A non-degenerate martingale - MathOverflow

A non-degenerate martingale

2011-12-24T09:38:46Z

Let $(\Omega, \mathcal{F}, P)$ be a probability space, on which $\mathcal{F}_t$ is filtration satisfying general conditions. $W_t$ is a standard Brownian motion. Let $Y_t$ be a martingale given by $$Y_t = \int_0^t \sigma_r d W_r$$ where $\sigma_t$ is a bounded $\mathcal{F}_t$ measurable process.

The question is, assume $\sigma_t>0$ almost surely for all $t$, then can we prove $$P(Y_1 = c) = 0$$ for all constant $c$?

Answer by James Martin for A non-degenerate martingale

2011-12-26T00:50:52Z

Here's an explicit construction that gives a counterexample. For simplicity let $c=0$ (not important).

First, let $\alpha>0$ and consider the probability that a standard Brownian motion started at 0 hits 0 at some time in the interval $(\alpha t, \alpha t+t)$. Then (1) this probability does not depend on $t$ (by Brownian scaling) -- call it $p_\alpha$; and (2) $p_\alpha\to 1$ as $\alpha\to 0$ (because with probability 1, the standard Brownian motion hits 0 at some time in the interval $(0,t)$).

Now we'll use this and Borel-Cantelli to show that with positive probability, we can construct $\sigma$ and a sequence of times $t_n\uparrow 1$ such that $Y_{t_n}=1$ for all $n$.

Let $\alpha_n$ be some sequence decreasing to 0 quickly enough that $\sum (1-p_{\alpha_n}) <\infty$.

Let $t_0=0$ and recursively define $t_1, t_2, \dots$ as follows.

Given $t_n<1$ and $Y(t_n)=0$, let $\sigma_t=\sqrt{\alpha_n}$ for $t\in(t_n, (1+t_n)/2)$, and let $\sigma_t=1$ for $t\in((1+t_n)/2, t_{n+1})$, where $t_{n+1}$ is defined by

$t_{n+1}=\inf \big[ t>(1+t_n)/2: Y_t=0 \big]$.

The idea of this definition: given $t_n<1$, we divide the remaining time interval $(t_n, 1)$ into two halves, and run BM at speed $\alpha_n$ on the first half and at speed 1 on the second half, stopping as soon as we hit 0 during the second half. Since we start at 0, the probability that we DO hit 0 at some point during the second half is easily seen to be $p_{\alpha_n}$ as defined above. Hence $P(t_{n+1}<1 | t_n<1)=p_{\alpha_n}$.

Now using Borel-Cantelli (and reasoning straightforwardly about independence) we get that there is positive probability that $t_n<1$ for all $n$. In that case also $t_n\uparrow 1$ (since $1-t_{n+1}<(1-t_{n})/2$). Also $Y_{t_n}=0$ for all $n$ by construction. But the process $Y_t$ is continuous (since $\sigma_t$ is bounded). So then also $Y_1=0$ as desired.

Answer by Yuri Bakhtin for A non-degenerate martingale

2011-12-26T01:07:15Z

I have not thought about such questions for a while, but I do not see an immediate mistake in the following reasoning:

Let $Y_t=E[\mathrm{sign}(W_1)| \mathcal{F}_t]$. Then it is a bounded martingale, and it also has a continuous modification since $Y_t=E[\mathrm{sign}(W_1) | W_t]$ and $E[\mathrm{sign}(W_1) | W_t=x]$ depends on $t$ and $x$ continuously. Therefore, it admits a representation via stochastic integral, but $P(Y_t=\pm1 )=1/2$.

This does not contradict the stochastic representation since $\sigma_t$ in it becomes increasingly large if $t$ is close to 1, but the process is far from $\pm1$. This diffusion pushes the process closer to the boundaries of $[-1,1]$.

Is this true, or am I missing something?

Answer by John Dawkins for A non-degenerate martingale

2011-12-27T02:53:26Z

Another counter-example: Let $f:R\to R$ be a non-constant $C^2$ funtion with $f, f', f''$ all bounded, and such that (i) $f$ vanishes in a non-empty open interval $I$, (ii) $f'>0$ outside the closure of $I$, and (iii) $E[f(W_1)]=0$. Consider the random variable $Y_1:= f(W_1)$. Clearly $P[Y_1=0]= P[W_1\in I]>0$. The process $$ \sigma_t(\omega):=h(t,W_t(\omega)), $$ where $$ h(t,x) :=\cases{\int_R {1\over\sqrt{2\pi(1-t)}} \exp[-(y-x)^2/2(1-t)] f'(y) dy,& 0\le t<1,\cr 0,&t\ge 1,\cr} $$ is predictable and bounded, and $\sigma_t(\omega)>0$ fo all $t\in[0,1)$ and all $\omega\in \Omega$. Moreover, by Ito's formula, $$ Y_1=\int_0^1 \sigma_t dW_t, $$ almost surely. In fact, $$ E[f(W_1)|{\mathcal F}_s]=\int_0^s \sigma_t\,dW_t,\qquad\forall s>0, $$ almost surely.