Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $W_t$ be a standard Brownian motion with $W_0 = 0$ and let $Z_t$ solve the stochastic differential equation $dZ_t = 2 Z_t W_t \mathrm{d}W_t$. This has solution

$$ Z_t=\exp\Big\{W_t^2-\int_0^t{(2W_s^2+1)ds}\Big\} \> . $$

It is easy to show that $Z_t$ is a local martingale since $P(\int_0^T{(Z_sW_s)^2ds}<\infty)=1.$

Could we show that $E[\int_0^T{(Z_sW_s)^2ds}]<\infty$, which implies $Z_t$ is a martingale in the interval $[0,T]?$

share|cite|improve this question
Btw, I think there's an error in your expression for $Z$. It should be $$ Z_t=\exp\left(W_t^2-\int_0^t(2W_s^2+1)\,ds\right).$$ – George Lowther Mar 13 '11 at 22:17
Yes. Sorry about that. typo. – Jun Deng Mar 13 '11 at 22:23
@George Ah! that makes more sense. If $Z_t$ is a martingale, then it converges pointwise as $t\to\infty$. But I couldn't figure out why that was true. Now I guess that the positive time integral in the exponent goes to infinity quickly enough to swamp the contribution from $W^2_t$. So $Z_t\to 0$ as $t\to \infty$. – Byron Schmuland Mar 13 '11 at 22:25
@Byron: I would expect the $\int_0^tW_s^2\,ds$ to dominate, as it has mean $\frac12t^2$ and the other terms have mean proportional to $t$. So it should tend to zero regardless of whether you have $+1$ or $-1$ in the integral. – George Lowther Mar 13 '11 at 22:31
@Byron: I think $Z_t$ does not go to $0$ as $t\rightarrow +\infty$, if $Z_t$ is a martingale, since $E[Z_t]=E[Z_0]=1.$ Right? – Jun Deng Mar 13 '11 at 22:43

Yes, $Z$ is a proper martingale. However, $\int_0^T(Z_sW_s)^2\,ds$ is not integrable for large $T$. As the quadratic variation of $Z$ is $[Z]_t=4\int_0^t(Z_sW_s)^2\,ds$, Ito's isometry says that this is integrable if and only if $Z$ is a square-integrable martingale, and you can show that $Z$ is not square integrable at large times (see below).

However, it is conditionally square integrable over small time intervals.

$$ \begin{align} \mathbb{E}\left[Z_t^2W_t^2\;\Big\vert\;\mathcal{F}_s\right]&\le\mathbb{E}\left[W_t^2\exp(W_t^2)\;\Big\vert\;\mathcal{F}_s\right]\\ &=\frac{1}{\sqrt{2\pi(t-s)}}\int x^2\exp\left(x^2-\frac{(x-W_s)^2}{2(t-s)}\right)\,dx \end{align} $$

It's a bit messy, but you can evaluate this integral and check that it is finite for $s \le t < s+\frac12$. In fact, integrating over the range $[s,s+h]$ (any $h < 1/2$) with respect to $t$ is finite. So, conditional on $W_s$, you can say that $Z$ is a square integrable martingale over $[s,s+h]$.

This is enough to conclude that $Z$ is a proper martingale. We have $\mathbb{E}[Z_t\vert\mathcal{F}_s]=Z_s$ (almost surely) for any $s \le t < s+\frac12$. By induction, using the tower rule for conditional expectations, this extends to all $s < t$. Then, $\mathbb{E}[Z_t]=\mathbb{E}[Z_0] < \infty$, so $Z$ is integrable and the martingale conditions are met.

I mentioned above that the suggested method in the question cannot work because $Z$ is not square integrable. I'll elaborate on that now. If you write out the expected value of an expression of the form $\exp(aX^2+bX+c)$ (for $X$ normal) as an integral, it can be seen that it becomes infinite exactly when $a{\rm Var}(X)\ge1/2$ (because the integrand is bounded away from zero at either plus or minus infinity). Let's apply this to the given expession for $Z$.

The expression for $Z$ can be made more manageable by breaking the exponent into independent normals. Fixing a positive time $t$, then $B_s=\frac{s}{t}W_t-W_s$ is a Brownian bridge independent of $W_t$. Rearrange the expression for $Z$ $$ \begin{align} Z_t&=\exp\left(W_t^2-\int_0^t(2(\frac{s}{t}W_t+B_s)^2+1)\,ds\right)\\ &=\exp\left(W_t^2-2\int_0^t\frac{s^2}{t^2}W_t\,ds+\cdots\right)\\ &=\exp\left((1-2t/3)W_t^2+\cdots\right) \end{align} $$ where '$\cdots$' refers to terms which are at most linear in $W_t$. Then, for any $p > 0$, $$ Z_t^p=\exp\left(p(1-2t/3)W_t^2+\cdots\right). $$ The expectation $\mathbb{E}[Z_t^p\mid B]$ of $Z_t^p$ conditional on $B$ is infinite whenever $$ p(1-2t/3){\rm Var}(W_t)=p(1-2t/3)t \ge \frac12. $$ The left hand side of this inequality is maximized at $t=\frac34$, where it takes the value $3p/8$. So, $\mathbb{E}[Z_{3/4}^p\mid B]=\infty$ for all $p\ge\frac43$. The expected value of this must then be infinite, so $\mathbb{E}[Z^p_{3/4}]=\infty$. It is a standard application of Jensen's inequality that $\mathbb{E}[\vert Z_t\vert^p]$ is increasing in time for any $p\ge1$ and martingale $Z$. So, $\mathbb{E}[Z_t^p]=\infty$ for all $p\ge 4/3$ and $t\ge3/4$. In particular, taking $p=2$ shows that $Z$ is not square integrable.

share|cite|improve this answer
You say "you can show that $Z$ is not square integrable at large times". I am sorry but I don't see it. Could you elaborate a little more about this ? Best Regards (sorry if this is obvious) – TheBridge Mar 17 '11 at 16:22
@TheBridge: I elaborated on this. – George Lowther Mar 19 '11 at 23:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.