Variance formula in terms of the CDF for a continuous nonnegative random variable.

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?

• If you are not too fussy about rigor, you can find $E(X)$ and $E(X^2)$ by integration by parts. – André Nicolas Aug 31 '15 at 1:55

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

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

• 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? – Felipe Schoemer Jardim Aug 31 '15 at 12:34

Another potential derivation for $$E[X^2]$$ uses the tail-probability equality.

$$Y = X^2$$ is a nonnegative random variable, so the tail-probability equality applies:

$$E[Y] = \int_{y=0}^{\infty} Pr[Y \geq y] dy$$

Replace $$Y = X^2$$ and $$y$$ with appropriate expressions of $$X$$ and $$x$$:

$$E[X^2] = \int_{x^2=0}^{\infty} Pr[X^2 \geq x^2] dx^2$$

Because we are assuming that $$X$$ is nonnegative, $$Pr[X^2 \geq x^2] = Pr[X \geq x]$$. We also change our variable of integration from $$x^2$$ to $$x$$ (essentially $$u$$-substitution) by noting that $$dx^2 = 2xdx$$. Making these changes, we get

$$E[X^2] = 2\int_{x=0}^{\infty}x Pr[X \geq x] dx$$

If you want it in CDF form, make the replacement $$Pr[X \geq x] = 1 - F_X(x)$$:

$$E[X^2] = 2\int_{x=0}^{\infty}x (1 - F_X(x)) dx$$