# Random variables $X_n\xrightarrow{p}X, Y_n\xrightarrow{d}Y \implies X_nY_n \xrightarrow{d}XY$

Suppose we have $$X_n\xrightarrow{p}X$$ and $$Y_n\xrightarrow{d}Y$$. I would like to show that the product $$X_nY_n$$ converges in distribution to $$XY$$.

I'm trying to decompose and bound the sequence $$\mathbb{P}(X_nY_n)$$ by sequences converging to $$\Bbb P(XY.

I'm thinking about something similar to $$\mathbb{P}(XY\delta)+?\leq\mathbb{P}(X_nY_n\delta)+?$$ but I can't see how to go about it exactly. Some hints would be much appreciated.

• The property that you intend to prove is wrong in general. Let $X_n=X$ for each $n$ and $X$ is Bernoulli distributed with parameter $p=0.5$. Let $Y_{2k-1}=X$ and $Y_{2k}=1-X$. Then $Y_n \xrightarrow{d} Y=X$ but the sequence $X_nY_n$ does not converge in distribution.
– NCh
Feb 24, 2019 at 14:21
• I agree with @Nch. Note that the assertion is true if we assume additionally that $X=c$ with probability $1$ for some constant $c$, see this question.
– saz
Feb 24, 2019 at 14:36

Forgetting for a while the convergence by letting $$X_n$$ independent of $$n$$ and $$Y_n$$ identically distributed, the question reduces to the following: if $$Y\overset{\mbox{law}}{=}Y'$$, are $$XY$$ and $$XY'$$ equal in distribution? As suggested by NCh, the answer is no: if $$X$$ takes the values $$0$$ and $$1$$ with probability $$1/2$$, $$Y=X$$ and $$Y'=1-X$$, then $$XY=X^2$$ has the same law as $$X$$ but $$XY'=X(1-X)=0$$.
In the context of the opening post, the best we can say is that the sequence $$\left(X_nY_n\right)_{n\geqslant 1}$$ is tight hence admits a convergent subsequence.