Distribution of the integral of a diffusion process

Suppose $X(t)$ is a diffusion process with $E[X(t)]=0$ and variances $\sigma^2_t$ concave in time. If $X$ is also a Brownian motion, then the distribution of $\int_0^T X(t) dt$ is known to be $N(0,\frac{T^3}{3})$.

I want to find the distribution in the more general case, when $X$ can be any diffusion process with mean zero every where. In an even more general case, what is the distribution if $X$ is a Markov process with mean zero?

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Is $\sigma$ deterministic, isn't it ? – Paul Feb 15 '13 at 1:14
Yes, for instance, $\sigma^2_t =t$ when $X$ is a Brownian motion. – afshi7n Feb 15 '13 at 1:23
in your general case it's still a deterministic function I think, because you suppose it's concave so it can't be a stochastic process itself – Paul Feb 15 '13 at 1:25

First case : Assuming that $\sigma$ is a deterministic function, it's known that the diffusion process is normally distributed, $X_t \sim \mathcal N (0, \int_0 ^t \sigma _s ^2 ds )$, since $X_t = \int _0^t \sigma_s dW_s$ wich is a Wiener integral.

Hence, $\int _0 ^T X_t dt \sim \mathcal ( 0, \frac {1}{3} (\int_0 ^T \sigma_s ^2 ds )^3 )$

Second case: Now we assume that $X$ is an arbitrary markov stochastic process.

We can conclude by Fubini-Tonelli theorem that $\mathbb E \left \{ \int _0 ^T X_t dt \right \} =0$

Also, we cans roght the integral as a limit of a Rieman's sum. By effect,

$$\int _0 ^T X_t dt = \lim _ {n \rightarrow \infty} \sum_{i=1} ^{n} \frac {T}{n} X_{\frac{i}{n}}$$ Then we can probably calculate it's variance using this fact and the markov propertie, but in order to effectivelly characterize the law of $Z = \int _0 ^T X_t dt$ we must compute it's characteristic function $\psi_Z(u)=\mathbb E \left \{ e ^{iu Z}\right\}$ and then identifie the law

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Very helpful, Thanks a lot! – afshi7n Feb 19 '13 at 0:01