RMSC 4005 Stochastic Calculus for Finance and Risk. 1 Exercises. (c) Let X = {X n } n=0 be a {F n }-supermartingale. Show that.

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1 1. EXERCISES RMSC 45 Stochastic Calculus for Finance and Risk Exercises 1 Exercises 1. (a) Let X = {X n } n= be a {F n }-martingale. Show that E(X n ) = E(X ) n N (b) Let X = {X n } n= be a {F n }-submartingale. Show that E(X ) E(X 1 ) E(X 2 ) (c) Let X = {X n } n= be a {F n }-supermartingale. Show that E(X ) E(X 1 ) E(X 2 ) 2. Let X be an integrable random variable defined on (Ω, F, P), and G is a sub-σ-field of F. Show that the random variable E(X G) is also integrable. 3. Given a filtered probability space (Ω, F, {F n } n=, P), and an integrable random variable X. Show that Y n E(X F n ) is a {F n }-martingale. 4. Let N(t) be a Poisson process with intensity λ. Show that M(t) N(t) λt is a martingale. 5. Let Y, Y 1,... be a Markov Chain process with transition probability matrix P and state space {1, 2,..., s}. Suppose f is a bounded function defined on {1, 2,..., s} and satisfies s f(i) = p ij f(j). j=1 Set X n = f(y n ) and F n = σ{y, Y 1,..., Y n } for n =, 1, 2,.... Show that {X n } is a {F n }-martingale. 1

2 1. EXERCISES 6. Show that T is a stopping time if and only if {T = n} F n n 7. If T 1 and T 2 are two stopping times on (Ω, F, {F n } n=, P), show that is also a stopping time. T min {T 1, T 2 } = T 1 T 2 8. Given a filtered probability space (Ω, F, {F n } n=, P). Suppose a stochastic process X = {X n } n= is adapted. Let B be a Borel set and define Show that τ is a stopping time. τ = min {n X n B}. 9. Suppose t is a positive constant (deterministic). Show that T t is a stopping time. 1. Let ξ 1, ξ 2,... be a sequence of independent random variables with P(ξ n = 1) = P(ξ n = 1) = 1/2. Define X =, X n = n i=1 ξ i, and F n = σ{x 1, X 2,..., X n } for n = 1, 2,... (i.e. X is a symmetric random walk). Let τ = min {n X n = K} where K is a positive integer and Show that (a) τ is a stopping time, (b) {Y n } n= is a martingale, Y n = ( 1) n cos [π(x n + K)] n. (c) E(( 1) r ) = ( 1) K, using the version of Optional Stopping Theorem stated in Question Let {W t } t be a P-Brownian motion. Show that (a) Cov(W s, W t ) = s t, (b) {cw t/c 2} t is also a P-Brownian motion. 2

3 1. EXERCISES 12. Let {W t } t be a Brownian motion, and define a random variable T a inf {t W t = a}, which is the hitting time of level a, where a is positive. Show that P(T a t) = 2P(W t a). 13. Let {W t } t be a Brownian motion, and < t < t 1 <. Show that the probability that W t has at least one zero (i.e. the path of the Brownian motion intersect the t-axis) in the interval (t, t 1 ) is given by 2 π arccos t. t Show that B sdb s = 1 2 B2 t 1 2 t. 15. Show that (a) sdw s = tw t W sds (b) W s 2 dw s = 1 3 W t 3 W sds 16. For constant c and α, define Prove that X t = e ct+αw t. (1) dx t = (c α2 )X t dt + αx t dw t (2) 17. Use Itô lemma to prove that the following stochastic processes are martingales: (a) X t = e 1 2 t cosw t (b) X t = e 1 2 t sinw t (c) X t = (W t + t)exp( W t.5t) 18. Show that if θ is a fixed constant, then the process L t = exp( 3 θdw s 1 2 θ 2 ds)

4 is a martinglae. Hence show that. E(L t ) = 1 t 19. Derive the Black-Scholes Option Pricing Formula for a European Call Option, S Φ(d 1 ) Ke rt Φ(d 2 ), where d 1 = ln(s K ) + (r + σ2 /2) σ, d 2 = d 1 σ T ; T with exercise price K, expiration date T, current stock price S, and the stated market dynamics. 2 Solutions 1. (a) Since X is a martingale, E(X n+1 F n ) = X n, n =, 1, 2,.... Taking expectation on both sides yields E(E(X n+1 F n )) = E(X n ). Hence E(X n+1 ) = E(E(X n+1 F n )) = E(X n ), n =, 1, 2,... This means that E(X ) = E(X 1 ) = E(X 2 ) =.... (b) Since X is a submartingale, E(X n+1 F n ) X n, n =, 1, 2,.... Taking expectation on both sides yields E(E(X n+1 F n )) E(X n ). Hence E(X n+1 ) = E(E(X n+1 F n )) E(X n ), n =, 1, 2,.... This means that E(X ) E(X 1 ) E(X 2 ).... (c) The proof is similar to that of part (b). 4

