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Martingale Theory Problem Set 3, with Solutions Martingales

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Martingale Theory Problem Set 3, with Solutions Martingales Martingale Theory Problem set 3, with solutions Martingales The solutions of problems 1,2,3,4,5,6, and 11 are written down. The rest will come soon. 3.1 Let ξj, j = 1; 2;::: be i.i.d. random variables with common distribution P ξi = +1 = p; P ξi = −1 = q := 1 − p; and , , their natural ltration. Denote Pn , . Fn = σ(ξj; 0 ≤ j ≤ n) n ≥ 0 Sn := j=1 ξj n ≥ 0 Sn (a) Prove that Mn := (q=p) is an (Fn)n≥0-martingale. (b) For λ > 0 determine C = C(λ) so that λ n Sn Zn := C λ be an (Fn)n≥0-martingale. SOLUTION: (a) ξn+1 ξn+1 E Mn+1 Fn = E Mn(q=p) Fn = MnE (q=p) Fn ξn+1 = MnE (q=p) = Mn (p(q=p) + q(p=q)) = Mn: (b) −1 C = C(λ) = Eλξ = λp + λ−1q: 3.2 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 a pounds or she lost b pounds. (a) What is the probability that she will be winning when she stops playing further. 1 (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 Pξ = +1 = = Pξ = −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: On the other hand, P T = TL + P T = Tr = 1: Solving the last two equations we get a b PT = T = ; PT = 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) ET = ab: 2 3.3HW Gambler's Ruin, 2 Answer the same questions as in problem 2 when the probability of winning or loosing one pound in each round is p, respectively, q := 1 − p, with p 2 (0; 1). Hint: Use the martingales constructed in problem 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; 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 PT = T = ; PT = 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 ET = (p − q)−1ES = (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 3 3.4 Let ξj, j = 1; 2; 3;::: , be independent and identically distributed random variables and Fn := σ(ξj; 0 ≤ j ≤ n), n ≥ 0, the natural ltration generated by them. Assume that γξ for some γ 2 R the exponential moment m(γ) := E e j < 1 exists. Denote S0 := 0, Pn , . Prove that the process Sn := j=1 ξj n ≥ 1 −n Mn := m(γ) expfγSng; n 2 N; is an (Fn)n≥0-martingale. SOLUTION: Very much the same as problem 1 (b). 3.5 Let (Ω; F; (Fn)n≥0; P) be a ltered probability space and Yn, n ≥ 0, a sequence of absolutely integrable random variables adapted to the ltration (Fn)n≥0. Assume that there exist real numbers un; vn, n ≥ 0, such that E Yn+1 Fn = unYn + vn: Find two real sequences an and bn, n ≥ 0, so that the sequence of random variables Mn := anYn + bn, n > 1, be martingale w.r.t. the same ltration. SOLUTION: Write down the martingale condition for Mn: E Mn+1 Fn = E an+1Yn+1 + bn+1 Fn = an+1unYn + an+1vn + bn+1 = anYn + bn: We get the recursions −1 an+1 = anun ; bn+1 = bn − an+1vn: The solution is n−1 !−1 Y a0 = 1; an = uk ; k=0 n X b0 = 0; bn = − akvk−1: k=1 3.6HW We place N balls in K urns (in whatever way) and perform the following discrete time process. At each time unit we choose one of the balls uniformly at random (that is : each ball is chosen with probability 1=N) and place it in one of the urns also uniformly chosen at random (that is: each urn is chosen with probability 1=K). Denote by Xn the number 4 of balls in the rst urn at time n and let Fn := σ(Xj; 1 ≤ j ≤ n), n ≥ 0, be the natural ltration generated by the process n 7! Xn. (a) Compute E Xn+1 Fn . (b) Using the result from problem 5, nd real numbers an; bn, n ≥ 0, such that Zn := anXn + bn be martingale with respect to the ltration (Fn)n≥0. SOLUTION: (a) N − Xn 1 Xn K − 1 N − Xn K − 1 Xn 1 E Xn+1 Fn = (Xn + 1) + (Xn − 1) + Xn + N K N K N K N K N − 1 1 = X + : n N K (b) Apply the result from problem 5 with N − 1 1 u = ; v = : n N n K 3.7 Let Xj, j ≥ 1, be absolutely integrable random variables and Fn := σ(Xj;; 1 ≤ j ≤ n), n ≥ 0, their natural ltration. Dene the new random variables n−1 X Z0 := 0;Zn := Xj+1 − E Xj+1 Fj : j=0 Prove that the process n 7! Zn is an (Fn)n≥0-martingale. 3.8 A biased coin shows HEAD=1 with probability θ 2 (0; 1), and TAIL=0 with probability 1 − θ. The value θ of the bias is not known. n For t 2 [0; 1] and n 2 N we dene pn;t : f0; 1g ! [0; 1] by Pn Pn xj n− xj pn;t(x1; : : : ; xn) := t j=1 (1 − t) j=1 : We make two hypotheses about the possible value of θ: either θ = a, or θ = b, where a; b 2 [0; 1] and a 6= b. We toss the coin repeatedly and form the sequence of random variables pn;a(ξ1; : : : ; ξn) Zn := ; pn;b(ξ1; : : : ; ξn) where ξj, j = 1; 2;::: , are the results of the successive trials (HEAD=1, TAIL=0). Prove that the process n 7! Zn is a martingale (with respect to the natural ltration generated by the coin tosses) if and only if the true bias of the coin is θ = b. SOLUTION: 5 3.9 Bonus Bellman's Optimality Principle We model a sequence of gambling as follows. Let ξj, j = 1; 2;::: , be independent random variables with the following identical distribution; P ξj = +1 = p; P ξj = −1 = 1 − p := q; 1=2 < p < 1: We denote α := p log2 p + q log2 q + 2; the entropy of the distribution of ξj. ξj is the return of unit bet in the jth round. A gambler starts playing with initial fortune Y0 > 0 and her fortune after round n is Yn = Yn−1 + Cnξn where Cn is the amount she bets in this round. Cn may depend on the values of ξ1; : : : ; ξn−1, and 0 ≤ Cn ≤ Yn−1. The expected rate of winnings within n rounds is: rn := E log2(Yn=Y0) : The gambler's goal is to maximize rn within a xed number of rounds. (a) Prove that no matter what strategy the gambler chooses (that is: no matter how she chooses Cn = Cn(ξ1; : : : ; ξn−1) 2 [0;Yn−1]) Xn := log2 Yn − nα is a supermartingale and hence it follows that rn ≤ nα. This means that she will not be able to make her average winning rate, over any number of rounds, larger than α. (b) However, there exists a gambling strategy which makes Xn dened above a martingale and thus realizes the maximal average winning rate. Find this strategy. That is: determine the optimal choice of Cn = Cn(ξ1; : : : ; ξn). SOLUTION: 3.10 Let n 7! ηn be a homogeneous Markov chain on the countable state space S := f0; 1; 2;::: g and Fn := σ(ηj; 0 ≤ j ≤ n), n ≥ 0, its natural ltration.
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