# Limit of conditional expectation

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

Consider $$(𝑌_𝑛)_{𝑛\in\mathbb N}$$ 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!

• Please always use MathJax for typesetting math. Jul 16, 2020 at 6:54

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 $$Y_n=2^{-n}\lceil 2^nY\rceil$$.