5 2. The random variable X is integrable means that E( X ) <. Define X + = max(x, ), X = max( X, ). Observe that both X + and X are positive random variables, and X = X + X, X = X + + X. To show that E(X G) is integrable, we have to show that This can be done as follows: E( E(X G) ) <. E( E(X G) ) = E( E(X + X G ) = E( E(X + G) E(X G) ) E( E(X + G) E(X G) ) = E(E(X + G) + E(X G)) = E(E(X + + X G)) = E(E( X G)) = E( X ) < 3. To show that the process Y is a {F n }-martingale, we need to check the three defining properties of a martingale. By the properties of conditional expectation, Y n = E(X F n ) is F n -measurable. Since X is integrable, Y n = E(X F n ), which is a conditional expectation of X, is also integrable by Exercise 2. Finally, we show the martingale identity: E(Y n+1 F n ) = E(E(X F n+1 ) F n ) = E(X F n ) = Y n for n = 1, 2,..., in which the second equality follows from the tower property of conditional expectation. 5

6 4. Let F t = σ{n(s); s t} = σ{m(s); s t}. The adaptability and integrability of the process M are trivial, so we only show the martingale identity. For s t, we have E(M(t) F s ) = E(N(t) λt F s ) = E(N(t) F s ) E(λt F s ) = E(N(t) N(s) + N(s) F s ) λt = E(N(t) N(s) F s ) + E(N(s) F s ) λt = E(N(t) N(s)) + N(s) λt = λ(t s) + N(s) λt = N(s) λs = M(s) 5. Again, we have to show the three defining properties of a martingale. Since Y n is F n -measurable and X n = f(y n ), X n is also F n -measurable. This means that the process X is {F n }-adapted. Since f is bounded, there exists a positive constant C such that f(i) < C for i = 1, 2,..., s. Then E( X n ) = E( f(y n ) ) < E(C) = C < for n =, 1, Hence X n is integrable. Now we show the martingale identity. For n =, 1, 2,..., E(X n+1 F n ) = E(f(Y n+1 ) F n ) = E(f(Y n+1 ) σ{y, Y 1,..., Y n }) s = f(j)p Yn j j=1 = f(y n ) = X n where the third equality follows from the fact that Y is a Markov Chain with transition probability matrix P = (p ij ). 6. ) Observe that {T = n} = {T n}\{t n 1} = {T n} {T n 1} c 6

7 As T is a stopping time, we have {T n} F n and {T n 1} F n 1 F n. This implies that {T = n} F n. ) Now assume that {T = n} F n for n =, 1, 2,.... Observe that {T n} = {T = } {T = 1} {T = n 1} {T = n}. For any k n, we have by assumption that Hence {T n} F n. {T = k} F k F n. 7. To show that T = T 1 T 2 is a stopping time, we only need to show that {T n} F n for any positive integer n. This can be seen easily from the following relationship: {T n} = {T 1 T 2 n} = {T 1 n} {T 2 n} 8. By Exercise 6, we only have to show that {T = n} F n for any positive integer n. First we can decompose the set {T = n} as: {T = n} = {X / B} {X 1 / B} {X n 1 / B} {X n / B}. Next, we make the following observations: For k =, 1, 2,..., n 1, we have {X k / B} = {ω Ω X k (ω) / B} = {ω Ω X k (ω) B c } = X 1 k (Bc ) F k F n since B c is a Borel set and X k is F k -measurable (X is {F n }- adapted). Again, as X is {F n }-adapted, X n is F n -measurable. This implies that {X n B} = {ω Ω X n (ω) B} = Xn 1 (B) F n Combining the above two points, we know that {T = n} F n, hence T is a stopping time. 7

