MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question
Cross-posted – Ilya Apr 4 '13 at 9:20

I assumed $\gamma=\sigma$.

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

Indeed, take $\alpha=0$, $X_0=1$, 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 $\alpha(0)>0$). For me, your question is a bit too wide in order to give a more precise answer.

share|cite|improve this answer

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)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.