# derivative of conditional expectation

Suppose $H:\Omega\times X\mapsto Y$ for some borel subset $X\subset \mathbf{R}$, Euclidean space $Y$, and probability space $(\Omega, \mathcal{F},P)$. Further suppose that $H$ is $C^1$ for each fixed $\omega\in\Omega$, and measurable for each $x\in X$.

Let $\mathcal G\subset \mathcal{F}$ be a sub-$\sigma$-algbebra, and let $\hat E$ denote the regular conditional expectation in the sense of Dynkin and Evstigneev or the conditional expectation of random integrands in the sense of Castaing and Ezzaki.

How do we know

1) If $\frac{\partial H}{\partial x}\hat E[H(\cdot, x)|\mathcal G]$ exists for each $x$ a.s.

2)When is $$\frac{\partial H}{\partial x}\hat E[H(\cdot, x)|\mathcal G]=\hat E[\frac{\partial H}{\partial x}H(\cdot, x)|\mathcal G]$$ a.s.

3) When is $$\frac{\partial H}{\partial x}\hat E[H(\cdot, x)|\mathcal G]_{x=u(\cdot)}=\hat E[\frac{\partial H}{\partial x}H(\cdot, u(\cdot))|\mathcal G]$$ a.s whenever $u:\Omega\rightarrow X$ is $\mathcal G$-measurable?

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If the random variables $\{|\partial H/ \partial x(\cdot,x)| : x \in X\}$ are uniformly integrable, then this follows quickly from the dominated convergence theorem for conditional expectations. The details are essentially the same as here: math.stackexchange.com/questions/94628/… – Dan Dec 7 '12 at 14:17
I am not convinced. Say that we consider continuity of $x↦E[H(x)|G]$ instead of differentiability, and $H(x)$ uniformly integrable. Then we know that if $x_n\rightarrow x$ then $E[H(x_n)|\mathcal{G}]\rightarrow E[H(x)|\mathcal{G}]$, P-a.s. Let $A_{x_n\rightarrow x}$ be the set of $\omega$ where this convergence does not hold. Now, there is an uncountable number of sequences converging to each $x$, and an uncountable number of $x$′s, and to have $x\mapsto E[H(x)|\mathcal G](\omega)$ contiuous for P−a.e.$\omega$, we also need that $P(\cup_x\cup_{x_n\rightarrow x} A_{\{x_n\},x} )=0$. – ern Dec 12 '12 at 20:04