8 9. If T always equals to a positive constant t, then for any n N, we have { Ω if t n; {T n} = {t n} = if t > n. which is F n -measurable since the empty set and the whole sample space Ω belong to the σ-field F n. 1. (a) We apply the result of Exercise 8 by letting the Borel set B be the { K, K}. (b) To show that Y is a {F n }-martingale, we have to check the three defining conditions for a martingale: Since we are dealing with the natural filtration, obviously X n is F n -measurable for any positive integer n. As Y n is a function of X n, it is also F n -measurable, i.e. the process Y is adapted. Since the cosine function is bounded by 1, we have E( Y n ) = E( ( 1) n cos(π(x n + K)) ) 1 <. This means that each Y n is integrable. For the martingale equality: E[Y n+1 F n ] = E[( 1) n+1 cos(π(x n+1 + K)) F n ] = ( 1) n+1 E[cos(π(X n + K + ξ n+1 )) F n ] = ( 1) n+1 E[cos(π(X n + K)) cos(πξ n+1 ) sin(π(x n + K)) sin(πξ n+1 ) F n ] = ( 1) n+1 cos(π(x n + K))E[cos(πξ n+1 ) F n ] ( 1) n+1 sin(π(x n + K))E[sin(πξ n+1 ) F n ] = Y n E[cos(πξ n+1 )] ( 1) n+1 sin(π(x n + K))E[sin(πξ n+1 )] = Y n (cos(π) + cos( π)) 2 ( 1)n+1 sin(π(x n + K))(sin(π) + sin( π)) 2 = Y n 8

9 11. (a) (c) In order to apply the version of Optional Stopping Theorem, we first show that the martingale Y and the stopping time T in the current question satisfy the three conditions in OST: i. It is a well-known fact that τ < a.s. (Recall what you have learned in the PM course!) ii. It is obviously that Y τ 1 and hence it is integrable. iii. Using the fact that E(X) E( X ), we have E(Y τ I {τ>n} ) E( Y τ I {τ>n} ) E( I {τ>n} ) = E(I {τ>n} ) = P(τ > n) P(τ = + ) Since τ <, P-a.s., we have P(τ = + ) =. squeezing principle, we obtain the desired result: E(Y τ I {τ>n} ). Now by OST, the following is true: E(Y r ) = E(Y ). As X =, we have Y = cos(πk) = ( 1) K, and hence ( 1) K = E(( 1) τ cos(π(x r + K))) { E(( 1) τ cos(2kπ)) if X τ = K, = E(( 1) τ cos()) if X τ = K. = E(( 1) r ). Cov(W s, W t ) = E(W s W t ) E(W s )E(W t ) = E(W s W t ) = E(W s (W t W s ) + W 2 s ) = E(W s )E(W t W s ) + E(W 2 s ) = E(W 2 s ) = s = t s 9 By the

10 (b) Let Y t = cw t/c 2. (i) Y = cw = (ii) For each ω Ω Y t (ω) = cw t/c 2(ω). Since W t (ω) is continuous in t, therefore Y t (ω) is continuous in t. (iii) For t t 1... t n, Y t Y = cw t /c 2 cw Y t1 Y t = cw t1 /c 2 cw t /c 2. Y tn Y tn 1 = cw tn /c 2 cw t n 1 /c 2 Since t t 1... t n, we have t c t 1 2 c... t n 2 c 2. Therefore, Y t Y, Y t1 Y t,..., Y tn Y tn 1 are independent. (iv) For t >, s, as W t c 2 + s c Note that W s c 2 Y t+s Y s N(, t c 2 ). = cw t+s c ( 2 = c W t c 2 + s c 2 N(, t) cw s c 2 W s {W t a} {T a t} by path continuity of W T a is stopping time Y t = W t+ta W Ta motion W Ta = a P(W t a) = P(W t a and T a t) c 2 ) = P(W t W Ta and T a t) is also a P-Brownian = P(T a t)p(w t W Ta T a t) = 1 2 P(T a t) 1

11 13. (i) Given W t = a >, Since P(W s = for some s (t, t 1 )) ( ) = P min (W t +t W t a) < t < t 1 t ( ) = P min Wt a < t < t 1 t ( ) = P max Wt a < t < t 1 t = P (T a t 1 t ) = 1 t P(T a t) = 2P(W t a) = f Ta (t) = a 2π t 3/2 e a2 /2t f Ta (s)ds P(W s = for somes (t, t 1 ) W t = a) = a 2 2 e x2 2t dx = 2πt π a/ e y2 2 dy t a 2π 1 t s 3/2 e a2 /2s ds We have, by symmetry, a similar result for a <. In this case, we have to replace a 2π in the above equation by a 2π. for a R, P(W t has at least one zero W t = a) = g(a) a 2π 1 t s 3/2 e a2 /2s ds 11

