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I am working with conditional expectations and am trying to derive a limit property.

Consider $(๐‘Œ_๐‘›)_{๐‘›\in\mathbb N}$ a sequence of square integrable random variables, that converge in $L^2$ to a square integrable random variable $Y$. Additionally assume that $\mathbb E[Y_n\mid Y]=Y_n$ (for example, $Y_n$ is a sequence of discrete quantizers of $Y$).

Is there anyway at all of guaranteeing that for some other $X$ in $L^2$, and for some form of convergence ($L^2, \mathbb P$ etc.) : $$\lim_{n\to+\infty}\mathbb E[X\mid Y_n]=\mathbb E[X\mid Y].$$

I am aware of the following similar question :

Conditional expectation of asymptotically independent random variables

but in that case $\mathbb E[Y_n\mid Y]=Y_n$ does not hold...

With a $L^2$-projection approach to conditional expectation, and with $Y_n$ converging in $L^2$ to $Y$, I keep thinking there must be some way of getting this to work... But maybe it just won't.

Thank you for any suggestions!

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  • $\begingroup$ Please always use MathJax for typesetting math. $\endgroup$ Commented Jul 16, 2020 at 6:54

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This is true if $\{\sigma(Y_n)\}$ is an increasing sequence of $\sigma$-fields s.t. $\sigma(Y_n)\nearrow \sigma(Y)$. In this case (see Theorem 4.6.8 on page 247 here), $$ \mathsf{E}[X\mid Y_n]\to \mathsf{E}[X\mid Y]\quad\text{a.s. and in $L^1$}. $$ For example, this holds when $Y_n=2^{-n}\lceil 2^nY\rceil$.

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  • $\begingroup$ Thanks ! Unfortunately, in my setting it is not the case... $\endgroup$
    – blreht
    Commented Jul 16, 2020 at 14:16

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