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Title says it all: if $X$ is a $\pm 1$-valued random variable with mean $\mu$, 'why' is $\mathbf{E}[|X - \mu|] = \mathbf{E}[|X-\mu|^2]$? Obviously one can do a couple of lines of trivial algebra to verify it, but is there a simple-to-state reason? Something you can say to make me say, "Oh yeah, of course" without any calculation?

Thanks!

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If $x=\pm1$ and $|\mu|\leqslant1$, $\mathrm{sgn}(x-\mu)=x$ hence $|x-\mu|=x\cdot(x-\mu)=(x-\mu)^2+\mu\cdot(x-\mu). $ Thus, $|X-\mu|=(X-\mu)^2+\mu\cdot(X-\mu)$ almost surely. Since $\mathrm E(X-\mu)=0$, this yields $\mathrm E(|X-\mu|)=\mathrm E((X-\mu)^2)$.

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    Good, I think this is perhaps the most satisfactory phrasing.2012-09-06