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Let $Y_t=\int_0^t X_sdW_s$. I know that if $\mathbb E\int_0^tX_s^2ds<\infty $, then $(Y_t)$ is a martingale, and $\mathbb EY_t=0$ for all $t$. But what if we only have $\int_0^t X_s^2ds<\infty $ ?

I know that in this case, $(Y_t)$ is a local martingale only, but do we have that $\mathbb E Y_t=0$ or not ? What I know, is thata there is a sequence of stopping time $\tau_n\nearrow \infty $ s.t. $(Y_{\tau_n\wedge t})$ is a martingale. Therefore, if $\mathbb E Y_{t\wedge \tau_n}=0$ for all $t$ and all $n$. But I think that we can't switch $\mathbb E$ and limit, so I think that that $\mathbb EY_{t\wedge \tau_n}$ won't hold, but I can't find a concrete example. Any idea ? And if it doesn't work, is there a condition on $Y=(Y_t)$ to have that $\mathbb E Y_t=0$ ?

Bruce
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2 Answers2

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There is the so-called Dudley representation theorem: for any distribution, there exists an adapted process $X$ such that $\int_0^t X_s^2 ds<\infty$ and $\int_0^t X_s dW_s$ has this distribution. So $\mathbb E Y_t$ can be anything (and can be even undefined, as another answer suggests).

A sufficient condition in terms of $Y$? I doubt it exists. Actually, it is possible that $\sup_{0\le s\le t} \mathbb E e^{|Y_s|}<\infty$ so that $Y$ is uniformly integrable in a very strong sense, but still $\mathbb E Y_t\neq 0$.

zhoraster
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Obviously, $\int_0^t e^{2B_s^4}ds<\infty $ a.s. According to Itô formula, $$\int_0^te^{B_s^4}\,\mathrm d B_s=\int_0^t e^{B_s^4}\,\mathrm d s-\int_0^t 2B_s^3e^{B_s^4}\,\mathrm d s.$$

Since the RHS is not integrable, $\mathbb E\left[\int_0^t e^{B_s^4}\,\mathrm d s\right]$ doesn't exist (see Wolfram Alpha).

Surb
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