Given a probability space $(\Omega,\mathcal F,\mathbf P)$. Let $\{X_n\}, \{Y_n\}$ be two sequences of real-valued random variables on $(\Omega,\mathcal F,\mathbf P)$, with $X_n\to X$ in distribution and $Y_n\to y$ in probability as $n\to\infty$, where $y\in\mathbb R$ is a constant. Let $f:\mathbb R\to\mathbb R$ be a continuous function. The question is if it's enough to show that \begin{equation} \mathbf E[f(X_n)Y_n]\to\mathbf E[f(X)]y, \quad n\to\infty? \end{equation}
What I know is that $f(X_n)\to f(X)$ in distribution as $n\to\infty$.
Do I need some uniform integrability assumptions? For instant,
- if the sequence $\{f(X_n)\}$ is uniformly integrable, then $\mathbf Ef(X_n)\to\mathbf Ef(X)$;
- if the sequence $\{Y_n\}$ is uniformly integrable, then $Y_n\to y$ in $L^1(\Omega,\mathcal F,\mathbf P)$.
But I don't know what to do next. I have no idea then.
Any hints or comments will be appreciated! TIA!
The given assumptions are not enough: in the case where $Y_n=1$ for all $n$, we have to be sure that the random variables $f\left(X_n\right)$ and $f\left(X\right)$ are integrable, which may not be the case. Even if they are, when $f$ is non-negative, we need $$ \tag{A.1} \left\{f\left(X_n\right),n\geqslant 1\right\}\mbox{ is uniformly integrable}. $$ So in order to avoid more or less standard counter-examples, assumption (A.1) is reasonable. Under this assumption, we can assume that $y=0$.
In order to avoid the counter-example where $X_n\to X$ and $Y_n\to 0$ almost surely, we need the assumption $$ \tag{A.2} \left\{f\left(X_n\right)Y_n,n\geqslant 1\right\}\mbox{ is uniformly integrable}. $$ Assume that (A.1) and (A.2) hold. We can assume that $Y_n\to 0$ almost surely (otherwise pass to subsequences) and use Egoroff's theorem and get $\mathbf E[f(X_n)Y_n]\to 0, \quad n\to\infty $. Indeed, fix a positive $\varepsilon$. By (A.2), there exists a positive $\delta$ such that if $\Pr(A)\lt \delta$, then $\mathbb E\left[\left\lvert f(X_n)Y_n \mathbf 1_A\right\rvert\right]\lt \varepsilon$. By Egoroof's theorem, there exists a set $E$ such that $a_n:=\sup_{\omega\in E}\left\lvert Y_n(\omega)\right\rvert\to 0$ and the measure of $\Omega\setminus E$ do not exceed $\delta$. Then $$ \mathbb E\left[\left\lvert f(X_n)Y_n \right\rvert\right]\leqslant a_n\mathbb E\left[\left\lvert f(X_n) \right\rvert\right]+\mathbb E\left[\left\lvert f(X_n)Y_n\mathbf 1_{\Omega\setminus E} \right\rvert\right]\leqslant a_n\sup_{k\geqslant 1}\mathbb E\left[\left\lvert f(X_k) \right\rvert\right]+\varepsilon. $$