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Martingale Theory Problem Set 4, with Solutions Stopping

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Martingale Theory Problem Set 4, with Solutions Stopping Martingale Theory Problem set 4, with solutions Stopping 4.1 Let (Ω; F; (Fn)n≥0;P ) be a ltered probability space and S and T two stopping times. Prove (without consulting the lecture notes) that S ^ T := minfS; T g, S _ T := maxfS; T g, and S + T are also stopping times. SOLUTION: f! 2 Ω: S(!) ^ T (!) ≤ ng = f! 2 Ω: S(!) ≤ ng \ f! 2 Ω: T (!) ≤ ng; f! 2 Ω: S(!) _ T (!) ≤ ng = f! 2 Ω: S(!) ≤ ng [ f! 2 Ω: T (!) ≤ ng; n [ f! 2 Ω: S(!) + T (!) ≤ ng = (f! 2 Ω: S(!) = lg \ f! 2 Ω: T (!) ≤ n − lg) : l=0 Since the events on the right hand side of these equations are all Fn-measurable, the statement of the problem follows. 4.2 Martingales for simple symmetric random walk on Z. Let n 7! Xn be a simple symmetric random walk on the one-dimensional integer lattice Z and (Fn)n≥0 its natural ltration. (a) Prove that and 2 are both -martingales. Xn Yn := Xn − n (Fn) (b) Find a deterministic sequence such that 3 be an -martingale. an 2 R Zn := Xn+anXn (Fn) (c) Find a deterministic sequences such that 4 2 be an bn; cn 2 R Vn := Xn + bnXn + cn (Fn)-martingale. SOLUTION: Let ξj, j = 1; 2;::: be i.i.d. random variables with the common distribution P (ξj = ±1) = 1=2, Fn := σ(ξj : 1 ≤ j ≤ n) and write n X X0 = 0;Xn := ξj: j=1 1 (a) E Xn+1 Fn = E Xn + ξn+1 Fn = Xn + E ξn+1 Fn = Xn: 2 E Yn+1 Fn = E Xn+1 − (n + 1) Fn 2 2 = E Xn − n + 2Xnξn+1 + ξn+1 − 1 Fn 2 2 = Xn − n + 2XnE ξn+1 Fn + E ξn+1 Fn − 1 = Yn: (b) Write rst 3 Zn+1 = (Xn + ξn+1) + an+1(Xn + ξn+1) 2 2 3 = Zn + 3Xnξn+1 + 3Xnξn+1 + ξn+1 + (an+1 − an)Xn + an+1ξn+1: Hence, E Zn+1 Fn = ··· = Zn + (an+1 − an + 3)Xn: It follows that Zn is a martingale if and only if an+1 − an + 3 ≡ 0: Thus, in order that n 7! Zn be a martingale we must choose an = a0 − 3n: That is, 3 Zn = Xn − 3nXn is a martingale. (c) Proceed similarly as in (b). Write 4 2 Vn+1 = (Xn + ξn+1) + bn+1(Xn + ξn+1) + cn+1 3 2 2 3 4 = Vn + 4Xnξn+1 + 6Xnξn+1 + 4Xnξn+1 + ξn+1 2 2 + (bn+1 − bn)Xn + 2bn+1Xnξn+1 + bn+1ξn + (cn+1 − cn): Hence, 2 E Vn+1 Fn = ··· = Vn + (bn+1 − bn + 6)Xn + (cn+1 − cn + bn+1 + 1): It follows that Zn is a martingale if and only if bn+1 − bn + 6 ≡ 0; cn+1 − cn + bn+1 + 1: 2 Thus, in order that n 7! Zn be a martingale we must choose bn = b0 − 6n; n X 2 cn = c0 − bk − n = c0 + (b0 − 1)n − 3n(n − 1) = c0 + (b0 + 2)n − 3n : k=1 That is, 4 2 2 Vn = Xn − 6nXn − 3n + 2n is a martingale. 4.3 Gambler's Ruin, 1 A gambler wins or looses one pound in each round of betting, with equal chances and independently of the past events. She starts betting with the rm determination that she will stop gambling when either she won n pounds or she lost m pounds. (a) What is the probability that she will be winning when she stops playing further. (b) What is the expected number of her betting rounds before she will stop playing further. SOLUTION: Model the experiment with simple symmetric random walk. Let ξj, j = 1; 2;::: be i.i.d. random variables with common distribution 1 P (ξ = +1) = = P (ξ = −1) ; i 2 i and Fn = σ(ξj; 0 ≤ j ≤ n), n ≥ 0, their natural ltration. Denote n X S0 = 0;Sn := ξj; n ≥ 1: j=1 Dene the stopping times TL := inffn > 0 : Sn = −bg;TR := inffn > 0 : Sn = +ag;T := minfTL;TRg: Note that fthe gambler wins a poundsg = fT = TRg; fthe gambler looses b poundsg = fT = TLg: (a) By the Optional Stopping Theorem E (ST ) = E (S0) = 0: Hence −bP (T = TL) + aP (T = Tr) = 0: 3 On the other hand, P (T = TL) + P (T = Tr) = 1: Solving the last two equations we get a b P (T = T ) = ; P (T = T ) = : L a + b R a + b (b) First prove that 2 is yet another martingale: Mn := Sn − n 2 E Mn+1 Fn = E Sn+1 Fn − (n + 1) 2 = E Sn + 2Snξn+1 + 1 Fn − (n + 1) = ··· = Mn: Now, apply the Optional Stopping Theorem 2 2 2 0 = E (MT ) = E ST − T = P (T = TL) b + P (T = TR) a − E (T ) : Hence, using the result from (a) E (T ) = ab: 4.4 Gambler's Ruin, 2 Let the lazy random walk Xn be a Markov chain on Z with the following transition proba- bilities 3 1 P Xn+1 = i ± 1 Xn = i = ; P Xn+1 = i Xn = i = : 8 4 Denote Tk := inffn ≥ 0 : Xn = kg: Let a; b ≥ 1 be xed integer numbers. Compute P Ta < T−b X0 = 0 and E Ta ^ T−b X0 = 0 . SOLUTION: Very similar to problem 3. 