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Let

  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $E$ be a $\mathbb R$-Banach space
  • $X,X_n:\Omega\to E$ be $\mathcal A$-measurable for $n\in\mathbb N$ with $$\left\|X_n-X\right\|_E\xrightarrow{n\to\infty}0\;\;\;\text{in probability}\tag1$$
  • $\mathcal F\subseteq\mathcal A$ be a $\sigma$-algebra on $\Omega$

Assume $$\left\|X_n\right\|_E\le Y\;\;\;\text{almost surely for all }n\in\mathbb N\tag2$$ for some $Y\in\mathcal L^1(\operatorname P)$.


By $(1)$ and $(2)$ we're able to conclude $X,X_n\in\mathcal L^1(\operatorname P;E)$ for all $n\in\mathbb N$ with $$\left\|X_n-X\right\|_{L^1(\operatorname P;\:E)}\xrightarrow{n\to\infty}0\tag3$$ by a simple generalization of the dominated convergence theorem. Now, since $\operatorname E\left[\;\cdot\;\mid\mathcal F\right]\in\mathfrak L(\mathcal L^1(\operatorname P;E))$, $$\left\|\operatorname E\left[X_n\mid\mathcal F\right]-\operatorname E\left[X\mid\mathcal F\right]\right\|_{L^1(\operatorname P;\:E)}\xrightarrow{n\to\infty}0\tag4.$$

But are we able to show that $$\left\|\operatorname E\left[X_n\mid\mathcal F\right]-\operatorname E\left[X\mid\mathcal F\right]\right\|_{E}\xrightarrow{n\to\infty}0\tag5\;\;\;\text{almost surely}?$$

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  • $\begingroup$ No, consider the case $\cal{F}=\{\emptyset, \Omega\}$. $\endgroup$ – JGWang Jun 4 '18 at 3:45
  • $\begingroup$ You can always extract a subsequence which converges almost surely though. $\endgroup$ – Marzikill Jun 5 '18 at 19:47
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The typewriter sequence (Example 4) is a counter example. Let $\Omega=[0,1]$, $\mathcal A= \mathcal B([0,1])$, $P=\text{Leb}$, $X_n=\mathbf 1([k/m^{-1},(k+1)/m^{-1}] )$, $ k\le m-1$, $m\in\mathbb N$, for appropriate $n(m,k)$, $X=0$, $\mathcal F= \mathcal A $, $Y=1$. If $\mathcal F$ is finite, then the conditional expectation converges a.s. by $L^1$-convergence with $X_n=|X_n|$.

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  • $\begingroup$ I think that follows by $(4)$ (Markov inequality). $\endgroup$ – Rgkpdx Jun 4 '18 at 0:11

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