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Suppose I have $X_n \sim F_n$, $X \sim F$ such that $\Vert F_n - F \Vert_\infty \rightarrow 0$, does this imply that $f(X_n) \xrightarrow{L^p} f(X)$ for any continuous $f$?

I know that, since we are dealing with probability measures/distribution functions, $X_n \xrightarrow{L^p} X$ holds.

I know that, for example, similar results hold for convergence in probability or weak convergence (using unif convergence implies $\xrightarrow{p}$, and then using CMT), but I was wondering if I can say something about $L^p$ convergence.

StubbornAtom
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jackson5
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1 Answers1

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No, you can't say anything about that convergence.
The reasoning behind that is the different nature of these two types of convergence.
While the condition that you give provides a lot of info on the convergence in law, it gives nothing on the spatial relation between $(X_n;n \ge 0)$.
A good thought counterexample is when $(X_n; n \ge 0)$ are independent and identically distributed.

Paresseux Nguyen
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  • But it is true that $X_n \xrightarrow{L^p} X$, right? Even when they are iid, so why can't this hold for $f(X_n)$? – jackson5 Nov 09 '20 at 17:03
  • No, that is not true. You can take Benoulli variables to have a more precise look on this matter. – Paresseux Nguyen Nov 09 '20 at 17:07
  • But it is true, I believe, since we are dealing with finite measure spaces. See here, for example: https://math.stackexchange.com/a/1529187/82654

    If I am misunderstanding something, could you please expand on your counterexample? Thanks!

     – jackson5 Nov 09 '20 at 17:11
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    There is nothing mysterious, if $(X_n)$ are idds with the law $\mathcal{B}(1/2)$ then for all $n,m$ distinct, we have: $$ \mathbb{E}| X_n-X_m|^p = 1/2$$ – Paresseux Nguyen Nov 09 '20 at 17:14
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    Ah I see, sorry my bad. The distribution functions converge in $L^p$ but not their random variables. Makes perfect sense, of course – jackson5 Nov 09 '20 at 17:24