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I'm looking for an elementary way of showing the following. If $(X_n)$ and $X$ are random variables such that $X_n \to X$ in distribution and such that $\{X_n\mid n\geq 1\}$ are uniformly integrable, then $E[X_n]\to E[X]$.

I've seen another topic on this, but the solution given there is using Skorokhod's theorem stating that convergence in distribution is equivalent to almost-sure convergence of copies of the random variables in some abstract probability space. I would like to do without that if possible. Thanks in advance!

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Isn't the probability space for which almost-sure convergence holds pretty explicit and concrete here, i.e., $((0,1),\mathcal B, \mathbb P)$ where $\mathbb P$ is Lebesgue measure? –  cardinal Feb 24 '12 at 8:05
    
To me isn't not obvious that this is the probability space to work on, and furthermore I would like to this exercise without using that equivalence. –  Stefan Hansen Feb 24 '12 at 8:42
    
That's fine, I understand (and appreciate!) that you are looking for an exercise not using this theorem. I just wanted to point out that the space is really not exotic. The construction is achieved just as you'd expect, by letting $Y_n$ be the generalized inverse of $F_n$ (the distribution of $X_n$). –  cardinal Feb 24 '12 at 9:34

1 Answer 1

up vote 8 down vote accepted

For any random variable $Z$ and any real number $x\geqslant0$, let $Z^x=\max\{-x,\min\{Z,x\}\}$. Let $\mathcal X=\{X_n\mid n\geqslant1\}$. Here are some steps of a proof:

  1. If $X_n\to X$ in distribution, then, for every $x\geqslant0$, $\mathrm E(X_n^x)\to \mathrm E(X^x)$.
  2. If $\mathcal X$ is uniformly integrable and $X_n\to X$ in distribution, then $X$ is integrable and $\mathcal X\cup\{X\}$ is uniformly integrable.
  3. If $\mathcal Y$ is uniformly integrable, then for every $\varepsilon\gt0$, there exists a finite $x\geqslant0$ such that $\mathrm E(|Y-Y^x|)\leqslant\varepsilon$ for every $Y$ in $\mathcal Y$.
  4. Prove the triangular inequality, valid for every $n\geqslant1$ and $x\geqslant0$: $$ |\mathrm E(X_n)-\mathrm E(X)|\leqslant\mathrm E(|X_n-X_n^x|)+|\mathrm E(X_n^x)-\mathrm E(X^x)|+\mathrm E(|X-X^x|). $$
  5. Conclude.
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If any of these steps is still a problem, please say so. –  Did Feb 24 '12 at 7:14
    
If 2. holds why is the following not true? We have that $X_n-X \to 0$ in probability and as $\{X_n-X\mid n\geq 1\}$ is uniformly integrable, we have that $X_n-X\to 0$ in $L^1$. –  Stefan Hansen Feb 24 '12 at 7:20
    
Thanks alot for the help. I think I got through everything except 2. Could you give a hint on showing that $X$ is integrable? –  Stefan Hansen Feb 24 '12 at 9:11
    
Re 2.: assume without loss of generality that $X_n\geqslant0$ almost surely for every $n$. By step 1., for every $x$, $\mathrm E(X^x)\leqslant\sup_n\mathrm E(X_n^x)\leqslant K$ with $K=\sup_n\mathrm E(X_n)$ which is finite since $\mathcal X$ is UI. Hence $\mathrm E(X)\leqslant K$. (Re your other comment, it is quite possible that some shortcuts exist, which allow to bypass one step or another.) –  Did Feb 24 '12 at 16:23

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