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 $[0,T]?$

  • $\begingroup$ Btw, I think there's an error in your expression for $Z$. It should be $$ Z_t=\exp\left(W_t^2-\int_0^t(2W_s^2+1)\,ds\right).$$ $\endgroup$ – George Lowther Mar 13 '11 at 22:17
  • $\begingroup$ Yes. Sorry about that. typo. $\endgroup$ – Jun Deng Mar 13 '11 at 22:23
  • $\begingroup$ @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 $t\to \infty$. $\endgroup$ – user940 Mar 13 '11 at 22:25
  • $\begingroup$ @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. $\endgroup$ – George Lowther Mar 13 '11 at 22:31
  • $\begingroup$ @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? $\endgroup$ – 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.

  • 1
    $\begingroup$ 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) $\endgroup$ – TheBridge Mar 17 '11 at 16:22
  • $\begingroup$ @TheBridge: I elaborated on this. $\endgroup$ – George Lowther Mar 19 '11 at 23:43

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