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Nov 18, 2018 at 21:16 comment added Did @JosephGarvin You say you only assume the former -- but then you introduce $\lim\limits_{n\to\infty}X_n$, which assumes de facto the existence of the pointwise limit. So, no, not right at all, precisely for the reason explained in my previous comment.
Nov 18, 2018 at 21:14 comment added Joseph Garvin @Did the former. Expectation only depends on the CDF, which if $X_n \to X$ in distribution by definition means $X_n$ in the limit becomes a random variable with an equal CDF, right?
Nov 18, 2018 at 21:12 comment added Did @JosephGarvin Sorry but what is your assumption? That $X_n\to X$ in distribution or that $X_n\to X$ almost surely? 'Cause nobody assumed the latter while you seem to do...
Nov 18, 2018 at 20:53 comment added Joseph Garvin I see. Am I right to think $E(\lim_{n\to\infty}X_n) = E(X)$ though? Since two variables with the same CDF must have the same expectation.
Nov 18, 2018 at 10:13 comment added Did @JosephGarvin Of course there is, replace $2^n$ by $7n$ in the example of this answer.
Nov 18, 2018 at 2:53 comment added Joseph Garvin Answering my own question: $E(X_n) = (1/n)2^n + (1-1/n)0 = (1/n)2^n$. Then taking the limit the numerator clearly grows faster, so the expectation doesn't exist. This begs the question though if there is example where it does exist but still isn't equal?
Nov 18, 2018 at 2:46 comment added Joseph Garvin So in the limit $X_n$ becomes a point mass at 0, so $\lim_{n\to\infty} E(X_n) = 0$. Then $E(X) = 0$. I don't see a problem?
Oct 11, 2015 at 23:57 comment added Did @WittawatJ. About what? Please explain your problem.
Oct 10, 2015 at 14:25 comment added wij Could you please give a bit more explanation?
Jun 3, 2012 at 16:20 vote accept LeafGlowPath
Jun 3, 2012 at 15:16 history answered Did CC BY-SA 3.0