$|X_k| \leq Y \in L^1,$ if $X_k \Rightarrow X$ then $E[X_k|\mathcal{F}] \Rightarrow E[X|\mathcal{F}]$

Question

Consider on a probability space $(\Omega,\mathcal{W},P),$ a sequence $(X_k)_k$ of random variable such that $|X_k| \leq Y \in L^1,$ consider a sub $\sigma$-algebra $\mathcal{F}.$

• If $X_k$ converges a.s to $X$ then $E[X_k|\mathcal{F}]$ converges a.s to $E[X|\mathcal{F}]$
• If $X_k$ converges in probability to $X$ then $E[X_k|\mathcal{F}]$ converges in probability to $E[X|\mathcal{F}].$

Does this hold for weak convergence? In other word, is it true that if $X_k$ converges in distribution to $X$ then $E[X_k|\mathcal{F}]$ converges in distribution to $E[X|\mathcal{F}]$?

Answer 1

No. Let $X, X_k$ be independent non-constant random variables drawn from the same distribution and let $\mathcal{F} = \sigma (X)$. Then $X_k$ converge in distribution to $X$; also, $E[X | \sigma (X)] = X$, but, by independence, $E[X_k | \sigma (X)] = E[X_k]$, which is a constant.