ST3180 UNIVERSITY OF WARWICK Third Year EXAMINATIONS: SUMMER 2020 Probability Theory Time allowed: 2 hours Approved calculators may be used. Full marks may be obtained by correctly answering 3 complete questions. Candi- dates may attempt all questions. Marks will be awarded for the best 3 answers only. 1 1: (a) State and prove the first and second Borel-Cantelli lemmas. [6] For the rest of this question you may use, without proof, Kolmogorov’s zero one law. (b) Let (Xn)n≥1 be a sequence of independent random variables on probability space (Ω,F ,P) taking values in {1, 2, 3, . . .}. Suppose that P(Xn ≥ i) = 1/i for each n and each i ∈ {1, 2, 3, . . .}. (i) Calculate P(Xn ≥ nα i.o.) for each fixed α > 0. (ii) Show that the random variable lim sup n→∞ log(Xn) log(n) is almost surely constant, and find the value of this constant. [8] (b) Suppose (Xn)n≥1 is a sequence of independent identically distributed random variables such that P(Xn = 1) = P(Xn = −1) = 1 2 for each n ∈ N . Let Sn = ∑n k=1Xk, and define B− = { lim inf n→∞ Sn = −∞ } and B+ = { lim sup n→∞ SN =∞ } . (i) Show that both B− and B+ belong to the tail σ-algebra of (Xn)n≥1, and that P(B+) = P(B−) ∈ {0, 1}. (ii) Using the Borel-Cantelli lemmas, show that for each k ≥ 1 lim sup n→∞ ( Sn+k − Sn ) = k a.s. [Hint: Consider An = {Sn+k − Sn = k}.] (iii) Deduce that P(Bc+ ∩Bc−) = 0 and hence P(B+) = P(B−) = 1. [6] [TOTAL: 20] Continued... 2 2: (a) Suppose that (Xn)n≥1 is a sequence of random variables with |Xn| ≤ Y for each n, where Y ∈ L1 (i.e. E[|Y |] <∞). Show that {Xn}n≥1 is uniformly integrable. [4] (b) Suppose that (Xn)n≥1 is a sequence of random variables converging to zero in probability. Fur- ther, suppose that there exists a constant K ∈ (0,∞) such that E[|Xn|3] ≤ K for each n. Show that Xn → 0 in L2. (You may use results about uniform integrability so long as you state them clearly) [4] (c) Suppose that (Xn)n≥1 is a sequence of independent normally distributed random variables, with common mean µ and common variance 1. Let Zn = exp(X1 + . . .+Xn). (i) Show that Zn → 0 in L2 if and only if µ < −1. (Hint: If W is a standard normal random variable then E(eθW ) = eθ2/2.) (ii) Show that Zn → 0 in probability if µ < 0. [5] (d) Let (Xn)n≥2 be a sequence of independent random variables such that for each n ∈ {2, 3 . . .} P(Xn = n) = P(Xn = −n) = 1 2n log n , and P(Xn = 0) = 1− 1 n log n . Let Sn = X2 + . . .+Xn. Show that Sn n → 0 in probability but not almost surely. (Hint: You may use without proof that ∑ n 1 n logn =∞.) [7] [TOTAL: 20] Continued... 3 3: (a) Suppose (Xn)n≥1 is a sequence of random variables and X is a random variable all on probability space (Ω,F ,P). Let µn, Fn be the law and distribution function of Xn respectively and µ, F be the law and distribution function of X respectively. (i) Suppose that µn → µ weakly; show that this implies Xn → X in distribution. (ii) Suppose that Xn → X in distribution, and let (an)n≥1 be a sequence of real numbers such that an → 0 as n→∞. Show that anXn converges in distribution to zero. [8] (b) Let µX be the law of a random variable X. Show that the distribution of X is symmetric, i.e. µX(−∞, x] = µX [−x,∞) for all x ∈ R, if and only if the characteristic function of X is real. [4] (c) In the following, identify if the sequence of random variables (Yn)n≥1 converges weakly and if so identify the limit. (You should explain your reasoning, and state clearly any results from lectures that you use). (i) Yn = max{U1, . . . , Un} where U1, U2, . . . are independent Uniform(−1, 1) random variables. (ii) Yn = n(1 − max{U1, . . . , Un}) where U1, U2, . . . are independent Uniform(−1, 1) random variables. (iii) Yn = √ 3 n(U1 + . . .+Un) where U1, U2, . . . are independent Uniform(−1, 1) random variables. (iv) Yn = nmin{U1, . . . , Un} where U1, U2, . . . are independent Uniform[0, 1] random variables. (v) Yn is an exponential random variable with mean λn > 0 for each n, i.e. P(Yn > y) = e−y/λn , where λn → 0 as n→∞. [8] [TOTAL: 20] Continued... 4 4: (a) Let X and Y and Z be integrable random variables on a probability space (Ω,F ,P), and let G ⊆ F be a sub-σ-algebra. Show that if Y and Z are both versions of E[X | G] then Y = Z almost surely. [3] (b) Suppose that (Xn)n≥1 is a random process adapted to (Fn)n≥1. Define the hitting time τA of (Xn)n≥1 on a Borel set A. Show that τA is a stopping time for this filtration. [3] (c) Supose (Xn)n≥1 is a symmetric simple random walk on Z started from 0. Use the Optional Stopping Theorem for Bounded Stopping Times to evaluate the probability (Xn)n≥1 hits b > 0 before −a < 0. [4] (d) Let (Xn)n≥1 be a sequence of independent random variables on probability space (Ω,F ,P), with E[Xn] = 1 for each n. Let F0 = {∅,Ω}, Fn = σ(X1, . . . , Xn) for n ∈ N, and M0 = 1, and Mn = n∏ k=1 Xk for n = 1, 2, . . . . (i) Show that (Mn)n≥0 is a martingale with respect to (Fn)n≥0. (ii) Now suppose ϕ(t) = E[etX1 ] <∞ for all t ∈ R. Let S0 = 0, Sn = ∑n k=1Xk, and Yn = etSn ϕ(t)n , for n = 0, 1, 2, . . . . Show, using part (i), that (Yn)n≥0 is a martingale with respect to (Fn)n≥0. [4] (c) Let (εn)n≥1 be independent random variables with P(εn = +1) = p, P(εn = −1) = q, where 1/2 < p = 1− q < 1 . Let Fn = σ(ε1, . . . , εn) and define Xn inductively by X0 = 1 and for n ≥ 1 Xn = Xn−1 + Vnεn . Assume Vn is a Fn−1-measurable random variable for each n, and that Vn is strictly between 0 and Xn. (i) Prove that if Xn > 0 then E[log(Xn+1/Xn) | Fn] = f(Vn+1/Xn) where f(x) = p log(1 + x) + q log(1− x) . (ii) Deduce that (log(Xn)− nα)n≥0 is a supermartingale, where α = p log p+ q log q + log 2 . [6] [TOTAL: 20] End. 5 ST3180 UNIVERSITY OF WARWICK Third Year EXAMINATIONS: SUMMER 2020 Probability Theory Time allowed: 2 hours Approved calculators may be used. Full marks may be obtained by correctly answering 3 complete questions. Candi- dates may attempt all questions. Marks will be awarded for the best 3 answers only. 1 1: (a) State and prove the first and second Borel-Cantelli lemmas. [6] Answer: (Applied Bookwork:) (Borel-Cantelli Lemmas) Let (An)n≥1 be a sequence of events on probability space (Ω,F ,P). Then, (BC1) If ∑ n P(An) <∞ then P(lim supAn) = P(An i.o.) = 0, (BC2) If (An)n≥1 are also independent then ∑ n P(An) = ∞ implies P(lim supAn) = P(An i.o.) = 1 (i.e. the converse of (BC1) holds under independence). (1 marks) For (BC1) we assume that ∑ n P(An) < ∞. Let Gn = ⋃ m≥nAm, which is a decreasing sequence of sets such that Gn ↘ G where G = lim supn→∞An. Fix k ∈ N and observe P(lim sup n→∞ An) = P ( ⋂ n≥1 Gn ) ≤ P(Gk) ≤∑ n≥k P(An) , where the final inequality holds by σ-subadditivity. By assumption the right hand side converges to zero as k →∞, which concludes the proof of (BC1). (2 marks) For (BC2) we assume that (An)n≥1 are independent and ∑ n P(An) =∞. The main idea of the proof is to take compliments and use the standard bound 1− x ≤ e−x for x ∈ R. Fix m, r ∈ N, then by independence P ( ⋂ m≤n Acn ) ≤ P( ⋂ m≤n≤r Acn ) = ∏ m≤n≤r P(Acn) = ∏ m≤n≤r ( 1− P(An) ) , Applying the bound 1− x ≤ e−x, we find P ( ⋂ m≤n Acn ) ≤ e−∑m≤n≤r P(An) → 0 as r →∞ , where convergence follows by assumption on ∑ n P(An). It follows that, since {Acn ev.} is a countable union of null sets, that P(An i.o.) = 1− P(Acn ev.) = 1, as required. (3 marks) For the rest of this question you may use, without proof, Kolmogorov’s zero one law. (b) Let (Xn)n≥1 be a sequence of independent random variables on probability space (Ω,F ,P) taking values in {1, 2, 3, . . .}. Suppose that P(Xn ≥ i) = 1/i for each n and each i ∈ {1, 2, 3, . . .}. (i) Calculate P(Xn ≥ nα i.o.) for each fixed α > 0. (ii) Show that the random variable lim sup n→∞ log(Xn) log(n) is almost surely constant, and find the value of this constant. [8] Answer: (Unseen (seen similar example):) (i) Apply (BC1) and (BC2) above with An = {Xn ≥ nα}. Observe ∑ n P(An) = ∑ n 1 dnαe { =∞ if α ≤ 1 , <∞ if α > 1 . 2 So (since the Xn are independent) the Borel-Cantelli lemmas imply P(Xn ≥ nα i.o.) = { 1 if α ≤ 1 , 0 if α > 1 . (3marks) (ii) Let Fn = σ(Xn), then (Fn)n≥1 is an independent sequence of σ-algebras. Let Ln = supm≥n logXmlogm and L = limn→∞ Ln. Since Ln is σ (⋃ m≥n Fm ) -measurable we have L is T -measurable, in particular L is almost surely constant by Kolmogorov’s zero-one law. Observe {L ≥ 1} ⊇ {Ln ≥ 1 i.o.} ⊇ {Xn ≥ n i.o.}. It follows that P(L ≥ 1) = 1. Now fix k ∈ N, {L > 1 + 2 k } ⊆ {L ≥ 1 + 2 k } ⊆ ⋂ m {Ln > 1 + 2 k − 1 m i.o.} ⊆ {Ln > 1 + 1 k i.o.} . By definition {Ln > 1 + 1k i.o.} ⊆ {Xn ≥ n1+1/k i.o}, and by part (i) P(Xn ≥ n1+1/k i.o) = 0. The conclusion follows, since {L > 1} = ⋃ k { L > 1 + 1 k } , and a countable union of null sets is null. (5 marks) (c) Suppose (Xn)n≥1 is a sequence of independent identically distributed random variables such that P(Xn = 1) = P(Xn = −1) = 1 2 for each n ∈ N . Let Sn = ∑n k=1Xk, and define B− = { lim inf n→∞ Sn = −∞ } and B+ = { lim sup n→∞ SN =∞ } . (i) Show that both B− and B+ belong to the tail σ-algebra of (Xn)n≥1, and that P(B+) = P(B−) ∈ {0, 1}. (ii) Using the Borel-Cantelli lemmas, show that for each k ≥ 1 lim sup n→∞ ( Sn+k − Sn ) = k a.s. [Hint: Consider An = {Sn+k − Sn = k}.] (iii) Deduce that P(Bc+ ∩Bc−) = 0 and hence P(B+) = P(B−) = 1. [6] Answer: (Unseen (seen similar):) (i) Notice that, for each N ∈ N fixed lim inf n→∞ Sn = −∞ if and only if lim infn→∞ (XN + . . . Xn) = −∞ , and hence B− is measurable with respect to σ(XN , XN+1, . . .) for each N ∈ N, and therefore B− is in the