12 (ii) = = 2 = 2 = = = = 2 π P(W t has at least 1 zero in (t, t 1 )) 1 π t 1 π t P(W t = a)g(a) da P(W t = a)g(a) da 1 e a2 /2t a t1 t [ s 3/2 e a2 /2s ds] da 2πt 2π 1 t 1 t 1 t t π (t 1 t )/t as 3/2 e a 2 2 ( 1 t + 1 s ) ds da s 3/2 ds (t + s) s dv 1 + v ) 2 = 2 π tan 1 ( t1 t t 14. (a) We want to compute ae a 2 2 ( 1 t + 1 s ) da ds = 2 π cos 1 t t 1 B s db s Let Π = { = t, t 1, t 2,..., t n = t} be a partition of interval [,t], the above integral can be approximated by n B tj 1 (B tj B tj 1 ). j=1 (Note: In Itô integral, we ALWAYS use left point, that is n n B tj 1 (B tj B tj 1 )instead of B tj (B tj B tj 1 )) j=1 As δ (π), B sdb s. j=1 n j=1 B t j 1 (B t j B tj 1 ) will converge in L2 to 12

13 Observe that B 2 t j B 2 t j 1 = (B tj B tj 1 ) 2 + 2B tj B tj 1 2B 2 t j 1 = (B tj B tj 1 ) 2 + 2B tj 1 (B tj B tj 1 ) B tj 1 (B tj B tj 1 ) = 1 2 [(B2 t j Bt 2 j 1 ) (B tj B tj 1 ) 2 ] n B tj 1 (B tj B tj 1 ) = 1 n [(Bt 2 2 j Bt 2 j 1 ) (B tj B tj 1 ) 2 ] j=1 j=1 = 1 2 B2 t 1 2 n (B tj B tj 1 ) 2 j=1 Since as δ (π), n (B tj B tj 1 ) 2 t in L 2 j=1 (Quadratic Variation of {B t } is [B] t = t!) As δ (π), n j=1 B t (B j 1 t j B ) 1 tj 1 2 B2 t 1 2t in L2 B sdb s = 1 2 B2 t 1 2 t 15. (a) d(tw t ) = W t dt + tdw t tw t = ( W ) + ie sdw s = tw t (g(t, x) = tx) W s ds + W s ds sdw s 13

14 (b) d(w 3 t ) = 3W 2 t dw t (6W t)(dw t ) 2 (g(t, x) = x 3 ) = 3Wt 2 dw t (6W t)dt Wt 3 = W 3 + ie 3W 2 s dw s + W 2 s dw s = 1 3 W 3 t W s ds 1 2 (6W s)ds (a) X t = e ct+αw t (g(t, x) = tx) dx t = ce ct+αw t dt + αe ct+αw t dw t α2 e ct+αw t (dw t ) 2 = cx t dt + αx t dw t α2 X t dt = (c + α2 2 )X tdt + αx t dw t X t = e 1 2 t cosw t (g(t, x) = e 1 2 t cosx) dx t = 1 2 e 1 2 t cosw t dt e 1 2 t sinw t dw t 1 2 e 1 2 t cosw t (dw t ) 2 ie X t = 1 2 e 1 2 t cosw t dt e 1 2 t sinw t dw t 1 2 e 1 2 t cosw t dt = e 1 2 t sinw t dw t = X X t is a martingale. (b) Similar to (a). (c) Similar to (a). e 1 2 s sinw s dw s Remarks: In general, if we want to check whether a process X t = g(t, W t ) 14

15 is a martingale, we use Itô lemma to write down dx t = dt + dw t. if is zero, then X t is a martingale!! 18. (a) Let By Itô s lemma, Y t = θdw s 1 θ 2 ds 2 { dyt = 1 2 θ2 dt θdw t = e Y t L t dl t = e Y t dy t ey t dy t dy t = L t ( 1 2 θ2 dt θdw t ) L tθ 2 dt = L t θdw t which is driftless. L t is a martingale E(L t ) = E(L ) = E(e Y ) = 1 Problem: Is L t θ H T? We don t know! Hence the above proof is not correct, strictly speaking. That s why we need the Novikov s condition. Another Method: L t = exp( θdw s 1 θ 2 ds) 2 = exp( θw t 1 2 θ2 t) 15

