# Show $X_n \to 0$ in probability

I am asked to show :

Let $$X$$ be a real-valued random variable on $$(\Omega, F , P)$$ and define

$$X_n(\omega) = nX(\omega)$$ if $$n and $$0$$ if else. Prove that $$X_n \to 0$$ in probability.

• (+1) Welcome to Stats.SE. You might be interested to know that you can use math typesetting via Mathjax. More information: math.meta.stackexchange.com/questions/5020/… – Sycorax Jan 14 at 19:41
• Are you sure you stated the problem as intended? By writing "$(\Omega,F,P),$" you suggest $\Omega$ is an abstract set, but in writing "if $n\lt w\le n+1$" you imply the positive real numbers are in $\Omega.$ – whuber Jan 14 at 19:47
• That's not the issue: the issue is that the argument of $X,$ which you call "w," must be a real value for this problem statement to make sense. At the very least, your book/exam/professor is making assumptions or restrictions that you haven't put in evidence here. – whuber Jan 14 at 19:53
• I don't see any use of indicator functions in your question. You might benefit from drawing graphs of some of the $X_n$ and contemplating what happens to the chances that their values exceed zero. What I wonder is whether the clause "$n\lt w\le n+1$" should instead read "$n\lt X(w)\le n+1.$" That would make sense for any abstract probability space; it would imply the same conclusion; but would require a different solution. – whuber Jan 14 at 20:09
• Please, then, edit your post to reflect that and add the self-study tag after reading information about it. – whuber Jan 14 at 21:18

## 1 Answer

I think the easiest way is to show that $$X_n(\omega)\stackrel{as}{\to}0$$ and note that almost sure convergence implies convergence in probability.

$$X_n(\omega)=0$$ for all $$n>X(\omega)$$, so for every fixed $$\omega$$, $$X_n(\omega)\to 0$$.

Therefore $$X_n(\omega)\stackrel{as}{\to}0$$, and it is a standard fact that this implies $$X_n(\omega)\stackrel{p}{\to}0$$