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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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    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 $X_n$).2012-02-24

1 Answers 1

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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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    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.)2012-02-24