16 i. W t F t and L t is a function of W t, L t F t ii. E( L t ) = E(L t ) = E(e θw t )e 1 2 θ2 t = e 1 2 ( θ)2t e 1 2 θ2 t = 1 < iii. L t is integrable. E(L t F s ) = E(e θ(w t W s +W s ) F s )e 1 2 θ2 t t s = e 1 2 θ2t e θw s E(e θ(w t W s ) F s ) e θw s F s = e 1 2 θ2t e θw s E(e θ(w t W s ) ) Indep. Increment = e 1 2 θ2t e θw s e 1 2 ( θ)2 (t s) W t W s = N(, s) = e θw s 1 2 θ2 s = L s {L t } is a martingale. 19. { dbt = rb t dt ds t = µs t dt + σs t dw t X = max{s T K, } Step 1: Find a probability measure Q the discounted stock price: Details: Z t = e rt S t. Z t = Bt 1 S t = e rt S t is a Q-martingale. 16

17 By Itô lemma, dz t = re rt S t dt + e rt ds t = rz t dt + e rt (µs t dt + σs t dw t ) Define Q by = rz t dt + µz t dt + σz t dw t = σz t [ µ r σ dt + dw t] dq dp = exp{ By Girsanov s Thm, T From W t = W t + µ r σ t, we obtain µ r σ dw s.5 T W t = W t + µ r t is a Q-B.M. σ ( µ r σ )2 ds} d W t dz t = dw t + µ r dt,hence σ = σz t [ µ r σ dt + dw t] = σz t d W t Z t is a Q - martingale. Step 2: Define E t e rt E Q (X F t ). Show that E t is a Q -martingale, and that de t = φ t dz t for some {φ t } Details: E t is a Q-martingale is easy to show! (c.f. Tutorial 4 Q3). By Martingale Representation Theorem, φ t s.t. ie de t = φ t d W t. E t = E + φ s d W s. 17

18 From Step 1, we know that dz t = σz t d W t de t = φ 1 t dz t σz t = φ t dz t Step 3: Find (φ t, θ t ) a self-financing, replicative portfolio Details: { θt : number of bonds φ t : number of shares For φ t, we use the φ t in step 2. For Q t, we use : Q t E t φ t Z t Check Self-Financing: V t = Q t B t + φ t S t = (E t φ t Z t )B t + φ t S t = E t B t φ t Z t B t + φ t S t = E t B t dv t = E t db t + B t de t (φ t, Q t ) is self-financing. = E t db t + B t φ t dz t = (θ t + φ t Z t )db t + B t φ t dz t = Q t db t + φ t (Z t db t + B t dz t ) = Q t db t + φ t d(z t B t ) = Q t db t + φ t ds t 18

19 Check replicate X : V T = E T B T = B T e rt E Q (X F T ) = E Q (X F T ) = X ( X F T ) Step 4: In step 3, we have found a SF, replicative portfolio. Hence, the value of option at time t is V t. Find V. Details: From Step 3, V t = B t E t V = B E = E = e rt E Q (X F ) = e rt E Q (X) = e rt E Q [max(s T K, )] We need to know the dist n of S T expectation: under Q in order to compute the ds t = µs t dt + σs t dw t = µs t dt + σs t (d W t µ r σ dt) = rs t dt + σs t d W t S t = S exp { (r- σ2 2 )t + σ W t } Hence S T = S exp { (r- σ2 2 )T + σ W t } and 19

20 W t is distributed as N(, T ) under Q. V = e rt E Q (max(s T K, )) = e rt σ2 (r max(s e 2 )T +σx K, ) 1 e x2 2T dx 2πT = e rt ln K S (r σ2 /2)T σ [S e (r σ2 /2)T +σx K] e x 2 2T dx 2πT d2 = e rt [S e (r σ2 /2)T +σy T K] e y 2 dy 2π = S d2 1 2π e.5(y σ T ) 2 dy Ke rt φ(d 2 ) = S φ(d 2 + σ T ) Ke rt φ(d 2 ) = S φ(d 1 ) Ke rt φ(d 2 ) 2 Remarks: where d 1 = ln s k + (r + σ2 /2)T σ T, d 2 = ln s k + (r σ2 /2)T σ T (a) The Q so constructed is called Equivalent Martingale Measure (EMM) or Risk-Neutral Measure. (b) A EMM turns the discounted stock price a martingale, ie fair game (c) For general X, we have P t = V t = E t B t = e r(t t) E Q (X F t ), ie P t is the expected discounted value of X, but not expected under P!! 2