# Time integral of a diffusion

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 $$dX_t=\alpha(X_t)dt+\gamma(X_t)dW_t\\\\ X_0=x_0$$ 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