# Example of product of sequence of random variables that does not converge in distribution to the product of the limits

One result of Slutsky's theorem is that when $$X_{n}$$ is a sequence of random variables converging to a random variable $$X$$ and $$Y_{n}$$ a sequence of random variables converging to a constant $$c$$ then it holds that in the limit $$n \rightarrow \infty$$ we have that $$X_{n}Y_{n}\ {\xrightarrow {d}}\ cX$$.

But apparently this doesn't hold if the sequence $$Y_{n}$$ converges to a non-constant random variable $$Y$$, i.e. generally it does not hold that $$X_{n}Y_{n}\ {\xrightarrow {d}}\ XY.$$

Does anybody have a good intuitive counter-example for this case?

Edit: Both $$X_{n}$$ and $$Y_{n}$$ converge in distribution to their respective limits.

• What mode of convergence is $Y_n$ following in the latter relation that you want a counter example of? Commented May 20, 2023 at 11:31

I am assuming you are looking for a counterexample of $$X_n \overset{d}{\to} X$$, $$Y_n \overset{d}{\to} Y$$ but $$X_nY_n \overset{d}{\not\to} XY$$. If so, let $$X, Y$$ be i.i.d. $$N(0, 1)$$ random variable, then let $$X_n \equiv Y_n = X$$. Then the condition $$X_n \overset{d}{\to} X$$, $$Y_n \overset{d}{\to} Y$$ is clearly satisfied. However, $$X_nY_n = X^2 \sim \chi^2_1$$ obviously does not converge in distribution to $$XY$$. One way to see this is by noting $$P[X^2 \leq 0] = 0 \not\to P[XY \leq 0] = \frac{1}{2}$$.