1
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.