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Let $B$ and $W$ be independent symmetric random walks starting at $0$ with filtration $\mathcal F_0=\{\emptyset,\Omega\}$ and $\mathcal F_n=\sigma(B_1,...,B_n,W_1,...,W_n)$.

Define $\mathcal E(B)_n=e^{B_n}(\cosh 1)^{-n}$ and $\mathcal E(W)_n=e^{W_n}(\cosh 1)^{-n}$ for all $n\ge 0$.

Also let $\tau=\inf\{n:\mathcal E(B)_n\le\frac 12\},\sigma=\inf\{n:\mathcal E(W)_n\ge 2\}$.

(a) Calculate $\Pr(\tau<\infty)$ and show that $\Pr(\sigma<\infty)\in (0,1)$.

(b) Which of the following three processes are uniformly integrable martingales: $$\mathcal E (B)^{\sigma\wedge\tau},\mathcal E(W)^{\sigma\wedge\tau},\mathcal E (B)^{\sigma\wedge\tau}\mathcal E(W)^{\sigma\wedge\tau}.$$


We can write \begin{align} \Pr(\tau<\infty)&=1-\Pr(\mathcal E(B)_n>\frac 12\text{ for all }n). \end{align} But then how to proceed. I have no idea why we are trying to compute $\mathcal E(B)_n$ and neither do I know how to compute this probability. Can someone give me a hint? Thank you!


Edit:

I have a rough idea but it still does not work. We need to involve $\Pr(\tau<\infty)$.

Firstly, it is easy to show that $\mathcal E(B)_n,\mathcal E(W)_n$ are martingales. Secondly, it is easy to see that $\tau,\sigma$ are stopping times with respect to the filtration $\mathcal \{F_n\}_{n\in\mathbb Z_{>0}} $ . So I want to use the Optimal Stopping Theorem. Note that \begin{align} \mathbb E[\mathcal E(B)_\tau]&=\mathbb E[\lim_{n\to\infty}\mathcal E(B)_{n\wedge\tau}]. \end{align} By Fatou's lemma, $$\mathbb E[\lim_{n\to\infty}\mathcal E(B)_{n\wedge\tau}]\le\liminf_{n\to\infty}\mathbb E[\mathcal E(B)_{n\wedge\tau}].$$ Since $\mathcal E(B)_n$ is a martingale, we have $$ \mathbb E[\mathcal E(B)_{n\wedge\tau}]=\mathbb E[\mathcal E(B)_{0\wedge\tau}]=1. $$

So $1\ge \mathbb E[\mathcal E(B)_\tau].$

Write $\mathbb E[\mathcal E(B)_\tau]=\mathbb E[\mathcal E(B)_\tau 1_{\tau<\infty}+\mathcal E(B)_\tau 1_{\tau=\infty}]=\mathbb E[\mathcal E(B)_\tau;\tau<\infty]+\mathbb E[\mathcal E(B)_\tau;\tau=\infty].$


Remark: Note that we have applied Fatou's lemma, but we cannot apply the dominated convergence to get equality, since the random walk can visit any big number before it returns to some point such that $\mathcal E(B)_n\le\frac 12$.

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I will write $M_n$ for $\mathcal{E}(B)_n$ to save typing. I claim $M_n \to 0$ almost surely.

Note that $M_n$ is a positive martingale so it converges almost surely to some limiting nonnegative random variable $M$. On the event $\{M > 0\}$, we have $\frac{M_{n+1}}{M_n} \to 1$ a.s. But $\frac{M_{n+1}}{M_n}$ can only have the values $\frac{1}{\cosh 1} e^{\pm 1}$ so it cannot converge to 1. Therefore $P(M > 0) = 0$ and so $M_n \to 0$ almost surely.

In particular, almost surely we have $M_n \le 1/2$ for all sufficiently large $n$, and thus $\tau < \infty$.

As for $\sigma$ (which I will think of as a stopping time for $B$ instead of $W$), by optional stopping we have $E[M_{\sigma \wedge n}] = 1$, and so by dominated convergence $E[\lim_{n \to \infty} M_{\sigma \wedge n}] = 1$ as well. If $\sigma < \infty$ a.s., we have $M_{\sigma \wedge n} \to M_\sigma \ge 2$ and this is absurd, as you noted. If $\sigma = \infty$ a.s., we have $\lim_{n \to \infty} M_{\sigma \wedge n} = \lim_{n \to \infty} M_n =0$ which is also absurd.


