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Let $ (X_t) $ be a stochastic process, and define a new stochastic process by $ Y_t = \int_0^t f(X_s) ds $. Is it true in general that $ \frac{d} {dt} \mathbb{E}(Y_t) = \mathbb{E}(f(X_t)) $? If not, under what conditions would we be allowed to interchange the derivative operator with the expectation operator?

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    A sufficient condition is that $E\left(\int_0^tf(X_s)ds\right)=\int_0^tE(f(X_s))ds$ and for that, some regularity of $(X_t)$ and $f$ and the finiteness of $\int_0^tE(|f(X_s)|)ds$ suffice. Keyword: Fubini.2016-10-26

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Interchanging a derivative with an expectation or an integral can be done using the dominated convergence theorem. Here is a version of such a result.

Lemma. Let $X\in\mathcal{X}$ be a random variable $g\colon \mathbb{R}\times \mathcal{X} \to \mathbb{R}$ a function such that $g(t, X)$ is integrable for all $t$ and $g$ is differentiable w.r.t. $t$. Assume that there is a random variable $Z$ such that $|\frac{\partial}{\partial t} g(t, X)| \leq Z$ a.s. for all $t$ and $\mathbb{E}(Z) < \infty$. Then $\frac{\partial}{\partial t} \mathbb{E}\bigl(g(t, X)\bigr) = \mathbb{E}\bigl(\frac{\partial}{\partial t} g(t, X)\bigr).$

Proof. We have $\begin{align*} \frac{\partial}{\partial t} \mathbb{E}\bigl(g(t, X)\bigr) &= \lim_{h\to 0} \frac1h \Bigl( \mathbb{E}\bigl(g(t+h, X)\bigr) - \mathbb{E}\bigl(g(t, X)\bigr) \Bigr) \\ &= \lim_{h\to 0} \mathbb{E}\Bigl( \frac{g(t+h, X) - g(t, X)}{h} \Bigr) \\ &= \lim_{h\to 0} \mathbb{E}\Bigl( \frac{\partial}{\partial t} g(\tau(h), X) \Bigr), \end{align*}$ where $\tau(h) \in (t, t+h)$ exists by the mean value theorem. By assumption we have $\Bigl| \frac{\partial}{\partial t} g(\tau(h), X) \Bigr| \leq Z$ and thus we can use the dominated convergence theorem to conclude $\begin{equation*} \frac{\partial}{\partial t} \mathbb{E}\bigl(g(t, X)\bigr) = \mathbb{E}\Bigl( \lim_{h\to 0} \frac{\partial}{\partial t} g(\tau(h), X) \Bigr) = \mathbb{E}\Bigl( \frac{\partial}{\partial t} g(t, X) \Bigr). \end{equation*}$ This completes the proof.

In your case you would have $g(t, X) = \int_0^t f(X_s) \,ds$ and a sufficient condition to obtain $\frac{d}{dt} \mathbb{E}(Y_t) = \mathbb{E}\bigl(f(X_t)\bigr)$ would be for $f$ to be bounded.

If you want to take the derivative only for a single point $t=t^\ast$, boundedness of the derivative is only required in a neighbourhood of $t^\ast$. Variants of the lemma can be derived by using different convergence theorems in place of the dominated convergence theorem, e.g. by using the Vitali convergence theorem.

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    @jochen you seem to be describing a random function rather than a random variable. Random variable has one value per path.2017-08-14