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## Hot answers tagged martingales

3

Note that $\mathcal{F}_s^{XY} \subseteq \mathcal{F}_s^X \vee \mathcal{F}_s^Y$ for all $s \geq 0$; this follows from the fact that the product $X_s Y_s$ is measurable with respect to $\mathcal{F}_s^X \vee \mathcal{F}_s^Y$. Hence, by the tower property of conditional expectation, $$\mathbb{E}(X_t Y_t \mid \mathcal{F}_s^{XY}) = \mathbb{E} \bigg[ \underbrace{\... 3 A martingale is uniformly integrable if and only if it converges in L^1. A uniformly integrable martingale converges almost surely. For convergence in L^2 we need to know a bit more. (for example the martingale being bounded in L^2 is sufficient) 2 If X is independent of \mathcal{G} then f(X) is also independent of \mathcal{G} for every Borel function f. This is easy to verify from the fact that \sigma(X) is independent of \mathcal{G}, and that (f(X))^{-1}(B) = X^{-1}(f^{-1}(B)). In particular we have E[f(X) \mid \mathcal{G}] = E[f(X)] whenever E[f(X)] exists. Apply this with f(... 2 Re pointwise convergence: Kolmogorov three series theorem is somewhat overkill here. Just note that$$|S_n(\omega)-S_{n-1}(\omega)| = 1, \qquad n \geq 1,$$for almost all \omega \in \Omega. In particular, for any \epsilon \in (0,1) there does not exist N \in \mathbb{N} such that |S_n(\omega)-S_{n-1}(\omega)| \leq \epsilon for all n \geq N; this ... 2 Question 1: Yes. According to Ito product rule d(tW)=W dt+tdW, then we could get$$ \int_{0}^t W(s)ds=tW(t)-\int_{0}^{t}s dW(s) $$For the right-hand side of first term, it is \mathcal{F}_t-measurable;for the second term, it is also \mathcal{F}_t-measurable. For how to determine whether a random variable is \mathcal{F}_t-measurable, there is an ... 1 I would rather use the \pi - \lambda theorem (and not monotone class theorem). Sets of the type A\cap B with A \in \mathcal F^{X}_s and B \in \mathcal F^{Y}_s form a \pi system and the class defined in that reference is a \lambda system containing this \pi system. Hence it contains the sigma algebra generated, which is \sigma(\mathcal F^{X}_s \... 1 If \ \mathcal{F}, \mathcal{G}\  are \sigma-algebras, then \mathcal{F}\vee\mathcal{G} is the \sigma-algebra generated by \ \mathcal{F}\cup\mathcal{G}\ , known as the join of \ \mathcal{F}\  and \ \mathcal{G}\ . 1 For two \sigma-algebras \mathcal{A} and \mathcal{B} on the same set, one usually denotes$$ \mathcal{A} \vee \mathcal{B} := \sigma(\mathcal{A}\cup\mathcal{B}).$$This is a pretty common notation when talking about filtrations. 1 If \mathbb{P}(S<\infty)=1 or if M_\infty is defined almost everywhere then obviously M_{S\wedge n}\to M_S almost surely. But if at some point M_n does not converge and S=\infty then at this point M_{S\wedge n}=M_n for all n and this sequence does not converge. So at such points M_S is not really defined. So the problem is when the set of ... 1 Formally you are right. What Williams means is that the process is bounded almost surely. This is enough to prove the theorem anyway. (he proves that a limit exists almost surely, so it's not a problem to ignore a set of probability zero here) 1 What you've described is the negative binomial distribution. We say X has a negative binomial distribution X \sim NB(r, p) if X is the number of independent bernoulli trials of parameter p until we get r 'successes'. So in your example, if X is the number of trials until we get 10 heads, X will have NB(10, 1/2) distribution. To calculate its ... 1 This is the probability that we get at most 9 heads in 19 tosses. That is,$$\sum_{j=0}^9\binom{19}j\frac1{2^{19}}$$1 The probability of getting at least m heads after n flips is just P(X \geq 10) where X ~ Bin(n,\frac{1}{2}). Does this help? 1 Your proof is correct. The boundness of the stopping times is needed in optional stopping theorem because the martingale M=(M_n) may not converge as n\to\infty. In other words, if the martingale M_n does not converge as n\to\infty and also the stopping time \tau is unbounded, then the random variable M_{\tau(\omega)}(\omega) cannot be well-... 1 As Brownian motion has independent increments, we have for s < t and \sigma > 0 that$$\mathbb{E}[e^{\sigma B_t}|\mathcal{F_s}] = \mathbb{E}[e^{\sigma (B_t - B_s)}\cdot e^{\sigma B_s} | \mathcal{F_s}] = e^{\sigma B_s} \mathbb{E}[e^{\sigma (B_t - B_s)}].$$This expectation can be computed with help of the moment generating function of the normal ... 1 This is not an answer, but is too long for comment. I will elaborate on my comment. We have for t\le T$$X_t = x + \int_0^t b(X_s)\,d s + \int_0^t \sigma(X_s) \, d W_s in \mathbb{R}. In the following C changes from line to line. For each t \le T then, \begin{align} E (X_t^2) &\le C \left[x^2 + E \left\{\left( \int_0^t b(X_s)\,d s\right)^2\... 1 For  0 <s<t, E(e^{\int_0^{t} W_u du}e^{\int_0^{s} W_u du}|\mathcal F_s)=e^{2\int_0^{s} W_u du}E(e^{\int_s^{t} (W_u-W_s) du}|\mathcal F_s) e^{(t-s)W_s}. The conditioning in the last expression is unnecessary by independence. Also, \int_s^{t} (W_u-W_s) du has the same distribution as \int_0^{t-s} W_u du (because (W_{r+s}-W_s)_{r \geq 0} is also ... 1 You are right that there might be overshooting, but we have some bounds on how much. First, since there might be overshooting now, we need to replace a and b with E(M_\tau\mid \tau = \tau_a) and E(M_\tau\mid \tau=\tau_b), and then we get P(M_\tau\ge b) = P(\tau=\tau_b) = \frac{1-E(M_\tau\mid \tau=\tau_a)}{E(M_\tau\mid \tau=\tau_b)-E(M_\tau\mid \...

