Additional questions for chapter 3 1. Let ξ 1, ξ 2,... be independent and identically distributed with φθ) = IEexp{θξ 1 }) <. Let S n = S 0 + ξ 1 +... + ξ n. Show that M n = exp{θs n} φθ) n is a martingale with respect to σs 0,..., S n ). Apply the result to the special case IP ξ 1 = 1) = p, and IP ξ 1 = 1) = 1 p. Use subsequently measurability and independence IEM n+1 F n ) = IE [ exp{θξ n+1 + S n )}/ϕθ) n+1 F n ] = exp{θs n)} ϕθ) n+1 IE [exp{θξ n+1 } F n ] = exp{θs n)} IE [exp{θξ ϕθ) n+1 n+1 }] = exp{θs n)} = M ϕθ) n n. The second part is a straightforward application.
2. i) Let ξ 1, ξ 2,... be independent with IEξ i ) = 0 and IEξ i 2 ) = σi 2. Let S n = S 0 + ξ 1 +... + ξ n, where S 0 is a constant, and let v n = n i=1 σ2 i be the variance of S n. Show that M n = Sn 2 v n is a martingale. ii) Suppose we are testing the hypothesis that observations ξ 1, ξ 2,... are independent and have density function f but the truth is that ξ 1, ξ 2,... are independent and have density function g where {x : fx) > 0} = {x : gx) > 0}. Let fx)/gx) when gx) > 0 hx) = 0 when gx) = 0 Show that M n = hξ 1 ) hξ n ) is a martingale. i) Using the definition of M n, the independence of ξ i ) and the property taking out what is known of conditional expectation we get: IEM n M n 1 F n 1 ) = IES n 1 + ξ n ) 2 S 2 n 1 σ 2 n F n 1 ) = IE2S n 1 ξ n + ξ 2 n σ 2 n F n 1 ) = 2S n 1 )IEξ n F n 1 ) + IEξ 2 n F n 1 ) IEσ 2 n F n 1 ) = 0. Thus IEM n F n 1 ) = M n 1 showing that M n ) is a martingale. ii) This is a special case of a product martingale. Set ζ i = hξ i ) = fξ i )/gξ i ). Then fx) IEζ i ) = gx) gx)dx = fx)dx = 1 and the claim follows from the second example in 3.3 in the book.
3. Let X = {X n } n N0 be an integrable stochastic process which is adapted to the filtration {F n } n N0. Show that X has a decomposition X n = X 0 + M n + A n where {M n } n N0 is a martingale with M 0 = 0 and {A n } n N0 is a predictable process with A 0 = 0. Show that the decomposition is unique. Also show that {A n } n N0 is monotonously increasing iff X is a submartingale. Hint:E[X n X n 1 F n 1 ] = A n A n 1 ) First we define {A n } n N0 recursively by A 0 = 0 and the hint A n = A n 1 + E[X n X n 1 F n 1 ] Then A n is F n 1 measurable by induction and the measurability of conditional expectation. Thus, {A n } n N0 is predictable. Now define M n = X n X 0 A n. Then we see that {M n } n N0 is clearly integrable and adapted. We only have to show that it is a martingale. E[M n F n 1 ] = E[X n F n 1 ] X 0 A n = E[X n F n 1 ] X 0 A n 1 E[X n X n 1 F n 1 ] = X n 1 X 0 A n 1 = M n 1 It remains to show uniqueness. Assume that X has a second decomposition X n = X 0 + M n + A n. Then we get M n M n = A n A n. We see that M n M n is predictable. Thus M n M n = E[M n M n F n 1 ] = M n 1 M n 1 Therefore M 1 M 1 = M 0 M 0 = 0 and then M 1 = M 1. By induction we see that M n = M n. Then we immediately have A n = A n and the uniqueness has been shown. It remains to show that {A n } n N0 is monotonously incresing iff X is a submartingale. This follows immediately by the definition of A n.
