0

If $Z = \frac{X+Y}{X}$ then $Y = X(Z-1)$. Then I believe I can do the following?

$\int_{-\infty }^{\infty } f_X(x)f_Y(x(z-1)) dx$

$=\int_{0 }^{z} \lambda e^{-\lambda x} \lambda e^{-\lambda x(z-1)}dx$

$=\lambda^2 \int_{0 }^{z} e^{-\lambda xz}dx$

$=\frac{-\lambda}{z} e^{-\lambda z^2}$

Is this the correct distribution of Z? If so is there an easier way?

Jokus
  • 95

0 Answers0