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I am solving the following problem.

$X_n\to \textrm{Exp}(5)$ and $Y_n\to 10$ both in distribution. To what does $E[X_n Y_n]$ converge?

My attempt:

From Wikipedia, I know that $(X_n, Y_n)\to (X, 10)$ in distribution, and also that $X_n Y_n \to 10X$ by combining the fact that “convergence to a constant in distribution implies convergence in probability” and Slutsky. However, I believe convergence in distribution does not say anything about the convergence of mean. So I do not know how to proceed.

Kaira
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  • one thing to note here is that you have something a liittle stronger, namely convergence in probability: https://math.stackexchange.com/questions/1716298/proof-for-convergence-in-distribution-implying-convergence-in-probability-for-co – Samael Manasseh Apr 25 '23 at 03:26
  • @SamaelManasseh Yes, so by using that fact and Slutsky we can prove that $X_nY_n$ converges to $10X$ in distribution. But again I’m back to convergence in distribution.(which was what I’m trying to say in the question). – Kaira Apr 25 '23 at 03:29
  • You know that $Y_n X_n \xrightarrow[]{d} 10 X $, take a look at https://en.wikipedia.org/wiki/Convergence_of_random_variables#Properties – jDAQ Apr 25 '23 at 04:00
  • @jDAQ I am not sure how that helps because the identity function $f(x)=x$ is not bounded. – Kaira Apr 25 '23 at 04:02
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    Does it need to converge? Let $Y_n = 10 + Z_n$ where $Z_n$ is some classic counterexample where $Z_n\to_p0$ but $E(Z_n)\to \infty$ (e.g. let it be an independent sequence that is $n^2$ with probability $1/n$, else $0$). – spaceisdarkgreen Apr 25 '23 at 04:10
  • @spaceisdarkgreen Although the problem seems to assume that it converges(and the professor said that it converges to 2), that was my suspect too. Thank you for providing some related counterexample. – Kaira Apr 25 '23 at 04:16
  • You can say absolutely nothing about $EX_nY_n$. You are misquoting Wikipedia. Are you assuming INDEPENDENCE? – Kavi Rama Murthy Apr 25 '23 at 05:00
  • @geetha290krm are you talking about me? – Kaira Apr 25 '23 at 11:26

1 Answers1

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It doesn't necessarily need to converge, and can converge to anything you like. Let $Y_n = 10 + Z_n$ where $Z_n$ is independent of $X_n,X$ and is some classic counterexample where $Z_n\to_p0$ but $E(Z_n)$ converges to something nonzero, or goes to infinity (e.g. let it be an independent sequence that is $n^2$ with probability $1/n).$

spaceisdarkgreen
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