1

If $X$ is independent of $\mathcal G$ then $X^{2}$ is also independent of $\mathcal G$ so the answer is YES.

1

Since you mention that you do not fully understand what "independent increments" means, I believe it is more appropriate to start with a simpler example that illustrates the idea. Instead of a Brownian motion, consider its discrete analogue: the random walk. I will start from a sequence $X_1,X_2,\ldots$ consisting of independent random variables. This means ...

1

Independent increments means that for $t>s$, the random variable $B_t-B_s$ is independent of the $\sigma$-algebra at time $s$, i.e. independent of $\mathcal{F}_s=\sigma(\cup_{v\leq s} B_v)$. In fact, for Brownian motion, $B_t-B_s\sim \mathcal{N}(0,t-s)$ and is independent of $\mathcal{F}_s$. Of course, this implies $(B_t-B_s)^2$ is also independent of $\... 1 Your answer is correct in the first two cases. If$t <r<s$then$W_s-W_r$is independent of$F_t$. [This can be proved from the fact that BM has independent increments]. Hence the answer in this case is$E(W_s-W_r)^{2}=s-r. 1 First of all, we notice that \begin{align} E[(X_t - X_s)(Y_t - Y_s)| \mathcal F_s] &= E[X_tY_t - X_tY_s -X_sY_t +\underbrace{X_sY_s}_{\mathcal{F}_s-\text{mesurable}}|\mathcal F_s]\\ &= E[X_tY_t|\mathcal F_s] - Y_sE[X_t|\mathcal F_s] - X_sE[Y_t|\mathcal F_s] + X_sY_s \\ &= E[X_tY_t|\mathcal F_s] - Y_sX_s - X_sY_s + X_sY_s \\ &=E[X_tY_t-... 1 There is Theorem 9 p.142 in Liptser and Shiryaev's "Theory of Martingales" of which a particular case is the following : LetM = (M_t)_{t \geq 0}$be a local martingale and define the process$B = (B_t)_{t \geq 0}$by : $$B_t = \sum_{0 < s \leq t}\frac{(\Delta M_s / (1+s))^2}{1 + |\Delta M_s / (1 + s)|},$$ where$\Delta M_t$is the jump of$M$at time$...

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