# Is all random variable finite (almost surely)?

I feel like what I've known are conflicting with each other, so I'd like to post it.

When some prove that if $$X_n \to X$$ in probability and $$Y_n \to Y$$ in probability, then $$X_nY_n \to XY$$ in probability (e.g. here), there is an argument $$\tag{1} P(|X|>M) \to 0, \quad \text{as } M\to \infty.$$

In fact, (1) is equivalent to $$P(|X|<\infty) = 1$$, since by continuity from above (in the probability space) in (*), we can derive $$P(|X|=\infty) = P(\cap_{n\ge 1}\{|X| \ge n\}) \overset{(*)}{=} \lim_{n\to \infty} P(|X|>n).$$

My question is that why (1) holds in this case? (I don't think this holds in general; for example, a sum of independent random variables with finite mean ($$>0$$))

• What would $X_n \to X$ in probability mean if $X$ were infinite somewhere? In particular, how would you interpret $|X_n-X|$? – kccu Apr 25 at 1:07
• @kccu so you mean that $|X|<\infty$ a.e. is implicitly assumed in the statement $X_n \to X$ in probability. – inmybrain Apr 25 at 1:16
• It's stronger than that. To say $X$ is a real random variable means it is finite a.e.; see en.wikipedia.org/wiki/… . And similarly for $X_n$ and so on. Equation (1) holds for all probability measures on the reals. – kimchi lover Apr 25 at 1:19
• @kimchilover You right there, but I'm thinking of the case when $X$ can be a limit of random variables. Isn't this case problematic here? – inmybrain Apr 25 at 1:24

If the $$X_n$$ are finite a.e. and $$|X_n \to X|$$ in probability, then this implies $$X$$ is finite a.e.
If not, then $$P(|X|=\infty )>0$$. Then for any $$\epsilon$$ and any $$n$$, \begin{align*} P(|X_n-X|>\epsilon) \geq P(|X_n|<\infty,|X|=\infty) = P(|X|=\infty) \end{align*} since $$P(|X_n|<\infty)=1$$. Then $$P(|X_n-X|>\epsilon)$$ cannot go to $$0$$ as $$n \to \infty$$.
• You are right, but your assumption is $X_n$ is finite for any $n$. How about the other cases? – inmybrain Apr 26 at 5:21