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 : Let $M = (M_t)_{t \geq 0}$ 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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