17
$\begingroup$

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

$\endgroup$

1 Answer 1

18
$\begingroup$

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).

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.