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?

  • $\begingroup$ @ Jonas : no it is not always true, but if you can interchange expectation and integral term then it is true so you only have to derive the conditions under which such operation is ok. Regards. $\endgroup$
    – TheBridge
    Commented Oct 22, 2012 at 20:58
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    $\begingroup$ Where could I find information about when such an operation is ok? $\endgroup$
    – jmbejara
    Commented Dec 4, 2013 at 20:04
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    $\begingroup$ 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. $\endgroup$
    – Did
    Commented Oct 26, 2016 at 19:42

2 Answers 2


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.

  • $\begingroup$ The uniform boundedness of $f$ seems to be a much too restrictive condition. $\endgroup$
    – Did
    Commented Oct 26, 2016 at 19:44
  • $\begingroup$ @Did yes, it's only a sufficient condition. In the lemma I showed, $Z$ is allowed to depend on $X$, so you can do much better, and if you use the Vitali convergence theorem you get the condition that the $f(X_t)$ are uniformly integrable. Do you know better results than this? $\endgroup$
    – jochen
    Commented Oct 26, 2016 at 20:25
  • $\begingroup$ @Did ah, yes, your Fubini solution is more elegant. $\endgroup$
    – jochen
    Commented Oct 27, 2016 at 8:03
  • $\begingroup$ @batman why not? You can have $X \in C\bigl( [0,\infty), \mathbb{R} \bigr)$ be the whole random path of the process $X$, and $g$ the function which integrates the path until time $t$. $\endgroup$
    – jochen
    Commented Aug 14, 2017 at 19:01
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    $\begingroup$ @jochen Well, Leibniz integral rule also applies to that case? I don't see how your comment answers my question. Am I wrong or the dominated convergence theorem is just a way to prove the Leibniz integral rule? In fact, the related Wikipedia page (https://en.wikipedia.org/wiki/Leibniz_integral_rule) seems to mention the dominated convergence theorem and, although I didn't look at the details, it looks similar to your answer. $\endgroup$
    – user168764
    Commented Jul 24, 2020 at 13:53

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


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


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