# Is this a valid method for time-integrating a stochastic process?

I have a stochastic process $X_t$, and I have a function $a(x | t)$ that reflects my beliefs about the value of $X_t$ ($a$ is a density function in its first parameter). I am studying the properties of the stochastic process $Y_t = \int_0^t X_s ds$. I am thinking of using the following method to find a density function $b(y | t)$ for $Y_t$:

Let $M_n$ be a function that returns the $n^{th}$ moment of a random variable. By Fubini's Theorem, $\int_0^t M_n(X_s) ds = M_n(\int_0^t X_s ds) = M_n(Y_t)$. Since this gives me a function that spits out all moments of $Y_t$, and since a random variable is uniquely determined by its moments, this is enough information to find $Y_t$.

My questions: (1) Have I applied Fubini's Theorem correctly? I'm having trouble formalizing the proof of that first equality, but I intuitively feel that it's true. (2) Are there any other obvious flaws with this method?

Thank you.

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(1) you have not, as $M_n(x) = x^n$ and $\int X_s^nds\neq(\int X_s ds)^n$ in many cases – Ilya Jan 3 '13 at 20:40
Ilya identifies the biggest issue here. It is also worth noting that not all random variables are identified by their moments. – Chris Janjigian Jan 3 '13 at 20:44
Okay, thank you. So this is true for the special case $n = 1$ (in other words, the integral of the expectation is the expectation of the integral), but not necessarily true for any other moment. Right? – GMB Jan 3 '13 at 20:50
If you assume that $X_t$ is positive or that the integral of the first absolute moments is finite, then yes. – Chris Janjigian Jan 3 '13 at 22:17
If you don't have any additional knowledge about $X_t$, i.e. if $a(x|t)$ is the only thing that you know, it may not be sufficient to find $b(y|t)$. – Ilya Jan 4 '13 at 8:48