Determine the probability distribution of a ratio of two random variables? Setting
You are given two independent random variables $X_0,X_1$ with common exponential density $f(x) = \alpha e^{-\alpha x}$. Let $R = \frac{X_o}{X_1}$. Determine $\Pr[R > t]$ for $t > 0$.
I got up to here
$$\Pr[R > t] = \Pr[X_o/X_1 > t] = \Pr[X_o > X_1 t] = 1 - \Pr[X_o \le tX_1]$$
I know how to express the last probability as a distribution, but it would have $X_1$ in it, making it a random variable. So how do I proceed?

One way to do this is to rescale: let $Y = t X_1$, so $X_0/X_1 > t$ is equivalent to $X_0 > Y$.  Now $X_0$ and $Y$ are independent exponential random variables with rate parameters $\alpha$ and $\alpha/t$ respectively.  Think of two independent Poisson processes with these rates.  One way to realize this is to have a combined Poisson process with rate $\alpha + \alpha/t$, and assign each occurrence to the $X_0$ or the $Y$ process with probabilities $\dfrac{\alpha}{\alpha + 1/t} = \dfrac{t}{t+1}$ and $\dfrac{\alpha/t}{\alpha + \alpha/t} = \dfrac{1}{t+1}$ respectively.  The probability that the next occurrence is from the $Y$ process, i.e. that $t X_1 = Y < X_0$, i.e. that $R = X_0/X_1 > t$, is then $\dfrac{1}{t+1}$.  
