calculating mean of a CDF

Question

I have this complex CDF:

F(x)=1-exp(-X^2/c)

when c is a constant.

How can I calculate mean?

In my calculation, I come to calculate this:

(2/c)(X^2)exp((-X^2)/c)

but calculating this for infinity is not possible

Answer 1

For $c>0$, the CDF of $X$ is $$F(x)=\left(1-\exp \left\{-\frac{x^2}{c}\right\}\right){\bf 1}_{x\geq 0}$$

implying a PDF $$f(x)=F'(x)=\left(\frac{2x}{c}\exp \left\{-\frac{x^2}{c}\right\}\right){\bf 1}_{x\geq 0}.$$

The mean of $X$ is given by $$\int_0^\infty xf(x)dx=\int_0^\infty \frac{2x^2}{c}\exp \left\{-\frac{x^2}{c}\right\}dx,$$

and substituting $x=t\sqrt{\frac{c}{2}}$ gives

$$\sqrt{\frac{c}{2}}\int_0^\infty t^2\exp \left\{-\frac{t^2}{2}\right\}dt,$$

which by evenness of the integrand is

$$\frac{1}{2}\sqrt{\frac{c}{2}}\int_{-\infty}^\infty t^2\exp \left\{-\frac{t^2}{2}\right\}dt,$$

which can be written as

$$\frac{\sqrt {c\pi}}{2}\left(\frac{1}{\sqrt {2\pi}}\int_0^\infty t^2\exp \left\{-\frac{t^2}{2}\right\}dt\right),$$

and the expression in parentheses is just the variance of a standard normal random variable, which is one, so the mean of $X$ is $\sqrt{c\pi}/2.$

Answer 2

It is not mentioned but I suspect that $F(x)=0$ for negative $x$.

In general for a nonnegative rv $X$:$$\Bbb EX=\int_0^{\infty}1-F_X(x)dx$$

Applying that here we find:$$\Bbb EX=\int_0^{\infty}e^{-\frac1cx^2}dx=\frac12\sqrt{2c}\int_0^{\infty}e^{-\frac12y^2}dy$$where the second equality appears by substituting $x=\frac12y\sqrt{2c}$.

This can also be written as:$$\sqrt{c\pi}\int_0^{\infty}\phi(y)dy$$where $\phi$ denotes the PDF of standard normal distribution.

That distribution is symmetric wrt $0$ so that the integral takes value $\frac12$ and our final conclusion is:$$\Bbb EX=\frac12\sqrt{c\pi}$$