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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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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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  • $\begingroup$ What is the probability distribution function? $\endgroup$ – user Mar 11 '14 at 4:37
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    $\begingroup$ @user How to deduce the PDF from the CDF? Tell me... $\endgroup$ – Did Mar 11 '14 at 6:34
  • $\begingroup$ @Did how did you arrive to $P(Y\geqslant X/t)=E(\mathrm{e}^{-\beta X/t})$. How is the probability and expectation related? $\endgroup$ – user35443 Nov 1 '18 at 13:57
  • $\begingroup$ @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. $\endgroup$ – Did Nov 3 '18 at 14:56
  • $\begingroup$ Makes sense, thanks! $\endgroup$ – user35443 Nov 4 '18 at 18:18
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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 $t\ge0.$

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

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  • $\begingroup$ Did you use u-substitutions, a lot of them? $\endgroup$ – mary Apr 19 '11 at 5:02
  • $\begingroup$ 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. $\endgroup$ – GWu Apr 19 '11 at 5:03

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