 # I have a question about the notation used in this problem please dont solve the problem for me. Sho cyfwelestoi 2022-05-29 Answered
I have a question about the notation used in this problem please dont solve the problem for me.
Show that $||XY|{|}_{1}=||X|{|}_{1}||Y|{|}_{1}$ for independent r.vs $X$ and $Y$. Show further that if $X$ and $Y$ are also integrable, then $E\left(XY\right)=E\left(X\right)E\left(Y\right)$
I don't understand what the difference between $||X|{|}_{1}$ and $E\left(|X|\right)$. I assume we're talking about the ${L}^{1}\left(\mathbb{P}\right)$ norm? But then
$||X|{|}_{1}={\int }_{\mathrm{\Omega }}|X\left(\omega \right)|d\mathbb{P}\left(\omega \right)=E\left(|X|\right)$
So if this is finite, then $E\left(X\right)$ exists therefore it is integrable already (so why "also"?) Therefore the first "show that" would follow from the second
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Under independence $||XY|{|}_{1}=||X|{|}_{1}||Y|{|}_{1}$ holds even if $E|X|$ or $E|Y|$ is infinity (by Tonelli's Theorem). But $E\left(XY\right)=\left(EX\right)\left(EY\right)$ holds if the expectations are finite.