Distribution of Ratio of Exponential and Gamma random variable

Solution 1:

As I claimed in the comments, $Y/X + 1$ follows the $\mathrm{Pareto}(1,n)$, not $Y/X$.

$$ \phi_{Y/X}(t)= \mathbb{E}\left(\exp\left(i t \frac{Y}{X}\right)\right) = \mathbb{E}\left( \phi_Y\left( \frac{t}{X} \right) \right) = \mathbb{E}\left( \frac{X}{X-i t} \right) $$ Writing the last expectation explicitly: $$ \begin{eqnarray} \phi_{Y/X}(t) &=& \frac{1}{(n-1)!} \int_0^\infty \frac{x^n}{x - i t} \mathrm{e}^{-x} \mathrm{d} x = \frac{1}{(n-1)!} \int_0^\infty \mathrm{d} u \int_0^\infty x^n \mathrm{e}^{-x} \mathrm{e}^{-u(x - i t)} \mathrm{d} x \\ &=& \frac{1}{(n-1)!} \int_0^\infty \mathrm{e}^{i t u} n! (1+u)^{-1-n} \, \mathrm{d} u = \mathrm{e}^{-i t} \int_0^\infty \mathrm{e}^{i t (u+1)} \, \frac{n}{(1+u)^{n+1}} \mathrm{d} u \\ &=& \mathrm{e}^{-i t} \phi_{\mathrm{Pareto}(1,n)}(t) \end{eqnarray} $$ This proves that $Y/X \stackrel{d}{=} Z - 1$, where $Z$ follows $\mathrm{Pareto}(1,n)$