CDF of a ratio of exponential variables

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.

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 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}