Convergence of $P(A|X_n)$

Question

I’m sure this is a basic question, but if you have a sequence of r.v. $X_n$ such that they have a distribution $F_n$ so that $X_n$ converge to $X$ with distribution $F$. When am I avail to write $lim_{n \to \infty} P(A|X_n) = P(A|X)$?

Answer 1

This is generally not true. Convergence in distribution is going to be much too weak. Even almost sure convergence is not enough; let $A$ be some event with probability $1/2$ and $X_n = \frac{1}{n} 1_A$. Then $P(A \mid X_n) = 1_A$, since $A \in \sigma(X_n)$, but $X_n \to 0$ almost surely (in fact, uniformly), and $P(A \mid 0) = P(A) = 1/2$.

If you could get that the generated $\sigma$-fields of the $X_n$ were increasing to that of $X$, in the sense that $\sigma(X_1) \subset \sigma(X_2) \subset \dots$ and $\sigma(X) = \sigma(X_1, X_2, \dots)$, then your conclusion would hold by the martingale convergence theorem. But none of the usual modes of convergence will imply that condition.