tail σ-algebra. The same argument applies to B+. It follows from Kolmogorov’s 0-1 law that P(B−),P(B+) ∈ {0, 1}. Finally, by symmetry (Xn)n≥1 and (−Xn)n≥1 have the same distribution and hence so do (Sn) and (−Sn). It follows that P(B+) = P(lim sup(−Sn) =∞) = P(− lim inf Sn =∞) = P(B−) . (2 marks) 3 (ii) Fix k ≥ 1, consider An as defined in the hint. Then P(An) = P(Xn+1 = 1, . . . , Xn+k = 1) = 1 2k for each n ∈ N . It follows that ∑ n≥1 P(Ank) =∞, since the {Ank}n≥1 are independent (BC2) implies P({Ank}n≥1 i.o.) = 1. Together with the fact that (Sn+k − Sn) ≤ k for each n we conclude that lim sup(Sn+k − Sn) = k almost surely. (2 marks) (iii) Since (Bc+ ∩Bc−) ⊆ {(Sn) bounded above and below} ⊆ ∞⋃ k=1 { lim sup(Sn+k − Sn) = k }c , and a countable union of null sets is null, we have P(Bc+ ∩Bc−) = 0. Combining with part (i) we get P(B+) = P(B−) = 1 as required. (2 marks) [TOTAL: 20] Continued... 4 2: (a) Suppose that (Xn)n≥1 is a sequence of random variables with |Xn| ≤ Y for each n, where Y ∈ L1 (i.e. E[|Y |] <∞). Show that {Xn}n≥1 is uniformly integrable. [4] Answer: (Applied Bookwork) Observer, by assumptions E(|Xn|1|Xn|≥K) ≤ E(|Y |1|Xn|≥K) ≤ E(|Y |1|Y |≥K) and the right hand side tends to zero (uniformly in n), by dominated convergence, since Y ∈ L1. For full marks it must be made clear that convergence is uniform in n. (4 marks) (b) Suppose that (Xn)n≥1 is a sequence of random variables converging to zero in probability. Fur- ther, suppose that there exists a constant K ∈ (0,∞) such that E[|Xn|3] ≤ K for each n. Show that Xn → 0 in L2. (You may use results about uniform integrability so long as you state them clearly) [4] Answer: (Applied Bookwork) We know from lectures that if (Yn)n≥1 is a sequence of random variables bounded in Lp for p > 1 then {Yn}n≥1 are uniformly integrable. Now let Yn = X2n for each n, this forms a sequence which is bounded in L3/2 and hence is U.I.. Also from lectures, a U.I. sequence which converges in probability converges in L1. This implies (Xn)n≥1 converges in L2. (Alternative proofs, for example using law of total expectation are also acceptable). (4 marks) (c) Suppose that (Xn)n≥1 is a sequence of independent normally distributed random variables, with common mean µ and common variance 1. Let Zn = exp(X1 + . . .+Xn). (i) Show that Zn → 0 in L2 if and only if µ < −1. (Hint: If W is a standard normal random variable then E(eθW ) = eθ2/2.) (ii) Show that Zn → 0 in probability if µ < 0. [5] Answer: (Unseen example) (i) Consider E(Z2n) = E[exp(2(X1 + . . .+Xn))] = ( E[exp(2(X1 − µ))] )n e2nµ , and apply the hint. (ii) Either by considering E[( 1n (X1 + . . . + Xn))] = 1/n → 0 as n → ∞ (convergence in L2 implies in prob.), or by the Weak Law of Large Numbers, 1n (X1 + . . . + Xn) to µ in probability as n → ∞. Hence P[X1 + . . .+Xn ≥ n(µ+ε)]→ 0 for each ε > 0. Fix ε > 0 such that µ+ε < 0 (possible since µ < 0). Now P[Zn ≥ en(µ+ε)]→ 0 and en(µ+ε) → 0 as n→∞, which implies convergence in probability. (5 marks) (d) Let (Xn)n≥2 be a sequence of independent random variables such that for each n ∈ {2, 3 . . .} P(Xn = n) = P(Xn = −n) = 1 2n log n , and P(Xn = 0) = 1− 1 n log n . Let Sn = X2 + . . .