CDF of a ratio of exponential variables Let $X$ and $Y$ be independent exponential variables with rates $\alpha$ and $\beta$, respectively. Find the CDF of $X/Y$.
I tried out the problem, and wanted to check to see if my answer of: $\frac{\alpha}{ \beta/t + \alpha}$ is correct, where $t$ is the time, which we need in our final answer since we need a cdf.
Can someone verify if this is correct?
 A: A different method may be of interest:
\begin{align}
\Pr(X/Y \le t) = {} & \Pr(X\le tY) \\[8pt]
= {} & \iint\limits_{(x,y)\,:\,x\,\le\, ty} e^{-\alpha x} e^{-\beta y} \big(\alpha\beta\,d(x,y)\big) \\[8pt]
= {} & \int_0^\infty \left( \int_0^{ty} e^{-(\alpha x+\beta y)} (\alpha\,dx) \right) (\beta\,dy) \tag 1
\end{align}
In the inside integral, $x$ goes from $0$ to $ty$ while $y$ remains fixed; thus we treat $y$ as constant in this substitution:
\begin{align}
& u = x/y \\[6pt]
& du = dx/y \\[6pt]
& \text{As $x$ goes from $0$ to $ty,$} \\
& \text{$u$ goes from $0$ to $t$.}
\end{align}
The integral $(1)$ becomes 
\begin{align}
& \int_0^\infty \left( \int_0^t e^{-(\alpha u+\beta)y} (\alpha y\,du) \right) (\beta\,dy)
\end{align}
The point is that the bounds of integration in the inside integral no longer depend on $y,$ so we can alter the order of operations, thus:
\begin{align}
& \int_0^t \left( \int_0^\infty e^{-(\alpha u+\beta)y}\alpha\beta y \, dy \right) \,du \\[8pt]
= {} & \int_0^t \Big( \text{a function of } u \Big)\, du \\[8pt]
\text{So } \Pr(X/Y \le t) = {} & \int_0^t \Big( \text{a function of } u \Big)\, du
\end{align}
Therefore that function of $u$ (which is readily found) must be the probability density function of the random variable $X/Y.$
Appendix: Evaluation of the integral:
\begin{align}
f_{X/Y}(u) = {} & \alpha\beta \int_0^\infty e^{-(\alpha u+\beta)y} y \, dy \\[8pt]
= {} & \frac{\alpha\beta}{(\alpha u+\beta)^2} \int_0^\infty e^{-(\alpha u+\beta)y} (\alpha u+\beta) y \big( (\alpha u + \beta) \, dy \big) \\[8pt]
= {} & \frac{\alpha\beta}{(\alpha u+\beta)^2} \int_0^\infty e^{-v} v \,dv \\[8pt]
= {} & \frac{\alpha\beta}{(\alpha u+\beta)^2}.
\end{align}
A: Here's a slightly different point of view.
\begin{align}
\Pr\left( \frac X Y \ge t \right) & = \iint\limits_{\{\,(x,y)\,:\, x\,\ge\,ty\,\ge\,0 \,\}} e^{-\alpha x} e^{-\beta y}   (\alpha\beta\,d(x,y)) \\[10pt]
& = \int_0^\infty \left( \int_{ty}^\infty e^{-\alpha x} (\alpha\,dx) \right) e^{-\beta y} (\beta\,dy) \\[10pt]
& = \int_0^\infty (e^{-\alpha ty}) e^{-\beta y} (\beta\,dy) \\[10pt]
& = \beta \int_0^\infty e^{-(\alpha t+\beta)y} \, dy = \frac \beta {\alpha t + \beta}.
\end{align}
This is $1$ minus the c.d.f. Find the c.d.f. and differentiate to get the p.d.f. on the interval $t\ge0.$
A: Recall one of the most important characterizations of the exponential distribution:

The random variable $Y$ is exponentially distributed with rate $\beta$ if and only if $P(Y\geqslant y)=\mathrm{e}^{-\beta y}$ for every $y\geqslant0$.

Let $Z=X/Y$ and $t\gt0$. Conditioning on $X$ and applying our characterization to $y=X/t$, one gets
$$
P(Z\leqslant t)=P(Y\geqslant X/t)=E(\mathrm{e}^{-\beta X/t}).
$$
Now, the density of the distribution of $X$ is $\alpha\mathrm{e}^{-\alpha x}$ on $x\geqslant0$, hence for every $\gamma\geqslant0$,
$$
E(\mathrm{e}^{-\gamma X})=\int_0^{+\infty}\alpha\mathrm{e}^{-(\alpha+\gamma) x}\mathrm{d}x=\frac{\alpha}{\alpha+\gamma}\left[-\mathrm{e}^{-(\alpha+\gamma) x}\right]_{0}^{+\infty}=\frac{\alpha}{\alpha+\gamma}.
$$
Substituting $\gamma=\beta/t$ yields the formula.
A: Here is a one-line proof.
$$
\mathbb P(X/Y \le t) = \mathbb P(Y \ge X/t) = \mathbb E[\exp(-\beta X/t)] = \text{MGF}_X(-\beta/t) = (1 - (-\beta/t)1/\alpha)^{-1} = \frac{\alpha}{\alpha + \beta/t} = \frac{\alpha t}{\alpha t + \beta}.
$$
N.B.: For the MGF of an exponential variable, see this table.
