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I need a counterexample for the problem: if $r>s\geq1$, convergence in $s^{\text{th}}$ mean does not imply convergence in $r^{\text{th}}$ mean.

The definition for convergence in mean is as follows: Let $r\geq1$ be a fixed number. A sequence of random variables $X_1, X_2, X_3,...$ converges in the $r^{\text{th}}$ mean to a random variable $X$, if

$$\lim_{n\to\infty}E(|X_n−X|^r)=0.$$

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    $\begingroup$ @BruceET I have added the definition now. $\endgroup$ Feb 10, 2022 at 14:38

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Try $P(X_n=n)=1/n^\alpha$ and $P(X_n=0)=1-P(X_n=n)$ with $X=0$.

Then, $$ E|X_n-X|^s=EX_n^s=n^{s-\alpha} $$

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  • $\begingroup$ Ok so basically, $E|X_n-X|^r=n^{(r-a)}$ and for $r>a$, it doesn't converge to 0. Is that how it works @Cristoph ? $\endgroup$ Feb 10, 2022 at 15:00
  • $\begingroup$ Yes, for $r\geq a$ actually, as $n^0=1\neq0$. $\endgroup$ Feb 10, 2022 at 15:09
  • $\begingroup$ Ok thank you so much $\endgroup$ Feb 10, 2022 at 15:11

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