# 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?

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## 2 Answers

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.

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

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 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