When can we interchange the derivative with an expectation?

Question

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?

Answer 1

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 continuously 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.

Answer 2

The lemma which is stated in jochen's answer is quite useful. However, there are cases in which the integrand is not differentiable with respect to the parameter. Here, there is a discussion about some results which can be made in a more general setup.

Let $\left(\mathbf{X},\mathcal{X},\mu\right)$ be a general measure space (e.g., a probability space) and let $\xi:\mathbf{X}\times[0,\infty)\rightarrow\mathbb{R}$ be such that:

(a) For every $s\geq0$, $x\mapsto\xi(x,s)$ is $\mathcal{X}$-measurable.

(b) For every $x\in\mathbf{X}$, $s\mapsto\xi(x,s)$ is right-continuous (This assumption can be weakened by letting it be valid just $\mu$-a.s. but then $\left(\mathbf{X},\mathcal{X},\mu\right)$ has to be a complete).

In particular, notice that (a) and the right-continuity assumption which is listed in (b) imply that $\xi\in\mathcal{X}\otimes\mathcal{B}[0,\infty)$ where $\mathcal{B}[0,\infty)$ is the Borel $\sigma$-field which is generated by $[0,\infty)$. For details see, e.g., Remark 1.4 on p. 5 of I. Karatzas, S.E. Shreve, Brownian Motion and Stochastic Calculus, Springer, 1988. Then, for every $(x,t)\in\mathbf{X}\times[0,\infty)$ define $g(x,t)=\int_0^t\xi(x,s)ds$ and note that $t\mapsto g(x,t)$ has a right-derivative which equals to $s\mapsto\xi(x,s)$. In addition, for every $t\geq0$ let

$$\varphi(t)\equiv\int_{\mathbf{X}}g(x,t)\mu(dx)=\int_{\mathbf{X}}\int_0^t\xi(x,s)ds\mu(dx)\,.$$

To make $\varphi(\cdot)$ be well-defined, let $m$ be Lebesgue measure on $[0,\infty)$ and assume that the pre-conditions of Fubini's theorem are satisfied, e.g., $\xi(x,s)$ is nonnegative (This assumption can be weakened by letting it be valid just $\mu$-a.s. but then $\left(\mathbf{X},\mathcal{X},\mu\right)$ has to be a complete) or integrable with respect to $\mu\otimes m$. Then, deduce that

$$\varphi(t)=\int_0^t\zeta(s)ds\ \ , \ \ \forall t\geq0$$

such that for every $t\geq0$, $\zeta(t)\equiv\int_{\mathbf{X}}\xi(x,t)\mu(dx)$. This means that if there is a right-continuous version of $\zeta(\cdot)$, then it equals to the right-derivative of $\varphi(\cdot)$. Moreover, if this version is continuous, then the fundamental theorem of calculus implies that it is the derivative of $\varphi(\cdot)$.

In particular, if some convergence theorem can be used in order to show that the right-continuity of $s\mapsto\xi(x,s)$ for every $x\in\mathbf{X}$ leads to a right-continuity of $\zeta(\cdot)$, then

$$\partial_+\varphi(t)=\zeta(t)\ \ ,\ \ \forall t\geq0$$

where $\partial_+$ is a notation for a right-derivative. For example, this happens when

$$|\xi(x,s)|\leq \psi(x) \ \ , \ \ \mu\text{-a.s.}$$

for some $\psi\in L_1(\mu)$.