Timeline for derivative of conditional expectation
Current License: CC BY-SA 3.0
5 events
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| Jan 8, 2013 at 14:49 | history | edited | ern | CC BY-SA 3.0 |
edited references.
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| Dec 12, 2012 at 20:04 | comment | added | ern | 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 $P(\cup_x\cup_{x_n\rightarrow x} A_{\{x_n\},x} )=0$. | |
| Dec 7, 2012 at 14:17 | comment | added | Dan |
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/…
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| Dec 5, 2012 at 20:09 | history | edited | ern | CC BY-SA 3.0 |
deleted 74 characters in body; deleted 1 characters in body; added 28 characters in body; added 6 characters in body
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| Dec 5, 2012 at 19:59 | history | asked | ern | CC BY-SA 3.0 |