Changing my question in light of Dan's answer. Thanks, Dan.

Consider a sequence of real random variables $X_i$ bounded in $L_1$, that is $\mathbb E\left|X_i\right|\leq M$ for all $i$. Suppose that they converge in distribution to $X$ (which by Fatou's lemma will also be in the $M$-ball of $L_1$). This is not enough to say that $\mathbb EX_i\to\mathbb EX$. We need uniform integrability (it is necessary for example if the RVs are nonnegative). http://en.wikipedia.org/wiki/Uniform_integrability

Boundedness in $L_1$ is not enough for uniform integrability. For example, the nonnegative RVs $X_i$ with CDF $(1-1/i)$ on $[0,i)$ and $1$ on $[i,\infty)$ are all in the 1-ball of $L_1$ but are not uniformly integrable. But their CDFs also do not converge uniformly.

So suppose we have $X_i$ bounded in $L_1$ converging in distribution to $X$ but also the CDFs converge uniformly $$\left|\left|F_i-F\right|\right|_\infty\to0.$$ Is it the case that $X_i$ are uniformly integrable and/or $\mathbb EX_i\to\mathbb EX$?


You need uniform integrability. To change notation a bit, suppose $X_i$ and $X$ have distributions $F_i$ and $F$, respectively. Suppose for now only that $F_i(x) \rightarrow F(x)$ at each continuity point $x$ of $F$. The following are from Billingsley's Convergence of Probability Measures, Theorems 3.5 and 3.6: (1) If $g(X_n)$ are uniformly integrable, then $E[g(X_n)] \rightarrow E[g(X)]$. (2) If $g \ge 0$, then the converse holds.

I doubt the uniform convergence or uniform continuity can gain you anything. For example, if $X$ has a Cauchy distribution and $X_n = (X \wedge n) \vee (-n)$, then the CDFs of $X_n$ converge uniformly to that of $X$ and $E[X_n] = 0$, but $E[X]$ fails to exist.

  • $\begingroup$ Thanks for your answer, Dan. Indeed I see from the converse (Thm 3.6) that I will definitely need uniform integrability for sure. But maybe uniform convergence will help get that. Suppose $g(X_i)$, $g(X)$ are $L_1$ bounded. I.e. $\mathbb E |g(X_i)|\leq M$, unlike your example. That's not enough for uniform integrability, true, but if the CDFs converge uniformly maybe it is? It feels right but I can't seem to show it. The converse of Vitali's convergence theorem seems maybe relevant (en.wikipedia.org/wiki/Vitali_convergence_theorem). $\endgroup$ – user30746 Jan 29 '13 at 20:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.