# Show the product of a sequence of random variables does not converge in mean square.

I have a question regarding mean square convergence in probability theory.

Given random variable $$X$$ such that E$$[X^2] < \infty$$ and E$$[X^4] = \infty$$. We are further told that given sequences of random variables $$X_n$$ and $$Y_n$$ = $$\frac{1}{n}X$$,

show that the mean square convergence of $$X_n$$ and $$Y_n = 0$$; that is ms-$$\lim_{n\to\infty} X_n = 0$$, ms-$$\lim_{n\to\infty} Y_n = 0$$ but ms-$$\lim_{n\to\infty} X_nY_n = \infty$$

I had no issue showing ms-$$\lim_{n\to\infty} X_n = 0$$ and ms-$$\lim_{n\to\infty} Y_n = 0$$. However, I have difficulties in formulating the proof for the last equality.

An idea I had was: Suppose for contradiction that ms-$$\lim_{n\to\infty} X_nY_n = f(X) < \infty$$. This can be rewritten as $$\lim_{n\to\infty}$$ $$E\left[\left(\frac{X^2}{n^2} - f(X)\right)^2\right] = 0$$.
However, upon expansion, this leads to a contradiction that $$\lim_{n\to\infty}$$ $$E\left[\left(\frac{X^2}{n^2} - f(X)\right)^2\right] \neq 0$$, using the fact that E$$\left[X^4\right] = \infty$$.

• I think we're allowed to expand the expression above since both sequences of random variables converge to $$0$$ in mean square.

• I am unsure if my argument is flawed.

Any help and guidance would be much appreciated! Sorry about the bad typesetting at the previous paragraph, still learning LaTex.

Thanks!