Limits of integral equation

Question

I have this integral equation where I need to find the limit $T\to 0$

\begin{equation} \lim_{T\to 0} \frac{\int_0^T E(\sigma_s^2)(T-s)^\alpha ds}{T} \end{equation}

where $1/2<\alpha<1$ and $E(\sigma_s^2)\to \sigma_0^2$ when $s\to 0$. Can anyone give me some direction to solve this limit function. L' Hopital's rule doesn't appear to be working... Although I can't evaluate the integral, I do "roughly" expect the answer to be something like \begin{alignat*}{2} \lim_{T\to 0} \frac{\int_0^T E(\sigma_s^2)(T-s)^\alpha ds}{T} &= \lim_{T\to 0} E(\sigma_0^2) \frac{\int_0^T(T-s)^\alpha ds}{T} \\ &=\lim_{T \to 0} \frac{\sigma_0^2}{\alpha+1}T^\alpha \end{alignat*} I am not sure if the above solution even makes sense. Can anyone guide me on this?

Is $\sigma$ a random variable and $\mathbf{E}$ its expectation? — Alex, Jul 29 '20 at 14:57
Just noticed: the upper limit on the integral converges to $0$ — Alex, Jul 29 '20 at 15:09
@Alex it's $\sigma^2_s$ unfortunately. That's what makes this confusing... — Crushh, Jul 29 '20 at 15:18
I think by Fubini's theorem (since $\sigma^2_s$ os everywhere positive) it's still fine — Alex, Jul 29 '20 at 15:21
@Alex hmmm, i'm still not quite sure how to evaluate the function :l — Crushh, Jul 29 '20 at 15:25

score 1 · Answer 1 · answered Jul 29 '20 at 15:13

Not a full solution, but I think there could be typos in the question. For the case when the upper limit on the integral is some $n \to \infty$, and the rv is a function of $s$ and $T$:

You have convergence in the expectation, $\mathbf{E} [X_T] \to_T \mathbf{E}X_0$, which implies Dominated Convergence (applies to continuous rvs rather than sequence too), you have: $$ \lim_{T \to 0} \int_{0}^{n}\mathbf{E}\sigma^2_{s, T}ds = \int_{0}^{n}\mathbf{E}\sigma^2_{0,s}ds $$

Limits of integral equation

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