When can we interchange the order of limits of the expected value of products of random variables? Suppose that the sequence of random variables $X_n$ converges to a limiting random variable $X$, and let $Y$ be an arbitrary random variable.
Is there any specific mode of convergence of the sequence $X_n$ that guarantees that
$$\lim_{n \to \infty} E[X_n Y] = E[X Y]?
$$
I'm pretty sure that convergence in distribution is not good enough, but I'm not sure about (e.g.) convergence in probability or convergence a.s.
What about if there are multiple sequences (i.e. $E[X_n Y_n]\stackrel{?}{\to} E[X Y]$), or if $X$ or $Y$ are constant or otherwise constrained? (I know that if $Y$ is constant, then it suffices for $X_n$ to converge in mean.)
 A: As it was mentioned by Dasherman, we can get some conditions, using Hölder's inequality.
Here is the result:
Statement 1: if $p, q \ge 1$ where $\frac1{p} + \frac1{q} = 1$,$X_n \to X$ in $L_p$ and $Y$ in $L_q$ then
$$ |E(X_n - X)Y| \le ||(X_n-X) Y ||_1 \le ||X_n -X||_p || Y||_q \to 0.$$
What may we say if conditions of statement 1 don't hold true?
Notice that $X_n$, $X$ and $Y$ should be defined in common probability space. Otherwise $EX_nY$ or $EXY$ are not defined.
Suppose that $X_n \to X$ in probability. Hence $(X_n,Y) \to (X,Y)$ in probability and $X_n Y \to X Y $  in probability.
Statement 2a:
If  $X_n \to X$ in probability and $|X_n Y| \le \xi$ for some $\xi$ such that $E\xi < \infty$ then $EX_n Y \to EXY$. Moreover, $E|X_n Y - XY| \to 0$.
Proof: as $X_n Y \to X Y $  in probab. then Statement 2 follows from Dominated convergence theorem.
With the help of notion of uniform integrability assumptions may be weakened.
Statement 2b: If  $X_n \to X$ in probability and if $\{X_n Y\}_{n\ge 1}$ is uniformly integrable class of r.v. then $EX_n Y \to EXY$. Moreover, $E|X_n Y - XY| \to 0$.
Statement 2a follows from statement 2b.
What may we say if we have only weak convergence?
Statement 3a: If  $X_nY \to XY$ in distribution and if $\{X_n Y\}_{n\ge 1}$ is uniformly integrable class of r.v. then $EX_n Y \to EXY$.
Proof. By Skorokhod's representation theorem there are $Z_n$ and $Z$ such that $Z_n \to Z$ a.s., distribution of $Z_n$ coinsides with distribution of $X_n Y$ and distribution of $Z$ coinsides with distribution of $X Y$. Obviously, $Z_n$ is uniformly integrable class of r.v. Thus $EX_n Y = EZ_n \to  EZ = EXY$.
First corollary 3a. If  $X_nY \to XY$ in distribution and  $\sup_n E G(|X_n Y|) < \infty$ for some non-negative increasing convex function $G(t)$ such that $\frac{G(t)}t \to \infty$ as $t \to \infty$ then $EX_n Y \to EXY$.
Proof: Corollary 3a follows immediately form de la Vallée-Poussin theorem and Statement 3a.
Second corollary 3a. If  $X_nY \to XY$ in distribution and  $\sup_n E |X_n Y|^{1 + \varepsilon} < \infty$ for some $\epsilon > 0$ then $EX_n Y \to EXY$.
Proof: follows immediately form First corollary 3a.
Statement 3b: If  $X_n \to X$ in distribution, $P(Y = const) =1 $  and if $\{X_n \}_{n\ge 1}$ is uniformly integrable class of r.v. then $EX_n Y \to EXY$.
Proof. As $X_n \to X$ in distribution and $Y = const$ a.s. then $X_n Y \to X Y$ and statement 3b follows from statement 3a.
First corollary 3b. If  $X_n \to X$ in distribution, $P(Y = const) =1 $  and  $\sup_n E G(|X_n|)< \infty$ for some non-negative increasing convex function $G(t)$ such that $\frac{G(t)}t \to \infty$ as $t \to \infty$ then $EX_n Y \to EXY$.
Proof: similarly to proof of First Corollary 3a.
Second corollary of 3b. If   $X_n \to X$ in distribution, $P(Y = const) =1 $  and  $\sup_n E |X_n|^{1 + \epsilon} < \infty$ for some $\epsilon > 0$ then $EX_n Y \to EXY$.
Proof: similarly to proof of Second Corollary 3  a.
Conclusion.
If $X_n \to X$ in $L_p$  then statement 1 works.
If $X_n \to X$ in probability (or, moreover, a.s.) then statements 2a and 2b work.
If $X_n \to X$ in distribution, last statements work.
A: Let $||\cdot||_p$ denote the $L^p$-norm. We want $X_nY\to XY$ in $L^1$. Note that
$$||X_nY - XY||_1 = ||(X_n-X)Y||_1 \leq ||X_n-X||_2 ||Y||_2,$$
by the Cauchy-Schwarz inequality. So $X_n\to X$ in $L^2$ suffices, if $||Y||_2<\infty$.
Using Hölder's inequality, you can also get different conditions (stricter for $X_n$ but weaker for $Y$ or vice versa).
