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We know that $\operatorname{Var}(X)=E[X^2]-E[X]^2$. I want to know that if $E[X^2]$ exists, does it guarantees then $E[X]$ will also exist?

Davide Giraudo
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UserA
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  • You might want to take a look at following question: https://math.stackexchange.com/questions/66029/lp-and-lq-space-inclusion –  Sep 06 '21 at 08:12
  • @KaviRamaMurthy More specifically, this is Lyapunov's Inequality which is indeed a direct consequence of Holder's inequality (for which, in turn, Cauchy-Schwarz is a particular case). – econbernardo Sep 08 '21 at 15:00

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$0\le |X| \le \max(1,X^2) \le 1+X^2$ so $0 \le E[|X|] \le 1+ E[X^2]$,

meaning that if $E[X^2]$ is finite then so too are $E[|X|]$ and $E[X]$

Henry
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  • It is against the site's rules to answer such questions: https://math.meta.stackexchange.com/questions/33508/enforcement-of-quality-standards?cb=1 – Kavi Rama Murthy Sep 08 '21 at 23:11