Scholes PDE wih boundary condiion V (T, S = S p, bu i is easier o compue condiional expecions. so ha The hedging porfolio is (2 Similarly, since ξ = E Q [e r(t S p T F ] = S p e (p 1(r+pσ2 /2(T V (, S = S p e (p 1(r+pσ2 /2(T. π = V S (, S = ps p 1 e (p 1(r+pσ2 /2(T. ξ = E Q [e r(t (log S T 2 F ] = e r(t {[log S + (r σ 2 /2(T ] 2 + σ 2 (T } we have (3 Finally, ξ = E Q [ e r(t π = 2e r(t [log S + (r σ 2 /2(T ]/S. ] S s ds F = e r(t 8 r(t 1 e S s ds + S. r

9 Unforunaely, he payou is no of he form g(s T so our heorem for calculaing he replicaion porfolio doesn apply. Sill, le and π = φ = ξ π S B 1 e r(t r = 1 B e rt S s ds. Noice ha φ B + π S = ξ by consrucion and ha by a calculaion dξ = φ db + π ds, so (φ, π is a self-financing replicaion sraegy. To check he self-financing condiion we did no need o use he specific dynamics of S. Indeed, we need only assume ha we can apply Iô s formula. Problem 9. (Black Scholes formula Le X N(, 1 be a sandard normal random variable, and v and m be posiive consans. Express he expecaion in erms of Φ, he disribuion funcion of X. Soluion 9. E[(e v/2+ vx m + ] = = = = F (v, m = E[(e v/2+ vx m + ] (e v/2+vx m + /2 e x2 dx log m/ v+ v/2 log m/ v+ v/2 log m/ v+ v/2 = Φ ( log m v + (e v/2+vx /2 m e x2 dx e v/2+ vx x 2 /2 dx m e s2 /2 ds m v 2 log m/ v+ v/2 log m/ v v/2 ( m Φ log m v v 2 e x2 /2 dx e 2 /2 d Problem 1. (sricly local maringale in finance Consider a marke wih zero ineres rae r = and sock price wih dynamics ds = S 2 dw. Consider a European claim wih payou ξ T = S T. (a Show ha here exiss a rading sraegy which replicaes he claim wih corresponding wealh ξ = V (, S where V (, S = S [ ( 2Φ 1 S T ] 1. (b Consider he sraegy of buying S claims and selling ξ shares. The ime wealh is V = and he ime T wealh is V T = (S ξ S T >. Is his sraegy an absolue arbirage? 9

10 Soluion 1. (a I is sraigh-forward, if a bi edious, o verify V V 2 S4 S = 2 and lim T V (, S = S. The replicaion sraegy is given by π = V (, S S as usual. (b This candidae arbirage is no admissible. Indeed, le V = S ξ ξ S. Noe ha V is a local maringale, as i is he linear combinaion of wo local maringales. Now if he sraegy were admissible, he wealh V would be bounded from below and hence a supermaringale. Therefore we would have o conclude = V E(V T = E(S ξ T ξ S T = (S ξ E(S T Bu his conradics S > ξ. So in his marke i is impossible o lock in he sure fuure profi a zero iniial cos, because doing so leaves open he possibiliy ha he wealh is goes negaive beween imes = and = T. Noe, however, ha here is an admissible arbirage relaive o he asse wih price S. Indeed, he sraegy of holding one share of he claim is a relaive arbirage, since he iniial discouned wealh is ξ /S < 1 and he erminal discouned wealh is ξ T /S T = 1. This is only a relaive arbirage since you do no shor S, and hence mus sar wih posiive iniial wealh. 1