+Xn. Show that Sn n → 0 in probability but not almost surely. (Hint: You may use without proof that ∑ n 1 n logn =∞.) [7] Answer: (Unseen example) Fix ε > 0 and consider P(|Sn/n| > ε), since Sn is mean zero we can apply Chebyshev’s inequality. Hence consider the variance of Sn, Var(Sn) = n∑ k=2 Var(Xk) = n∑ k=2 E(X2k) = n∑ k=2 k log k ≤ (n− 2)n log n + 2 log 2 , where in the final line we used x 7→ x/ log x is increasing for x ≥ 3 > e. So by Chebyshev’s inequality P(|Sn/n| > ε) ≤ E(S 2 n) ε2n2 ≤ 1 ε2 log n + 2 ε2n2 log 2 → 0 , 5 as n→∞, as required. For (lack of) almost sure convergence, consider An = {Xn = n} and Bn = {Xn = −n}, then ∑ n≥2 P(An) = ∑ n≥2 P(Bn) = ∑ n≥2 1 n log n =∞ , so by (BC2) P(An i.o.) = P(Bn i.o.) = 1, so in particular Sn can not converge almost surely. (7 marks) [TOTAL: 20] Continued... 6 3: (a) Suppose (Xn)n≥1 is a sequence of random variables and X is a random variable all on probability space (Ω,F ,P). Let µn, Fn be the law and distribution function of Xn respectively and µ, F be the law and distribution function of X respectively. (i) Suppose that µn → µ weakly; show that this implies Xn → X in distribution. (ii) Suppose that Xn → X in distribution, and let (an)n≥1 be a sequence of real numbers such that an → 0 as n→∞. Show that anXn converges in distribution to zero. [8] Answer: (i) (Applied Bookwork) Fix x ∈ R a continuity point of F , and δ > 0. Define hx ∈ Cb(R) by hx(y) = 1 if y ≤ x , 1− y−xδ if y ∈ (x, x+ δ) , 0 if y ≥ x+ δ . Then by assumption µn(hx)→ µ(hx) as n→∞, and by construction of hx we have Fn(x) ≤ µn(hx) and µ(hx) ≤ F (x+ δ) , which implies that lim sup n→∞ Fn(x) ≤ lim sup n→∞ µn(hx) = µ(hx) ≤ F (x+ δ) . Now take the limit δ → 0 and use continuity of F at x to get lim supn→∞ Fn(x) ≤ F (x). We use the same trick to get the desired upper bound on F (x). That is, define gx ∈ Cb(R) by gx(y) = 1 if y ≤ x− δ , 1− y−(x−δ)δ if y ∈ (x− δ, x) , 0 if y ≥ x . Then by the same argument, lim inf n→∞ Fn(x) ≥ lim infn→∞ µn(gx) = µ(gx) ≥ F (x− δ) . Now take the limit δ → 0 and use continuity of F at x to get lim infn→∞ Fn(x) ≥ F (x). It follows that Fn d−→ F . (4 marks) (ii) (Unseen) Let Yn = anXn. Fix ε > 0 and u a continuity point of F such that F (u) > 1− ε. If x > 0 then for all n sufficiently large x/an > u, and |Fn(u)− F (u)| < ε. It follows that FYn(x) = P(anXn ≤ x) = P(Xn ≤ x/an) ≥ Fn(u) > 1− 2ε . Thus limn→∞ FYn(x) = 1. By the same argument for x < 0 limn→∞ FYn(x) = 0. (4 marks) (b) Let µX be the law of a random variable X. Show that the distribution of X is symmetric, i.e. µX(−∞, x] = µX [−x,∞) for all x ∈ R, if and only if the characteristic function of X is real. [4] Answer: (Similar to exercise) ϕX(t) = E[eitX ] = E[ei(−t)(−X)] = E[eit(−X)] = ϕ−X(t). So by Le´vy’s in- version formula (i.e. one-to-one correspondence between probability measures on (R,B) and characteristic functions) X has the same law as −X if and only if ϕX(t) is real. (4 marks) (c) In the following, identify if the sequence of random variables (Yn)n≥1 converges weakly and if so identify the limit. (You should explain your reasoning, and state clearly any results from lectures that you use). (i) Yn = max{U1, . . . , Un} where U1, U2, . . . are independent Uniform(−1, 1) random variables. 