# Distribution of two independent exponential random variables

1. Aug 12, 2008

### lizzyb

Q: If $$X_1$$ and $$X_2$$ are independent exponential random variables with respective parameters $$\lambda_1$$ and $$\lambda_2$$, find the distribution of $$Z = X_1 / X_2$$.

Discussion

The best method to attack this problem apparent to me is coming up with a cumulative distributive function for $$Z$$ and then differentiating it.

Work Completed So Far
(note: I don't mean for anyone to check if I did my math right I'm just wondering if I'm going about solving the question right)

$$F_Z (a) = P\{ Z \leq a \} = P\{ X_1 / X_2 \leq a \} = P \{ X_1 \leq a X_2 \}$$
$$\int_0^\infty \int_0^{a x_2} \lambda_1 e^{- \lambda_1 x_1} dx_1 \lambda_2 e^{- \lambda_2 x_2} dx_2 = \lambda_1 \lambda_2 \int_0^\infty e^{\lambda_2 x_2} dx_2 \left[ -\frac{1}{\lambda_1} e^{-\lambda_1 x_1} \right]^{a x_2}_0 = - \lambda_2 \int_0^\infty e^{- \lambda_2 x_2} dx_2 ( e^{-\lambda_1 a x_2} - 1 )$$

$$= \lambda_2 \int_0^\infty e^{- \lambda_2 x_2} dx_2 -\lambda_2 \int_0^\infty e^{-\lambda_2 x_2 - \lambda_1 a x_2} dx_2 = \lambda_2 \int_0^\infty e^{- \lambda_2 x_2} dx_2 - \lambda_2 \int_0^\infty e^{-x_2(\lambda_2 + \lambda_1 a)} dx_2$$

$$= \lambda_2 ( - \frac{1}{\lambda_2} ) \left[ e^{- \lambda_2 x_2 } \right]_0^\infty - \lambda_2 \frac{-1}{\lambda_2 + \lambda_1 a} \left[ e^{-x_2(\lambda_2 + \lambda_1 a) \right]_0^{\infty} = 1 - \frac{\lambda_2}{\lambda_2 + \lambda_1 a}$$

So $$f_{X_1/X_2}(a) = f_Z(a) = \frac{d F_Z(a)}{da} = \frac{\lambda_1 \lambda_2}{(\lambda_1 a + \lambda_2)^2}$$

Does this look like I'm doing it right?

Thanks.

Last edited: Aug 12, 2008
2. Aug 13, 2008