# 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?

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.

• What is the probability distribution function? – user Mar 11 '14 at 4:37
• @user How to deduce the PDF from the CDF? Tell me... – Did Mar 11 '14 at 6:34
• @Did how did you arrive to $P(Y\geqslant X/t)=E(\mathrm{e}^{-\beta X/t})$. How is the probability and expectation related? – user35443 Nov 1 '18 at 13:57
• @user35443 For every $x\geqslant0$, $P(Y\geqslant x/t)=e^{-\beta x/t}$. Now, $X$ is independent of $Y$ hence $P(Y\geqslant X/t\mid X)=e^{-\beta X/t}$ almost surely. Finally, take expectations on both sides. – Did Nov 3 '18 at 14:56
• Makes sense, thanks! – user35443 Nov 4 '18 at 18:18

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}

This is correct. I did the calculation and got the same answer.

• Did you use u-substitutions, a lot of them? – mary Apr 19 '11 at 5:02
• No, I used Mathematica :) If you do it by hand and you are not familiar with $\int \alpha e^{-\alpha x} dx$ you might need a lot of substitutions. – GWu Apr 19 '11 at 5:03

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.