Convergence of conditional expectations $E[X|Y_i]$.

Question

Let $X$ and $Y$ be random variables and suppose that the sequence $Y_i$ converges $L^1$ to $Y$: $$Y_i\overset{L_1}{\longrightarrow}Y$$ Does $E[X|Y_i]$ also converge $L^1$ to $E[X|Y]$?: $$E[X\mid Y_i]\overset{L_1}{\longrightarrow} E[X\mid Y]$$

Answer 1

To answer the specific question here: no.

For example, with $X\in L^1$, define $Y_i=i^{-1}X$.

Then $Y_i\to 0$ in $L^1$, but $$E[X|Y_i]=E[X|i^{-1}X]=X\not\to E[X|0]=E[X]$$ except in the case when all the mass of $X$ is at a single point.

More generally: The fact that $Y_i$ converge to $Y$ in $L^1$ does not tell you anything useful about whether $E[X|Y_i]\to E[X|Y]$. The random variables $Y_i$, $Y$ only impact on the truth of the latter statement through their associated $\sigma$-algebras. That's why Lévy's Upward Theorem, the most relevant theorem to answering this question, is formulated in terms of a filtration of $\sigma$-algebras.

Lévy's Upward Theorem. Suppose $X\in L^1$ and $(\mathcal{F}_i)_{i\geq 1}$ is a filtration (increasing sequence) of $\sigma$-algebras. Then $M_i:=E[X|\mathcal{F}_i]$ is a uniformly integrable martingale and $M_i\to E[X|\mathcal{F}_\infty]$ almost surely and in $L^1$, where $\mathcal{F}_\infty=\sigma(\cup_{i\geq 1}\mathcal{F}_i)$.

So, if $\sigma(Y_i)\uparrow\sigma(Y)$ then, by this theorem, we can conclude that $E[X|Y_i]\to E[X|Y]$ in $L^1$.