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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.

Thank you in advance.

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    $\begingroup$ 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 $L^2$ $\endgroup$
    – Andrew
    Apr 20, 2023 at 15:56
  • $\begingroup$ 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. $\endgroup$
    – polemical
    Apr 20, 2023 at 16:03
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    $\begingroup$ So then you can apply DCT, no? Now your derivative is dominated by $c|Y-t|^2$, which is inetgrable. $\endgroup$
    – Andrew
    Apr 20, 2023 at 16:25
  • $\begingroup$ Sadly no, since DCT requires identifying a dominating function $g(x)$ s.t. $$ \left\lvert \frac{\partial f(x,t)}{\partial t} \right\rvert < g(x) ~~ \forall t ~~.$$ $\endgroup$
    – polemical
    Apr 20, 2023 at 16:45
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    $\begingroup$ Yes, you only need a local bound. The difference quotient is equal to the derivative by the mean value theorem. As long as you have a bound in some (non degenerate) neighborhood, no matter how small, you are good. $\endgroup$
    – Andrew
    Apr 20, 2023 at 17:05

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