Pareto distribution derivation

Question

If i try to proove derivation of

\begin{align} F_Y(y) &= \mathbb{E}\{\mathbb{P}[W\leq y/h]\}\\ &=\int_{0}^{\infty}\bigg( 1-\frac{(y/h)^{-2/\alpha}}{R^2} \bigg)\frac{h^{m-1}\mathrm{exp}(-h/\theta)}{\theta^{m}\Gamma(m)}\mathrm{d}h\\ &=1-\frac{\theta^{2/\alpha}y^{-2/\alpha}}{R^2\Gamma(m)}\int_{0}^{\infty}(h/\theta)^{2/\alpha+m-1}\mathrm{exp}(h/\theta)\mathrm{d}h/\theta\\ &=1-\frac{\Gamma(2/\alpha+m)}{\Gamma(m)}\frac{(my)^{-2/\alpha}}{R^2} \end{align} I've try the following

\begin{align} &=\int_{0}^{\infty}\bigg( 1-\frac{(y/h)^{-2/\alpha}}{R^2} \bigg)\frac{h^{m-1}\mathrm{exp}(-h/\theta)}{\theta^{m}\Gamma(m)}\mathrm{d}h\\ &=\int_{0}^{\infty}\bigg(\frac{h^{m-1}\mathrm{exp}(-h/\theta)}{\theta^{m}\Gamma(m)}-\frac{(y/h)^{-2/\alpha}}{R^2}\frac{h^{m-1}\mathrm{exp}(-h/\theta)}{\theta^{m}\Gamma(m)}\bigg)\mathrm{d}h\\ &=\frac{1}{\theta^m\Gamma(m)}\int_{0}^{\infty}\bigg(h^{m-1}\mathrm{exp}(-h/\theta)-\frac{(y/h)^{-2/\alpha}h^{m-1}\mathrm{exp}(-h/\theta)}{R^2}\bigg)\mathrm{d}h\\ &=\frac{1}{\theta^m\Gamma(m)}\bigg(\int_{0}^{\infty}h^{m-1}\mathrm{e}^{{-h}(1/\theta)}\mathrm{d}h-\int_{0}^{\infty}\frac{(y/h)^{-2/\alpha}h^{m-1}\mathrm{exp}(-h/\theta)}{R^2}\mathrm{d}h\bigg)\\ &=\frac{1}{\theta^m\Gamma(m)}\bigg(1^{1/\theta}\Gamma(m)-\frac{1^{1/\theta}\Gamma(m)}{R^2}\int_{0}^{\infty}(y/h)^{-2/\alpha}\mathrm{d}h\bigg)\\ \end{align}

In which i cannot progress more (I'm not even sure if my calculation even correct). Do i miss some properties? or anyone have a hint how to progress with this derivation?

Answer 1

I'm with you till the fifth line where you start integrating. For the first term, you have $$ \int_0^\infty h^{m-1}e^{-h/\theta}dh.$$ Substitute in $u=h/\theta$ and get $$\int_0^\infty h^{m-1}e^{-h/\theta}dh= \theta^m\int_0^\infty u^{m-1}e^{-u}du = \theta^m \Gamma(m)$$ where we used the formula$$ \Gamma(z) = \int_0^\infty x^{z-1}e^{-x}dx.$$

For the second term, pull the $y$ and $R$ stuff out front and combine the powers of $h$ in the integral to get $$ \frac{y^{-2/\alpha}}{R^2}\int_0^\infty h^{m-1+2/\alpha} e^{-h/\theta}dh.$$ Making the same substitution gives $$ \frac{y^{-2/\alpha}}{R^2}\int_0^\infty h^{m-1+2/\alpha} e^{-h/\theta}dh =\frac{y^{-2/\alpha}}{R^2}\theta^{m+2/\alpha}\int_0^\infty u^{m-1+2/\alpha}e^{-u}du = \frac{y^{-2/\alpha}}{R^2}\theta^{m+2/\alpha}\Gamma(m+2/\alpha).$$

Putting it all together and dividing by the denominator $\Gamma(m)\theta^m$ gives $$ 1- \frac{y^{-2/\alpha}}{R^2} \theta^{2/\alpha}\frac{\Gamma(m+2/\alpha)}{\Gamma(m)}$$

So I agree with the answer except for a factor of $m^{2/\alpha}$ in the second term.