Some hints on (b):

  • Suppose $\mathcal{E}(B)^{\sigma \wedge \tau}$ were uniformly integrable. Since $\sigma$ is independent of $B$ and $\tau$, for each $0 \le k \le \infty$ we have $E[\mathcal{E}(B)^{\sigma \wedge \tau}_\infty \mid \sigma=k] = E[\mathcal{E}(B)_\infty^{k \wedge \tau}]$. Compute this for each $k$ and write $$E[\mathcal{E}(B)_\infty^{\sigma \wedge \tau}] = \sum_{0 \le k \le \infty} E[\mathcal{E}(B)^{\sigma \wedge \tau}_\infty \mid \sigma=k] P(\sigma = k).$$

  • $\mathcal{E}(W)^\sigma$ is uniformly integrable, and $\sigma \wedge \tau \le \sigma$.

  • No ideas about the third one.

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  • $\begingroup$ Thank you. So this is basically what I got except the last sentence. Do you have any idea on part (b)? $\endgroup$ – Bach Dec 14 '20 at 0:40
  • $\begingroup$ I agree that $\mathcal E(W)^{\sigma\wedge\tau}$ is uniformly integrable. But why isn't $\mathcal E(B)^{\sigma\wedge\tau}$? If we verify directly from the definition of uniformly integrable martingale, I think $\mathcal E(B)^{\sigma\wedge\tau}$ satisfies the condition. Also, I am confused with your first hint. Is $\mathcal E(B)_\infty$ the limit of $\mathcal E(B)_n$? But doesn't $\mathcal E(B)_n$ converge to $0$ a.s.? $\endgroup$ – Bach Dec 14 '20 at 4:36
  • $\begingroup$ @Bach: What's your proposed proof of uniform integrability? And yes, $M_\infty$ is mean to be the limit of $M_n$ for a martingale $M$. I agree that $\mathcal{E}(B)_n$ converges to 0 a.s., but $\mathcal{E}(B)^{\sigma \wedge \tau}_n$, etc, does not; since $\tau < \infty$ it converges to $\mathcal{E}(B)_{\sigma \wedge \tau}$. $\endgroup$ – Nate Eldredge Dec 14 '20 at 4:42
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I made some progress, but still not there.

Let $X_n=B_n-B_{n-1}$, and we know $\Pr(X_n=1)=\Pr(X_n=-1)=0.5\ \forall n$, and they are independent.

Let $Y_n=e^{X_n} (\cosh 1)^{-1}>0$, it is easy to see that $\mathbb E Y_n=1$ and they are also independent since $Y_n$ is the conposition composition of a continuous function(hencec Borel measurable) and $X_n$.

Note that $\mathcal E(B)_0=1$ and for all $n>0$: $\mathcal E(B)_n=e^{B_n}(\cosh 1)^{-n}=\prod_{i=1}^n[e^{X_i} (\cosh 1)^{-1}]=\prod_{i=1}^nY_i$.

By Kakutani's theorem we conclude that $\mathcal E(B)_n\to 0$ a.s.(since $\mathbb E\sqrt{Y_i}<1$(generalized mean inequality or see:Prove inequality of generalized means) and it follows that $\prod_{i=1}^\infty Y_i=0 $)

Hence $\Pr(\tau<\infty)=1$ by definition.


Now we study $\Pr(\sigma<\infty)$.

Firstly, it is easy to show that $\mathcal E(B)_n,\mathcal E(W)_n$ are martingales.

Secondly, it is easy to see that $\tau,\sigma$ are stopping times with respect to the filtration $ \{\mathcal F_n\}_{n\in\mathbb Z_{>0}} $ . So I want to use the Optimal Stopping Theorem. Note that \begin{align} \mathbb E[\mathcal E(W)_\sigma]&=\mathbb E[\lim_{n\to\infty}\mathcal E(W)_{n\wedge\sigma}]. \end{align} Since $\mathcal E(W)_{n\wedge\sigma}\le 2e(\cosh 1)^{-1}\ \forall n$, by the bounded convergence theorem we have $$\mathbb E[\lim_{n\to\infty}\mathcal E(W)_{n\wedge\sigma}]=\lim_{n\to\infty}\mathbb E[\mathcal E(W)_{n\wedge\sigma}].$$ Since $\mathcal E(W)_n$ is a martingale, we have $$ \mathbb E[\mathcal E(W)_{n\wedge\sigma}]=\mathbb E[\mathcal E(W)_{0\wedge\sigma}]=1. $$

So $1= \mathbb E[\mathcal E(W)_\sigma].$

Write $$\mathbb E[\mathcal E(W)_\sigma]=\mathbb E[\mathcal E(W)_\sigma 1_{\sigma<\infty}+\mathcal E(W)_\sigma 1_{\sigma=\infty}]=\mathbb E[\mathcal E(W)_\sigma;\sigma<\infty]+\mathbb E[\mathcal E(W)_\sigma;\sigma=\infty].$$

If $\Pr(\sigma<\infty)=1$, then $1=\mathbb E[\mathcal E(W)_\sigma; \sigma<\infty]\ge 2$. Contradiction!

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