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 ($>0$))

  • $\begingroup$ What would $X_n \to X$ in probability mean if $X$ were infinite somewhere? In particular, how would you interpret $|X_n-X|$? $\endgroup$ – kccu Apr 25 at 1:07
  • $\begingroup$ @kccu so you mean that $|X|<\infty$ a.e. is implicitly assumed in the statement $X_n \to X$ in probability. $\endgroup$ – inmybrain Apr 25 at 1:16
  • $\begingroup$ 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. $\endgroup$ – kimchi lover Apr 25 at 1:19
  • $\begingroup$ @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? $\endgroup$ – 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 $n \to \infty$.

  • $\begingroup$ You are right, but your assumption is $X_n$ is finite for any $n$. How about the other cases? $\endgroup$ – inmybrain Apr 26 at 5:21
  • $\begingroup$ As kimchi lover noted, that is part of the definition of random variable. $\endgroup$ – kccu Apr 26 at 13:32

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