4. Assume that ξ n ) n N is a sequence of independent random variables with E[ξ n ] = 0 n and E[expξ n )] < n. Furthermore and F n = σξ 1,..., ξ n ) S 0 = 0 S n = ξ k. a) Show that S n is a martingale with respect to F n ) n 0. b) Show that P n = exps n ) is a submartingale with respect to F n ) n 0. c) Now assume ξ n N0, σ 2 n). Determine the Doob-decomposition of P n. a) We have E[ S n ] n E[ ξ k ] < and S n is F n -measurable by definition of F n. E[S n+1 F n ] = E = [ n+1 ] ξ k F n = ξ k + E[ξ n+1 ] = ξ k + E[ξ n+1 F n ] = ξ k = S n b) We define fx) = expx). As f is convex, by applying the conditional Jensen formula we get E[fS n+1 ) F n ] fe[s n+1 F n ]) = fs n ) In addition to this, P n is adapted, as S n is adapted and fx) = expx) is Borel-measurable. Apart from this, E[ P n ] = n E[expξ k )] <. Thus, P n = exps n ) is a submartingale with respect to F n ) n 0. c) The Doob-decomposition is A n = E[P k P k 1 F k 1 ] = P k 1 e 1 2 σ2 k 1). The martingale part M n of the decomposition is M n = P n A n P 0 = P n P k 1 e 1 2 σ2 k 1) 1 Then P n = M n + A n + P 0, where A n is predictable, M n is a martingale and P 0 a constant.
5. We assume that Ω, F, IF, IP ) is a standard filtered probability space with IF = F n ) n=0 a filtration. Let U = U n ) n=0 be an adapted sequence and consider the discrete stochastic exponential E n U) = n 1 + U k ), E 0 U) = 1, where U n = U n U n 1. Consider the difference equation X n = X n 1 U n, X 0 = 1. DE) i) Verify that E n U) is a solution of DE). ii) Assume E n U) 0. Show that E n U) is a martingale if U n ) is a martingale. iii) Let α n ) n=0 be a deterministic series of positive numbers and V = V n ) n=0 be an adapted sequence. Set Prove that A n = IE e α k V k 1 ) Fk 1. { } Z n = exp α k V k En 1 A), Z 0 = 1 is a martingale. Hint: You may use E n U) 1 = E n Y ), where Y n = U n U n ) 2 1+ U n ) ). i) Define X n := E n U). Then X n = E n U) E n 1 U) = n 1 1 + U n 1) 1 + U k ) = U n X n 1. ii) Let U n ) be a martingale. Adaptedness of E n U)) can be seen. Integrability of E n U)) is provided by assumption. Define X n := E n U). Then, according to i), X n ) solves DE). Thus, by DE) and the martingale-property of U n ) E X n F n 1 ) = EX n 1 U n F n 1 ) which shows that X n ) is a martingale. = X n 1 E U n F n 1 ) = X n 1 0 = 0, iii) Adaptedness of Z n ) follows from adaptedness of A n ), integrability of Z n ) is provided by assumption. Applying the hint to A n ) and simplifying yields E n A) 1 = n k. 1 + A k
With this result, we get { } Z n = exp α k V k E n A) 1 = e α n Vn = e αn Vn n 1 + A n n 1 + A n Now, with the definition of A n, we find 1 A n 1 + A n = Applying this gives Z n = Z n 1 e α n Vn = Z n 1 n 1 e α k Vk Z n 1. 1 Eexp{α n V n } F n 1 ). n 1 + A n e α n V n Eexp{α n V n } F n 1 ) 1 k 1 + A k ) 1 ) and thus E Z n F n 1 ) = ) e αn Vn Z n 1 E Eexp{α n V n } F n 1 ) F n 1 Z n 1 = Z n 1 Eexp{α n V n } F n 1 ) E ) e αn Vn F n 1 Zn 1 = 0, completing the proof that Z n ) is a martingale.,
6. Let σ and τ be two stopping times with respect to the filtration F n ) n=0. Show that F σ τ = F σ F τ. Also show that the events belong to F σ F τ. {τ < σ}, {σ < τ}, {τ σ}, {σ τ}, {σ = τ} From Proposition 3.5.2 in the book we know that as σ τ σ and also σ τ τ we have F σ τ F σ and F σ τ F τ. Thus F σ τ F σ F τ. For the other direction assume A F σ F τ. Therefore and We have A {σ n} F n A {τ n} F n n n. A {σ τ n} = A {σ n} {τ n}) = A {σ n} A {τ n}) F n n. From this it follows that A F σ τ. We now have to verify that the mentioned sets are in F σ τ. First show {τ σ} F σ : {τ σ} {σ n} = Analogously we get {τ σ} F τ by: {τ σ} {τ n} = n {σ = k} {τ k} F n. k=0 n {τ = k} {σ > k} F n. The result for {σ τ} follows by reversing the roles of σ and τ. We also have: n {τ = σ} {σ n} = {σ = k} {τ = k} F n. k=0 k=0 Thus, we have {σ = τ} F σ. By reversing the role of σ and τ, we get the result. The result for the last two sets follows by taking complements.
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