In probability, is it possible for some random variable $X$, that $E[|X|]=E[X^2]$? Can you explain why as well?
I believe the answer has something to do with symmetric distributions but I'm not 100% sure.
2026-03-30 06:32:15.1774852335
Is there a random variable $X$ such that $E[ |X| ] = E [ X^2 ]$?
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Cauchy-Schwarz inequality gives $$ \mathbb E\lvert X\rvert\leqslant \left(\mathbb E\left[X^2\right]\right)^{1/2} $$ so the condition in the question gives $\mathbb E\left[X^2\right]\leqslant 1$ and not so much information.
Random variable taking the values $0$ and $1$ satisfy $\mathbb E\lvert X\rvert=\mathbb E\left[X^2\right]$ since $\lvert X\rvert=X^2$, hence indicator functions are examples.