Convergence of random variable times function: $nX_n$

Question

If $X_n\xrightarrow[]{p}X$, can I prove that $n(X_n-X)\xrightarrow[]{p}0$ if $X$ is a natural number.

I know that if $Y_n$ is bounded in probability $Y_nX_n\xrightarrow[]{p}0$, or that if $n$ is a constant it works too, but I am not sure about the above.

Thank you all.

Answer 1

Your proposition is incorrect. What happens if $X_n = \frac{1}{n}$ ?

Answer 2

If $X_n$ can be anything then $X$ being natural number changes nothing, and your proposition would still be incorrect. But if you mean $X_n$ is also a natural number, then yes. Proof:

When $\epsilon<1$, $Pr(|X_n - X|<\epsilon)=Pr(X_n=X)$, so for any $\epsilon<1$(the $\epsilon\geq1$ scenario is trivial), $lim_{n\to\infty}Pr(X_n=X) = lim_{n\to\infty}Pr(|X_n-X|<\epsilon) = 1$. The last "$=$" is from the definition of Convergence in Probablity.

Therefore, for any $\epsilon$, $lim_{n\to\infty}Pr(|n(X_n-X)|<\epsilon) \geq lim_{n\to\infty}Pr(n(X_n-X)=0) = lim_{n\to\infty}Pr(X_n=X) = 1$. So $lim_{n\to\infty}Pr(|n(X_n-X)|<\epsilon) = 1. \space\space\square.$