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Is there a formula for the variance of a (continuous, non-negative) random variable in terms of its CDF?

The only place I saw such formula was is Wikipedia's page for the Variance (https://en.wikipedia.org/wiki/Variance).

Unfortunately, I was not able to prove the expression there. Please, can anyone help?

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  • $\begingroup$ If you are not too fussy about rigor, you can find $E(X)$ and $E(X^2)$ by integration by parts. $\endgroup$ – André Nicolas Aug 31 '15 at 1:55
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Here is a derivation of a formula for $E(X^2)$. The calculation is excessively informal. For "nice" density functions it is not difficult to justify. A similar calculation gives us $E(X)$. Then the variance is $E(X^2)-(E(X))^2$.

We find $\int_0^\infty x^2f(x)\,dx$ by integration by parts. Let $u=x^2$ and $dv=f(x)\,dx$. Then $du=2x\,dx$ and we can take $v=F(x)-1$. (Here we are being a little tricky.)

Then our integral is $$\left. x^2(1-F(x))\right|_0^\infty +\int_0^\infty 2x(1-F(x))\,dx.$$ The first part vanishes at both ends. So we find that

$$E(X^2)=\int_0^\infty 2x(1-F(x))\,dx.$$

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    $\begingroup$ Thank you very much! You helped me a lot! One more question: should the first part of the integral (the part which vanishes) be (F(x) - 1)x^2? $\endgroup$ – Felipe Schoemer Jardim Aug 31 '15 at 12:34

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