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Given $X_n\to X$ and $E[X_n]\to E[X]$, is it true that $E[X_n^+]\to E[X]$?
Here $X_n^+(\omega)=max(X_n(\omega),X(\omega))$
I see that it is true that $X_n^+\to X$ but I can't seem to apply DCT/MCT to prove the above

This question arose because I wanted to solve converse of this question Is is true that if $E|X_n - X| \to 0$ then $E[X_n] \to E[X] $? (also assuming that $X_n\to X$ of course)

Anvit
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  • Is it almost sure convergence ? $L^2$ convergence, convergence in probability, or may in distribution ? – P. Quinton Apr 29 '19 at 17:18
  • @P.Quinton I haven't heard any of the last three terms you used. This is in Probablity Theory II course and we just learnt about DCT/MCT/Fatou's Lemma. I'm not sure if that helps but I dont exactly know which type of convergence it is – Anvit Apr 29 '19 at 17:21
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    The type of convergence is crucial to understand. Can you supply a definition of "$X_n\to X$" as you are using it? – Greg Martin Apr 29 '19 at 17:28
  • @Anvit You start with "given $X_n\to X$.." What exactly does that mean? If you don't know that then actually you are not entitled to ask this question. You must know what you are talking about. – drhab Apr 29 '19 at 17:30
  • I read the wikipedia page, the definition we did was of Converegence in Probability – Anvit Apr 29 '19 at 17:32
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    @NateEldredge My definition of $X_n^+$ is a not the standard one. It is $max(X_n,X)$ – Anvit Apr 29 '19 at 17:50
  • Oh, I see. Maybe you ought to change the title. – Nate Eldredge Apr 29 '19 at 17:51

1 Answers1

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In this answer I preassume (on base of the comment of Anvit) that $X_n\to X$ stands for $X_n\stackrel{p}{\to} X$.

Let $P\left(X_{n}=n\right)=\frac{1}{4n}=P\left(X_{n}=-n\right)$ and $P\left(X_{n}=0\right)=1-\frac{1}{2n}$.

Further let $X=0$.

Then $X_{n}\stackrel{p}{\to}0$ and $\mathbb{E}X_{n}=0=\mathbb{E}X$ for every $n$.

But $X_{n}^{+}=\max\left(X_{n},0\right)$ has expectation $n\times\frac{1}{4n}=\frac{1}{4}$ which does not converge to $\mathbb EX=0$.

drhab
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