Stochastic Calculus and Black-Scholes Theory MTH772P Exercises Sheet 1

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1 Stochastic Calculus and Black-Scholes Theory MTH772P Exercises Sheet. For ξ, ξ 2, i.i.d. with P(ξ i = ± = /2 define the discrete-time random walk W =, W n = ξ ξ n. (i Formulate and prove the property of independence of increments of (W n, n. (ii Show that (W n, n is a discrete-time Markov chain. (iii Show that (W n, n is a discrete-time martingale. (iv Calculate the covariance Cov(W i, W j. (v Find the limit distribution of W nt / n as n, for t >. Solution (i For any integer times = j < j <... < j k the increments W j W j = ξ ξ j,..., W jk W jk = ξ jk ξ jk are independent. Because ξ, ξ 2,... are independent by the assumption, the vectors (ξ,..., ξ j,..., (ξ jk +,..., ξ jk are independent, hence the increments are independent as functions of the independent random vectors. (ii For any possible path w,..., w n of the random walk we have P(W n = w n W n = w n, W n 2 = w n 2,..., W = w = P(W n W n = w n w n W n = w n, W n 2 = w n 2,..., W = w. In this conditional probability the event {W n W n = w n w n } is the same as {ξ n = w n w n }, while the event {W n = w n, W n 2 = w n 2,..., W = w } can be written in terms of ξ,..., ξ n. It follows from the independence of ξ i s that P(W n = w n W n = w n, W n 2 = w n 2,..., W = w = P(W n W n = w n w n = P(W n W n = w n w n W n = w n = which is the Markov property for discrete-time processes. (iii We have using rules for conditional expectations P(W n = w n W n = w n, E(W n+ W,..., W n = E(W n + ξ n+ W,..., W n = E(W n W,..., W n + E(ξ n+ W,..., W n = W n + Eξ n+. Note that we could also write the conditional expectation E(W n+ W,..., W n as E(W n+ F n, where F n is the σ-algebra generated by the events {W = w,..., W n = w n } with arbitrary w,..., w n. Also, F n is the σ-algebra generated by the random variables ξ,..., ξ n. (iv Assume i j. Since EW i = we have Cov(W i, W j = E(W i W j = E(W i (W i + (W j W i = E(Wi 2 +E(W i E(W j W i = Var(W i = ivarξ = i. Thus Cov(W i, W j = i j for any i, j. (v The limit distribution of (ξ ξ nt / nt (for every fixed t > and n is N (, by the central limit theorem. From this, the limit distribution of W nt / n is N (, t.

2 2. Let B = (B(t, t be a standard Brownian motion. Show that the following processes are standard BM: (i X(t = B(t + s B(s, where s is constant. (ii X(t = B(ct/ c, for any c >, (iii X(t = B( t B(, where t [, ] (this BM is defined on [, ]. Solution (i The process has continuous paths and X( =. The increments of X(t i+ X(t i over intervals of partition = t < t <... < t n are the increments of the BM over the intervals between times s < t + s <... < t n + s, hence they are independent and N (, (t i+ t i -distributed. (ii B(ct is N (, ct-distributed, hence B(ct/ c is N (, t-distributed (check the mean and the variance. We have X( =. The increments of X over the intervals of partition = t < t <... < t n are the increments of BM over the intervals of partition = ct < ct <... < ct n, hence the X-increments are independent. (iii The increments of X(t j+ X(t j over = t < t <... < t n = are the increments of the BM over the intervals of partition t n <... < t. The independence of increments follows. The rest is obvious. 3. For X a random variable with density function f consider the event A = {X }. (i Define G to be the σ-algebra generated by A (i.e. the smallest σ-algebra containing event A. Write down the list of all elements of the σ-algebra G. (ii In terms of integrals with density f, describe the random variable E(X 3 G. (iii Using the formulas you derived in (ii show explicitly that E(E(X 3 G = E(X 3. (iv Make the calculations for (ii, (iii assuming that X has N (, distribution. Solution (i G = {, Ω, {X }, {X > }}. (ii The events {X }, {X > } are disjoint and their union is Ω. Hence we can write E(X 3 G = E(X 3 X (X + E(X 3 X (X <, where ( is the indicator random variable. In particular, E(X 3 G may take two values, depending on whether X or X <. In terms of the density, these values are E(X 3 X = x 3 f(xdx f(xdx, E(X3 X < = x3 f(xdx f(xdx. 4. Let (F t, t be a filtration for BM. That means that {ω Ω : B(s x} F t for s t and x R, and that the increments of BM after t are independent of F t. For < a < b < c show that B(c B(b is independent of F a. Solution We have B(c B(b independent of F b, which means that any event {B(c B(a < x} is independent of any event A F b. But F a F b, hence B(c B(b is also independent of F a. 5. (BM with drift Let X(t = B(t + tµ. Show that (X(t, t is a Markov process and find its transition density. Is the process a martingale? 2

