I understand that
$Var(X) = E(X^2) - E(X)^2 $
And that the second moment, variance, is
$E(X^2)$
How is variance simultaneously $E(X^2)$ and $E(X^2) - E(X)^2$?
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$$ \mathbb{E}(X^n) = \text{raw moment}\\ \mathbb{E}\left[\left(X-\mathbb{E}(X)\right)^n\right] = \text{central moment} $$ where the 2nd central moments represents the variance.
only equal when $\mathbb{E}(X) = 0$ as with $\mathcal{N}(0,1)$.
Chinny84
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Simple: $$\operatorname{Var}(X)\neq E(X^2)$$
The second moment is not, in general, equal to variance.
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Under which conditions is the second moment equal to variance? – slimydummy Nov 25 '16 at 17:32
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2@slimydummy If $E(X^2)-(E(X))^2=E(X^2)$, so use some algebra and you get... – Nov 25 '16 at 17:33
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That was dumb of me. Thank you – slimydummy Nov 25 '16 at 17:35
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5@slimydummy True to your name. :) But the thruth is that there are no stupid question but maybe stupid answers. – callculus42 Nov 25 '16 at 17:51