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Let $\{X_n\}_{n=1}^\infty$ be a sequence of random variables converging to a random variable $X$. Does this fact implies that $\lim\limits_{n\to \infty} EX_n = EX$?

If not, is there any other sufficient condition?

Ѕᴀᴀᴅ
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SBM
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3 Answers3

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Let $(\Omega,\mathcal{F},P)=([0,1],\mathcal{B}([0,1]),\lambda)$. Then the sequence $X_n=n\mathbf{1}_{(0,1/n)}$ converges almost surely and hence also in probability to $X=0$, but $\lim_{n\to \infty}{\rm E}[X_n]=1\neq {\rm E}[X]$.

However, if we require that $\{X_n\mid n\geq 1\}$ is uniformly integrable, then the result holds. In fact it only requires that $X_n$ converges to $X$ in distribution. See this answer for hints on the proof.

Community
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Stefan Hansen
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  • Your answer is wrong. Note that convergence in probability implies convergence in distribution , and $Xn \xrightarrow{d} X$ iff for all bounded continuous functions h(x), $E[h(Xn)] \to E[h(X)]$. Now if you take h(x)=1, the following should hold: $E[Xn] \to E[X]$ – Susan_Math123 May 16 '16 at 18:53
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    @Susan: Really? – Stefan Hansen May 16 '16 at 19:46
  • Yes, look at page 253 of A.N. Shiryaev, probability book. – Susan_Math123 May 16 '16 at 19:57
  • Also, note that $X_n=n1_{(0,1/n)}$ does not converge almost surely to 0. Just check it for convergence in probability to 0, you will see it is not true. – Susan_Math123 May 16 '16 at 19:59
  • @Susan: If $h(x)=1$ for all x, then $h(X_n)=h(X)=1$ pointwise. And yes, $X_n=n\mathbf{1}_{(0,1/n)}$ does converge almost surely to 0. – Stefan Hansen May 16 '16 at 20:16
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    You are right, I meant $h(x)=x$, but it is not bounded and does not lie in the theorem's condition, which I stated above. – Susan_Math123 May 16 '16 at 20:19
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Hint: try a sequence of random variables that are $0$ with high probability but large with low probability.

Robert Israel
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I searched the Internet to no avail, so I figured I should post my example inspired by the hint:

$P (X_n = 0) := 1 - \frac1n$, $P(X_n = n) := \frac1n$.

Then $X_n \xrightarrow P 0$ and $E[X_n] = 1$.

Kenny Lau
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