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There are at least 2 places in Wikipedia saying that $X_n$ converges to $X$ in mean in and only if $X_n$ converges to $X$ in probability and $X_n$ is uniformly integrable. See the following link for example:

However, I found a reference in Billingsley's book saying that we only need convergence in distribution along with uniform integrability to imply the convergence in mean. Is Wikipedia wrong, or am I missing something here?


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The Wikipedia statement is right, and the statement which you attribute to Billingsley is wrong, in some sense inherently so. Convergence in distribution is a notion of convergence of measures. A sequence of random variables can converge in distribution even if they live on totally different probability spaces, and their joint distribution never enters. In contrast, convergence in mean (i.e. $L^1$) only makes sense for a sequence of random variables defined on the same probability space (which is to say, you have to know all their joint distributions). So it doesn't make sense for the former to imply the latter.

Even if your random variables happen to be defined on the same measure space, it fails in general, as Jacob Katz pointed out. For a very explicit counterexample, let $\lbrace X_i \rbrace_{i=1}^\infty$ be iid Bernoulli. Trivially $X_i$ converges in distribution, and they are uniformly integrable because they are uniformly bounded. But $E|X_i - X_j| = 1/2$ for all $i \ne j$, so the $X_i$ are not Cauchy in mean.

It's possible that Billingsley is in error, but I think it's more likely that you have missed some context. Can you post the relevant text?

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Thanks, Nate. Yes, I think I was confused between convergence in mean and convergence of mean, as Jakob pointed out below. Billingsley's saying the later. BTW, the text is "Weak Convergence of Measures: Applications in Probability" by Billingsley, 1971. – cresmoon Aug 21 '10 at 18:33

Slight misunderstanding in KalEl's answer: convergence in mean - which is convergence in the $L^1$ norm associated with the probability measure, and as such is a feature of the sequence of random variables itself rather than only the sequence of distributions - isn't the same thing as convergence of mean.

Anyway, no, convergence in distribution as such isn't enough, even given uniform integrability - for instance, take two identically distributed nonzero random variables with a.s. disjoint support and alternate between them for your sequence, say (take $[0,1]^2$ with the usual Lebesgue measure for your probability space and use the indicator functions of the top and bottom half, for example), if KalEl's suggestion of an iid sequence doesn't immediately help you.

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Thanks, Jakob. I had the same misunderstanding as KalEl's: I thought convergence in mean is the same as convergence of mean. – cresmoon Aug 21 '10 at 18:29

I don't have Billingsley's book here, but I guess what he says is that uniform integrability and convergence in distribution imply convergence of the means (and not in mean), i.e. $E[X_n]$ converges to $E[X]$.

One simple (though not truely elementary) way to see it is to use Skorokhod's theorem stating that convergence in distribution is equivalent to almost-sure convergence of copies of the random variables in some abstract probability state. Then we just apply the usual Vitali theorem about uniform integrability for these copies. It gives convergence in mean of the copies of $X_n$ to the copy of $X$, and therefore convergence of the mean of (the copy of) $X_n$ to that of (the copy of) $X$. I put parentheses since the mean only depends on the law and hence is the same for $X_n$ or a copy of it. This is the result.

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Convergence of mean (or any other feature of the distribution for that matter) does not imply weak convergence of the random variables.

Take the example of an infinite sequence of iid random variables of any non degenerate distribution and see what is going wrong.

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Are you saying that Wikipedia is wrong? Otherwise, convergence in mean implies convergence in probability, and it in turn will imply convergence in distribution, right? – cresmoon Aug 21 '10 at 10:29
I take it back, I think I misunderstood as Jakob pointed out. – KalEl Aug 21 '10 at 14:15

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