3 Solution Let (F t, t be a filtration for the BM. The Markov property of the BM itself means that for s < t E(f(B(t F s = E(f(B(t B(s for any function f. (Note that if we take f(x = (x a the conditional expectation becomes the conditional probability E[(B(t a F s ] = P[B(t a F s ]. Thus from the Markov property of BM E(f(X(t F s = E(f(B(t + tµ F s = E(f(B(t + tµ B(s. But conditioning on B(s is the same as conditioning on X(s = B(s + sµ, because X(s and B(s uniquely determine one another. Hence the above becomes and so E(f(B(t + tµ B(s = E(f(B(t + tµ X(s = E(f(X(t X(s, E(f(X(t F s = E(f(X(t X(s, which is the Markov property for (X(t, t. To compute the transition probability function of X we should reduce to the BM, and there are various equivalent ways to do that. The probability that X moves from X(s = x to some value X(t y is P(X(t y X(s = x = P(B(t + tµ y B(s = x sµ = P(B(t y tµ B(s = x sµ = P(B(t B(s y x (t sµ B(s = x sµ = P(B(t B(s y x µ(t s = y x (t sµ exp{ u 2 /(2t 2s}du The transition density from x to y in time t s is obtained by differentiating this (conditional distribution function in y p(t s, x, y = e (y x (t sµ2 /(2t 2s Another possibility is to use the fact that the transition density satisfies E(f(X(t X(s = x = p(t s, x, yf(ydy for any function f. Using Lemma.9 from the lecture notes E(f(X(t X(s = x = E[f((B(t B(s + B(s + tµ B(s = x sµ] = E[f((B(t B(s + x sµ + tµ] = p(t s,, yf(u + x sµ + tµdu = p(t s, x, yf(ydy, where the last step used change of variable u + x sµ + tµ = y and the transition density p(t s,, y = exp( 2y 2 /(2t 2s/ of the BM. 3

4 Finally, somewhat heuristic but quick way is as follows. Process X moves from X(s = x to X(t [y, y + dy] when the BM moves from B(s = x sµ to B(t [y tµ, y + dy tµ]. The latter is an event of probability p(t s, x sµ, y tµdy = p(t s,, y x (t sµdy. Discarding dy yields the transition density p(t s, x, y = p(t s,, y x (t sµ for the process X. 6. (Geometric BM Let S(t = S( exp(νt + σb(t, where S(, σ, ν are positive constants. Show that (S(t, t is a Markov process and find its transition density. Solution The Markov property is shown as in Exercise 5: if S(s = x then B(s = (log x S νt/σ, so we can compute S(s from B(s and vice versa. Let for shorthand b = (log x S νt/σ. We have E[f(S(t S(s = x] = E[f(S(t B(s = b] = E[f(S(e νt+σ(b(t B(s+σB(s B(s = b] = E[f(S(e νt+σ(b(t B(s+σb ] = Using the change of variable (recall the definition of b f(s(e νt+σu+σb e u2 /(2t 2s du. y = S(e νt+σu+σb, u = log(y/x ν(t s, du = dy σ σy the above integral becomes exp ( (log(y/x ν(t s2 2σ f(y 2 (t s σy dy. Therefore the transition density function of (S(t, t is ˆp(t s, x, y = exp ( (log(y/x ν(t s2 2σ 2 (t s σy 7. (Black-Scholes formula Let S(t = S( exp((r σ 2 /2t + σb(t, where S(, σ, r are positive constants. For K > and T > show that E[e rt (S(T K + ] = S(Φ(d + (T, S( Ke rt Φ(d (T, S(, where Φ is the standard normal distribution function, and d ± (T, S( = ( σ log S( T K σ2 + (r ± 2 T Solution We need to compute the integral integrate E[e rt (S(T K + ] = e rt σ (log(k/s( (r σ2 /2T 4.. ( S(e (r σ2 /2T e +σx x 2 /(2T K dx. T

5 Substituting y = x/ T and using linearity of the integral the above becomes S(e σ2 T/2 S( σ T (log(k/s( (r σ2 /2T σ T σ T (log(k/s( (r σ2 /2T Ke rt σ T (log(k/s( (r σ2 /2T e y2 /2+σ T y dy e y2 /2 dy = e z2 /2 dz Ke rt Φ(d (T, S( = S(Φ(d + (T, S( Ke rt Φ(d (T, S(. 5