I understand that

$Var(X) = E(X^2) - E(X)^2 $

And that the second moment, variance, is


How is variance simultaneously $E(X^2)$ and $E(X^2) - E(X)^2$?


2 Answers 2


$$ \mathbb{E}(X^n) = \text{raw moment}\\ \mathbb{E}\left[\left(X-\mathbb{E}(X)\right)^n\right] = \text{central moment} $$ where the 2nd central moments represents the variance.

only equal when $\mathbb{E}(X) = 0$ as with $\mathcal{N}(0,1)$.


Simple: $$\operatorname{Var}(X)\neq E(X^2)$$

The second moment is not, in general, equal to variance.

  • $\begingroup$ Under which conditions is the second moment equal to variance? $\endgroup$
    – slimydummy
    Nov 25, 2016 at 17:32
  • 2
    $\begingroup$ @slimydummy If $E(X^2)-(E(X))^2=E(X^2)$, so use some algebra and you get... $\endgroup$
    – user223391
    Nov 25, 2016 at 17:33
  • $\begingroup$ That was dumb of me. Thank you $\endgroup$
    – slimydummy
    Nov 25, 2016 at 17:35
  • 3
    $\begingroup$ @slimydummy True to your name. :) But the thruth is that there are no stupid question but maybe stupid answers. $\endgroup$ Nov 25, 2016 at 17:51

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