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Hi,

Could anyone give an example such that: $$Y_i \rightarrow Y_{\infty}, \text{a.s.},$$ and $Y_i$'s are uniformly integrable. But $\mathbb{E}(Y_i|\mathcal{G})$ does not converge a.s. to $\mathbb{E}(Y_{\infty}|\mathcal{G})$ for some sub-$\sigma$ algebra $\mathcal{G}$.

Thanks,

John

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Construct (on a suitable product of probability spaces) two independent sequences $(X_n)_{n\in\mathbb N}$ and $(Z_n)_{n\in\mathbb N}$ of positive random variables such that $E(X_n) \to 0$ but $X_n$ does not converge a.s., $E(Z_n)=1$ and for each $\omega\in\Omega$ there is $n\in\mathbb N$ such that $Z_m(\omega)=0$ for $m\ge n$. Then $Y_n=X_nZ_n$ converges to $0$ (even point-wise) and in $\mathscr L_1$ (because of the positivity and $E(Y_n)=E(X_n)E(Z_n)$) but for the $\sigma$-algebra $\mathscr G =\sigma (X_n:n\in\mathbb N)$ you can pull out the measurable factor to obtain $E(Y_n | \mathscr G)= X_n E(Z_n|\mathscr G) =X_n E(Z_n)=X_n$.

Although this is known it is not as well-known as it should be (a colleague once showed me a book on probability theory where the a.s. convergence of the conditional expectations was claimed).

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  • $\begingroup$ Do you know if there is a counterexample if it is additionally assumed that $(Y_n)$ is a martingale? $\endgroup$
    – Iosif Pinelis
    Commented Apr 28 at 14:52
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    $\begingroup$ I don't know. Uniformly integrable martingales are of the form $Y_n=E(Y_\infty|\mathscr F_n)$ but if there is no relation between the filtration $\mathscr F_n$ and $\mathscr G$ I do not see how this helps. $\endgroup$
    – Jochen Wengenroth
    Commented Apr 28 at 15:20
  • $\begingroup$ Thank you for your response. $\endgroup$
    – Iosif Pinelis
    Commented Apr 28 at 15:53

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