# Interchanging Derivative and Expectation without Dominated Convergence

Let $$Y$$ be a continuous random variable in $$\mathcal{L}^p(\mathbb{R}, \mathcal{B})$$ for every $$p\in[1,\infty)$$ and let $$f(x,t)$$ be a continuous function, $$f: \mathbb{R}^2 \rightarrow \mathbb{R}$$, s.t. $$\frac{\partial f}{\partial t}(x,t)$$ is continuous and, for some $$c>0$$, $$\left \lvert \frac{\partial f}{\partial t}(x,t) \right\rvert < c \left \lvert x-t \right\rvert^2 ~.$$ Is there any condition or theorem that allows interchanging the derivative and the integral as follows $$\frac{\partial }{\partial t} \, \mathbb{E} \left[ f(Y,t) \right] \overset{?}{=} \mathbb{E} \left[ \frac{\partial }{\partial t} \, f(Y,t) \right] ~~ ?$$ The well-known dominated convergence theorem cannot be exploited because of the absence of a uniform bound on the derivative. I thought about Vitali's convergence theorem but, again, the uniform integrability hypothesis seems not to be satisfiable.

• This doesn't seem hopeful, at least to me. Your condition on the partial of $f$ won't even guarantee the expectation to be well-defined. $Y$ is only $L^1$ but the partial based on your condition means that $Y$ needs to be $L^2$ Apr 20, 2023 at 15:56
• Sure, I tried to make the statement as simple as possible, but I clearly missed the crucial point you point out. Actually, in my framework $Y$ is a Gaussian random variable, so moments of all orders exist and are well defined. I proceed to edit the question. Thank you. Apr 20, 2023 at 16:03
• So then you can apply DCT, no? Now your derivative is dominated by $c|Y-t|^2$, which is inetgrable. Apr 20, 2023 at 16:25
• Sadly no, since DCT requires identifying a dominating function $g(x)$ s.t. $$\left\lvert \frac{\partial f(x,t)}{\partial t} \right\rvert < g(x) ~~ \forall t ~~.$$ Apr 20, 2023 at 16:45