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I want to prove that if $\xi_n$ is sequence of non-negative random variables, $\xi_n\xrightarrow {a.e.}\xi$ and $\mathbb{E}\xi_n^p \to \mathbb{E}\xi^p $ than $$\mathbb{E}|\xi_n-\xi|^p\to0 \text{ (i.e. }\xi_n\xrightarrow {L^p}\xi)$$ I don't know where to start. I would be very grateful for help!

Nick Niceday
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1 Answers1

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This is the Brezis-Lieb lemma and goes as follows: suppose that $p>1$.

  • Prove first that there is a constant $C$ depending on $p$ such that for all $a,b\in\mathbb{R}$ you have $$\left||a+b|^p-|a|^p-|b|^p \right|\leq C\left(|a|^{p-1}|b|+|a||b|^{p-1} \right).$$

  • Assume in addition that $\|\xi_n\|_{L^p}\leq M$ for some $M\in\mathbb{R}$ and prove that $$\mathbb{E}\left[|\xi_n|^p-|\xi_n-\xi|^p\right]\to \mathbb{E}[|\xi|].$$

  • Now remove the above assumption, and conclude.

For the case $p=1$.

  • Prove that for all $a,b\in\mathbb{R}$ you have $$||a+b|-|a|-|b||\leq 2|b|.$$
  • Assume in addition that $\|\xi_n\|_{L^1}\leq M$ for some $M\in\mathbb{R}$ and prove that $$\mathbb{E}\left[|\xi_n|-|\xi_n-\xi|\right]\to \mathbb{E}[|\xi|].$$
  • Now remove the above assumption, and conclude.

Here is a way to prove the result without the Brezis-Lieb lemma. Indeed, for $p\geq 1$ we have the inequality $$|a-b|^p\leq 2^{p-1}(|a|^p+|b|^p),\;\;\text{ for all }a,b\in\mathbb{R}. $$ Therefore $$|\xi_n-\xi|^p\leq 2^{p-1}(|\xi_n|^p+|\xi|^p).$$ Also since $\xi_n \xrightarrow {a.e.} \xi$ then you clearly have that $$2^{p-1}(|\xi_n|^p+|\xi|^p)\xrightarrow {a.e.} 2^{p}|\xi|^p\in L^1\;\;\; (\text{since }\xi\in L^p),$$ and by the assumption $\mathbb{E}[|\xi_n|^p]\to \mathbb{E}[|\xi|^p]$ we have $$ \mathbb{E}[2^{p-1}(|\xi_n|^p+|\xi|^p)]\to \mathbb{E}[2^p\,|\xi|^p]. $$ Therefore by (a slightly stronger version of) dominated convergence applied to $$ f_n:=|\xi_n-\xi|^p, \;\;\; f:=0,\;\;\;g_n:=2^{p-1}(|\xi_n|^p+|\xi|^p),\;\;\;g:=2^p|\xi|^p,$$ you get that $\mathbb{E}[f_n]\to 0 $ as desired.

UserA
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