Is the derivative of the expected value equal to the expected value of the derivative

Question

Is the derivative of the expected value equal to the expected value of the derivative of a random function? That is, is $\partial E[f(.)]/\partial x = E[\partial f(.)/\partial x]$?

Answer 1

Suppose you have a function $f(x,\omega)$ (here $\omega$ is the sampling variable) and are interested in $\frac{\partial (Ef)}{\partial x}$. This is

$$\lim_{h \to 0} \int_\Omega \frac{f(x+h,\omega)-f(x,\omega)}{h} dP(\omega).$$

The pointwise limit of the integrand is $\frac{\partial f}{\partial x}$, so you just want to pass the limit under the integral sign. There are a number of theorems that can be used to do this. The most practical is the Lebesgue dominated convergence theorem. This says that you may pass the limit under the integral sign if (in this context) there is some $g = g(\omega) \geq 0$ with $Eg < \infty$ so that $\left | \frac{f(x+h,\omega)-f(x,\omega)}{h} \right | \leq g(\omega)$ for any $h$ in a neighborhood of zero. So in particular you can do it if $\frac{\partial f}{\partial x}$ is bounded, because then you can take $g=2|f'|$.

The most general is the Vitali convergence theorem, which is the most general in the sense that its hypotheses are necessary and sufficient.