4.5 Gambler's Ruin, 3 Answer the same questions as in problem 3 when the probability of winning or loosing one pound in each round is p, respectively, 1 − p, with p 2 (0; 1). Hint: Use the martingales constructed in problem 3.1. SOLUTION: Model the experiment with simple biased random walk. Let ξj, j = 1; 2;::: be i.i.d. random variables with common distribution P (ξi = +1) = p; P (ξi = −1) = q; 4 and Fn = σ(ξj; 0 ≤ j ≤ n), n ≥ 0, their natural ltration. Denote n X S0 = 0;Sn := ξj; n ≥ 1: j=1 Dene the stopping times TL := inffn > 0 : Sn = −bg;TR := inffn > 0 : Sn = +ag;T := minfTL;TRg: Note that fthe gambler wins a poundsg = fT = TRg; fthe gambler looses b poundsg = fT = TLg: (a) Use the Optional Stopping Theorem for the martingale (q=p)Sn : Sn b a 1 = E (q=p) = (p=q) P (T = TL) + (q=p) P (T = TR) : On the other hand, P (T = TL) + P (T = Tr) = 1: Solving the last two equations we get 1 − (q=p)a 1 − (p=q)b P (T = T ) = ; P (T = T ) = : L (p=q)b − (q=p)a R (q=p)a − (p=q)b (b) Now, apply the Optional Stopping Theorem to the martingale Sn − (p − q)n. Hence 1 − (p=q)b 1 − (q=p)a E (T ) = (p − q)−1E (S ) = (p − q)−1 a − b T (q=p)a − (p=q)b (p=q)b − (q=p)a a(1 − (p=q)b) + b(1 − (q=p)a) = (p − q)−1 : (q=p)a − (p=q)b 4.6 Let (Ω; F; (Fn)n≥0; P) be a ltered probability space and S and T two stopping times such that P (S ≤ T ) = 1. (a) Dene the process n 7! Cn := 11fS<n≤T g. Prove that (Cn)n≥1 is predictable process. That is: for all n ≥ 1, Cn is Fn−1-measurable. (b) Let the process (Xn)n≥0 be (Fn)-supermartingale and dene the process n 7! Yn as follows n X Y0 := 0;Yn := Ck(Xk − Xk−1): k=1 5 Prove that (Yn)n≥0 is also (Fn)-supermartingale. (c) Prove that if S and T two stopping times such that P (S ≤ T ) = 1 and (Xn)n≥0 is supermartingale then for all n ≥ 0, E (Xn^T ) ≤ E (Xn^S). SOLUTION: (a) f! : Cn(!) = 1g = f! : S(!) < n; T (!) ≥ ng = f! : S(!) ≤ n − 1g \ f! : T (!) ≤ n − 1gc: Since both events on the right hand side are Fn−1-measurable, the process n 7! Cn is predictable, indeed. (b) Yn is clearly Fn measurable. We check the supermartingale condition: E Yn+1 Fn = Yn + E Cn+1(Xn+1 − Xn) Fn = Yn + Cn+1E Xn+1 − Xn Fn ≤ Yn: In the second step we use the result from (a). In the last step we use Cn+1 ≥ 0. (c) Note that Xn^T − Xn^S = Yn; and use (b). 4.7 A two-dimensional random walk. HW Let Xn be the following two dimensional random walk: n 7! Xn is a Markov chain on the 2 two dimensional integer lattice Z with the following transition probabilities: 1 P Xn+1 = (i ± 1; j) Xn = (i; j) = ; 8 1 P Xn+1 = (i; j ± 1) Xn = (i; j) = : 8 1 P Xn+1 = (i ± 1; j ± 1) Xn = (i; j) = : 8 (a) Prove that 3 M := jX j2 − n n n 2 is a martingale with respect to the natural ltration of the process. (We denote by jxj the 2 Euclidean norm of x 2 Z .) (b) For R 2 R+ dene the stopping time 2 2 TR := inffn ≥ 0 : jXnj ≥ R g: 6 Give sharp lower and upper bounds for E TR X0 = (0; 0) : SOLUTION: Let ξj, j = 1; 2;::: be i.i.d. random two-dimensional vectors with the common distribution P (ξj = (±1; 0)) = P (ξj = (0; ±1)) = P (ξj = (±1; ±1)) = frac18; Fn := σ(ξj : 1 ≤ j ≤ n) and write n X X0 = 0;Xn := ξj: j=1 Note that 3 E (ξ ) = 0; E jξ j2 = : j j 2 (a) 2 3 2 2 3 E jXn+1j − (n + 1) Fn = E jXnj + 2Xn · ξn+1 + jξn+1j − (n + 1) Fn 2 2 2 3 2 3 2 3 = jXnj − n + E jξn+1j Fn = jXnj − n: 2 2 2 (b) Note rst that p 2 2 2 R ≤ jXTR j ≤ (R + 2) : Apply the Optional Stopping Theorem, 2 E (T ) = E jX j2 : R 3 TR From these two relations we get 2 2 p R2 ≤ E (T ) ≤ (R + 2)2: 3 R 3 4.8 We toss repeatedly a fair coin.
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