7 (ii) Yn = n(1 − max{U1, . . . , Un}) where U1, U2, . . . are independent Uniform(−1, 1) random variables. (iii) Yn = √ 3 n(U1 + . . .+Un) where U1, U2, . . . are independent Uniform(−1, 1) random variables. (iv) Yn = nmin{U1, . . . , Un} where U1, U2, . . . are independent Uniform[0, 1] random variables. (v) Yn is an exponential random variable with mean λn > 0 for each n, i.e. P(Yn > y) = e−y/λn , where λn → 0 as n→∞. [8] Answer: (Unseen example) (i) Weak convergence to 1. Check (equivalently) convergence in distribution. For u ≥ 1, P(Yn ≤ u) = 1, if u ≤ −1 then P(Yn ≤ u) = 0 and for u ∈ (−1, 1), P(Yn ≤ u) = ((1 + u)/2)n → 0. (ii) Weak convergence to an Exp(1/2) random variable. Check (equivalently) convergence in distribution. For u > 0, P(Yn ≤ u) = 1 − P(max{U1, . . . , Un} < 1 − u/n) = 1 − P(X1 < 1 − u/n)n = 1 − ((2 − u/n)/2)n → 1− e−u/2. (iii) Weak convergence to standard normal. This follows from the Central Limit Theorem, which states if X1, X2, . . . is a sequence of independent identically distributed random variables with mean µ and variance σ, then 1√ nσ2 (X1 + . . .+Xn − nµ) weakly−−−−→ N (0, 1) . Here µ = 0 and σ2 = E[X21 ] = 23 . (iv) P(nmin{U1, . . . , Un} ≤ x) = 1− P ( U1 > x n , . . . , Un > x n ) = 1− ( 1− x n )n → 1− e−x , as n→∞. So Yn converges in distribution to an Exp(1) random variable. (v) Fix f ∈ Cb(R) then E f(Yn) = ∫ R f(x) 1 λn e−x/λndx = ∫ R f(λny)e −ydy → f(0) , as n→∞, where the final limit follows from the dominated convergence theorem since the integrand is dominated by |f(λny)e−y| ≤ (sup |f |)e−y and converges to f(0)e−y. Hence Yn converges weakly to 0. (8 marks) [TOTAL: 20] Continued... 8 4: (a) Let X and Y and Z be integrable random variables on a probability space (Ω,F ,P), and let G ⊆ F be a sub-σ-algebra. Show that if Y and Z are both versions of E[X | G] then Y = Z almost surely. [3] Answer: (Applied bookwork:) Suppose Z1 and Z2 are both versions of E[X | G], let Aε = {Z2−Z1 > ε} ∈ G (because Z1 and Z2 are both G-measurable so Z1 − Z2 is). Therefore, by (3.), E[X ; Aε] = E[Z1 ; Aε] = E[Z2 ; Aε]. So by linearity of expectation, and construction of Aε, we have 0 = E[Z1−Z2] ≥ εP(Aε), hence P(Z2 − Z1 > ε) = 0. By the same argument (symmetry) P(Z1 − Z2 > ε) = 0, and since a union of null events is null P(|Z1 − Z2| > ε) = 0. Finally, by taking compliments and applying monotone convergence of measures P(Z1 = Z2) = P ( ⋂ n≥1 {|Z1 − Z2| ≤ 1 n }) = lim n→∞P (|Z1 − Z2| ≤ 1/n) = 1. (3 mark) (b) Suppose that (Xn)n≥1 is a random process adapted to (Fn)n≥1. Define the hitting time τA of (Xn)n≥1 on a Borel set A. Show that τA is a stopping time for this filtration. [3] Answer: (Applied bookwork:) (Fn)n≥1 is a filtration on (Ω,F ,P) if for all n ∈ N, Fn ⊆ Fn+1 ⊆ F . The hitting time is defined by τA = inf{n ∈ N : Xn ∈ A}. τA is a stopping time if and only if {τA ≤ n} ∈ Fn for each n ∈ N. This holds because {τA ≤ n} = ⋃ k≤n {Xk ∈ A} and since (Xn)n≥1 is adapted to (Fn)n≥1 we have {Xk ∈ A} ∈ Fk ⊆ Fn. Hence {τA ≤ n} ∈ Fn. (3 marks) (c) Supose (Xn)n≥1 