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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 $[0,T]?$

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    Btw, I think there's an error in your expression for $Z$. It should be $$ Z_t=\exp\left(W_t^2-\int_0^t(2W_s^2+1)\,ds\right).$$2011-03-13
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    Yes. Sorry about that. typo.2011-03-13
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    @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 $t\to \infty$.2011-03-13
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    @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.2011-03-13
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    @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?2011-03-13
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    @Jun Deng For a positive martingale, we have $Z_t\to Z_\infty$ almost surely, but only an inequality for the mean: $E[Z_\infty]\leq E[Z_0]$. Equality holds if the martingale is uniformly integrable.2011-03-13
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    @Jun: It does tend to zero. See also Gambler's ruin. http://en.wikipedia.org/wiki/Gambler%27s_ruin2011-03-13
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    @George You are quite right. I hadn't thought it through very carefully.2011-03-13
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    @Byron: Thanks. Now I get it.2011-03-13
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    @George: Thanks. With your comments, it is easy to prove $Z_t$ is square-integrable martingale over interval $(0,\frac{1}{2}).$ I will figure out the remaining parts.2011-03-13

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