Question

the burr distribution has a probability function for a continuous random variable x given by f(x)={(ckx^c-1)/(1+x^c)^k+1 x>o 0 otherwise} show that the E(x) = kB(k+1/c,1/c+1)

Answer 1

beta function:

$B(x,y)=\intop^1_0t^{x-1}(1-t)^{y-1}dt$

$E(x) =\int^{\infin}_0xf(x)dx=\int^{\infin}_0x\frac{ckx^{c-1}}{(1+x^c)^{k+1}}dx$

let $u=(1+x^c)^{-1}\implies x=(\frac{1-u}{u})^{1/c}$

$du=-cx^{c-1}(1+x^c)^{-2}dx=-cx^{c-1}u^2dx$

then:

$E(x) =k\int^{0}_1(\frac{1-u}{u})^{1/c}u^{k+1}(-u^{-2})du=k\int^{1}_0u^{(k-1/c)-1}(u-1)^{(k+1/c)-1}du=$

$=kB(k+1/c,1/c+1)$