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Define $\bar\sigma^2_t=\frac{1}{t}\int_0^t\sigma^2(X_s)ds$ where $\sigma(x)\geq0$ is a measurable function and $X_t$ a diffusion process defined by \begin{equation} dX_t=\alpha(X_t)dt+\gamma(X_t)dW_t\\\\ X_0=x_0 \end{equation} and $\sigma(x_0)>0$. Also assume $\mathbb{E}\left[|\sigma(X_t)-\sigma(X_0)|\right]=o(t^\beta)$ for some $1<\beta<2$.

My question is: What kind of conditions ensure $\displaystyle\lim_{t\to0}\mathbb{E}\left[\frac{1}{\bar\sigma_t}\right]<\infty$, or $\displaystyle\lim_{t\to0}\mathbb{E}\left[\frac{1}{\bar\sigma^2_t}\right]<\infty$?

Does it perhaps always hold as $\bar\sigma_t$ is a finite variation process? I'm not necessary looking for the minimal conditions, but rather something sufficient (but not trivial).

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Cross-posted –  Ilya Apr 4 '13 at 9:20

2 Answers 2

If $\sigma$ is bounded away from zero: $\sigma^2(X)>M$, then $\overline \sigma_t^2\ge M$ and your expectations are finite.

Also if $\sigma(x_0)\neq 0$ and if $\sigma$ is continuous then together with $X$'s continuity you can apply the fundamental theorem of calculus to show, that $\overline\sigma_t^2$ converges almost surely to $\sigma^2(x_0)$.

To interchange the limit and the expectation you would then need to show that $\{1/\overline\sigma^2_t\}_t$ is uniformly integrable, which might help you. (There are several sufficient conditions for uniform integrability and boundedness is not necessary)

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If $\sigma$ vanishes somewhere and is only measurable, then your conditions are clearly not enough to guarantee anything, even if $\gamma=0$.

Indeed, take $\gamma=0$, $X_0=1$, and $\sigma(x)=1_{x>0}$. It is not hard to see that your condition $\mathbb{E}\left[|\sigma(X_t)-\sigma(X_0)|\right]=o(t^\beta)$ is satisfied. On the other hand, the diffusion gets stuck at $x=0$ if it reaches that point, so $$\mathbb{E}\left[\frac{1}{\bar\sigma_t}\right]\geq \infty \cdot\mathbb{P}(\tau_0<t/2)=\infty.$$

The situation changes if $\sigma>0$ or if there is a drift that pushes you away from points where $\sigma$ vanishes (for example, if $\sigma$ vanishes only at $0$, is Lipshitz, and $\gamma(0)>0$). For me, your question is a bit too wide in order to give a more precise answer.

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