is a symmetric simple random walk on Z started from 0. Use the Optional Stopping Theorem for Bounded Stopping Times to evaluate the probability (Xn)n≥1 hits b > 0 before −a < 0. [4] Answer: (Seen in Exercises:) Let T = τ{−a,b}, then T is not bounded, but it is almost surely finite by a Borel-Cantelli argument, and Tk = T ∧ k is bounded for each k ∈ N. Hence, by dominated convergence, E[XT ] = limk→∞ E[XTk ], and E[XTk ] = 0 by OST. Let p = P(XT = b) = 1 − P(XT = −a), so that E[XT ] = p b − a(1 − p) and solving gives p = a/(b+ a). (4 marks) (d) Let (Xn)n≥1 be a sequence of independent random variables on probability space (Ω,F ,P), with E[Xn] = 1 for each n. Let F0 = {∅,Ω}, Fn = σ(X1, . . . , Xn) for n ∈ N, and M0 = 1, and Mn = n∏ k=1 Xk for n = 1, 2, . . . . (i) Show that (Mn)n≥0 is a martingale with respect to (Fn)n≥0. (ii) Now suppose ϕ(t) = E[etX1 ] <∞ for all t ∈ R. Let S0 = 0, Sn = ∑n k=1Xk, and Yn = etSn ϕ(t)n , for n = 0, 1, 2, . . . . Show, using part (i), that (Yn)n≥0 is a martingale with respect to (Fn)n≥0. [4] Answer: (Unseen example (seen similar):) (i) Fix n ∈ N. Firstly, by independence and integrability of the (Xn)n≥1 E[|Mn|] = n∏ k=1 E[|Xk|] <∞ , 9 and hence Mn is integrable. Since Mn depends only on X1, . . . , Xn it is Fn-measurable. Since Mn+1 = Xn+1Mn and Mn is Fn-measurable, we have E[Mn+1 | Fn] = E[Xn+1Mn | Fn] = MnE[Xn+1 | Fn] = MnE[Xn+1] = Mn a.s. , where we used ‘taking out what is known’ in the second equality and independence in the third. (ii) Using part (i), since etSn+1 = etXn+1etSn , we have Y0 = 1, , and Yn = n∏ k=1 etXk ϕ(t) for n = 1, 2, . . . .. Also E[etXk/ϕ(t)] = 1 and (etXn/ϕ(t))n≥1 is and independent sequence. Hence the result follows from part (i) with etXn/ϕ(t) in place of Xn. (4 marks) (c) Let (εn)n≥1 be independent random variables with P(εn = +1) = p, P(εn = −1) = q, where 1/2 < p = 1− q < 1 . Let Fn = σ(ε1, . . . , εn) and define Xn inductively by X0 = 1 and for n ≥ 1 Xn = Xn−1 + Vnεn . Assume Vn is a Fn−1-measurable random variable for each n, and that Vn is strictly between 0 and Xn. (i) Prove that if Xn > 0 then E[log(Xn+1/Xn) | Fn] = f(Vn+1/Xn) where f(x) = p log(1 + x) + q log(1− x) . (ii) Deduce that (log(Xn)− nα)n≥0 is a supermartingale, where α = p log p+ q log q + log 2 . [6] Answer: (Unseen example:) (i) Observe log ( Xn+1 Xn ) = log ( Xn + Vn+1 Xn ) 1εn+1=+1 + log ( Xn − Vn+1 Xn ) 1εn+1=−1 . Also E[1εn+1=+1 | Fn] = E[1εn+1=−1] = p and E[1εn+1=+1 | Fn] = q. Taking expectation conditioned on Fn, and ‘taking out what is known’ we get E[log(Xn+1/Xn) | Fn] = p log ( 1 + Vn+1 Xn ) + q log ( 1− Vn+1 Xn ) = f ( Vn+1 Xn ) . (3 marks) (ii) By part (i), E[log(Xn+1)− (n+ 1)α | Fn] = log(Xn)− nα+ f ( Vn+1 Xn ) − α , so we have to show f(Vn+1/Xn)− α ≤ 0. Notice f is differentiable, so to find the max of f on [0, 1] consider the derivative. For each x ∈ (0, 1) f ′(x) = p x+ 1 − q 1− x , so f is increasing on (0, p− q] and decreasing on [p− q, 1], hence attains its maximum at f(p− q) = p log(2p) + q log(2q) = α . It follows that (log(Xn)− nα)n≥0 is a supermartingale. (3 marks) 10 [TOTAL: 20] End. 11
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