Mean of a given distribution

Question

Find the mean of a distribution with given CDF: $$F(x)=1-{1\over{1+bx^a}}, t>0$$

By definition: $$EX=\int_0^\infty xdF(x)=\int_0^\infty\frac{abx^a}{(1+bx^a)^2}dx $$

How can I solve this integral or Is there a simplest method?

Answer 1

Let $b>0$ and $a>1$ (if $a\le 1$ the integral doesn't converge and the mean doesn't exist).

We have $$ \Bbb E(X)=\int_0^\infty\left(1-F(x)\right)\mathrm d x=\int_0^\infty\frac{1}{1+bx^a}\mathrm d x=\frac{1}{b^{1/a}}\int_0^\infty\frac{1}{1+u^a}\mathrm d u $$ after the substitution $b^{1/a}x=u$. Putting $u^a=z$, $\mathrm du=\frac{1}{a}z^{\frac{1}{a}-1}\mathrm dz $ we have $$ \Bbb E(X)=\frac{1}{b^{1/a}}\int_0^\infty\frac{1}{1+u^a}\mathrm d u=\frac{1}{ab^{1/a}}\int_0^\infty \frac{z^{\frac{1}{a}-1}}{1+z}\mathrm d z=\frac{1}{ab^{1/a}}B\left(\frac{1}{a},1-\frac{1}{a}\right) $$ where $B\left(x,y\right)$ is the Euler's Beta function. We have $$ B\left(\frac{1}{a},1-\frac{1}{a}\right)=\frac{\Gamma\left(\frac{1}{a}\right)\Gamma\left(1-\frac{1}{a}\right)}{\Gamma\left(1\right)}=\Gamma\left(\frac{1}{a}\right)\Gamma\left(1-\frac{1}{a}\right)=\frac{\pi}{\sin\frac{\pi}{a}} $$ and then $$ \Bbb E(X)=\frac{1}{b^{1/a}}\frac{\pi/a}{\sin(\pi/a)}=\frac{1}{b^{1/a}\,\mathrm{Sinc}\left(\frac{\pi}{a}\right)} $$