Say $U=\frac{X}{Y}$. X and Y are independent with each other. X is a Uniform distribution r.v. $X\sim \mathcal{U}(0,1)$. Y is an exponential distribution r.v., $Y\sim\mathcal{Exp}(\lambda)$, whose pdf is $f_Y(y)=\lambda e^{-\lambda y}$. What is the distribution of U? I tried to tackle this problem by caluculating the CDF, as follows:

$F_U(U)=Pr(\frac{X}{Y}\leq U)=Pr(Y\geq \frac{X}{U})=\int_{0}^{1}\int_{\frac{X}{U}}^{\infty}\lambda e^{-\lambda y}dy dx=\frac{U}{\lambda}(1-e^{\frac{\lambda}{U}})$. But this is clearly not right. Because when $u\rightarrow \infty$, $F_{U}(U)\rightarrow \infty$ instead of 1. I want to know what is the mistake that i have made.

  • $\begingroup$ Please include what you have tried. Furthermore, without assuming anything about the joint distribution of $(X,Y)$, e.g. independence, nothing can be said about $U$. $\endgroup$ – Stefan Hansen Jul 21 '14 at 15:44
  • $\begingroup$ What is your thought on this problem so far? Have you look up the definition of what it means for $U=\frac{Y}{X}$? If you did, what details have you managed to work out? $\endgroup$ – Gina Jul 21 '14 at 15:44
  • $\begingroup$ thanks for your comments. The question has been updated $\endgroup$ – DouglasKeuk Jul 21 '14 at 15:57

Remark: The question has changed. The solution below is for $X$ and $Y$ as described in the question as first posed, where we were looking at $\frac{Y}{X}$ with $Y$ uniform and $X$ exponential. We will not edit that solution, since the general idea can be used to deal with variants.

Solution to original problem:

We give the random variable another name, like $W$, so that $U$ will not do double duty. We find the cumulative distribution function $F(w)$ of $W$, and then differentiate to find the density. So for fixed $w\gt 0$, we find $\Pr(W\le w)$, that is, $\Pr(Y\le wX)$.

We will assume, as we are probably expected to assume, that $X$ and $Y$ are independent. The problem should really have specified this.

There is a typo in the density function of $X$. It is $\lambda e^{-\lambda x}$ for $x\gt 0$.

Draw the line $y=wx$. We want the probability that $(X,Y)$ ends up below that line. It is easier to find the probability of the complement, which is the probability that $(X,Y)$ ends up in the triangle with corners $(0,0)$, $(1/w,1)$, and $(0,1)$. Integrate. We get $$1-F(w)=\int_{x=0}^{1/w}\int_{y=wx}^1 \lambda e^{-\lambda x}\,dy\,dx.$$ The integration with respect to $y$ is trivial, and the second integration is not too bad. And we do not even have to integrate. If all we want is the density we can differentiate under the integral sign.

  • $\begingroup$ your result is the same as mine. But the result is not correct since when $w\rightarrow \infty$, $F(w)\neq 1$ $\endgroup$ – DouglasKeuk Jul 21 '14 at 16:44
  • $\begingroup$ Note that $1-F(w)$, as described in the displayed formula of the answer above, goes to $0$ as $w$ blows up. So $F(w)$ goes to $1$. The fact that the double integral goes to $0$ can be seen without evaluating, since the second interval of integration has length $1/w$. $\endgroup$ – André Nicolas Jul 21 '14 at 16:56
  • $\begingroup$ Ding it in the order you used, so integrating out $x$ first, which is easier than my approach, we get $e^{-\lambda/w}$. This approaches $1$ as $w\to\infty$. $\endgroup$ – André Nicolas Jul 21 '14 at 17:29
  • $\begingroup$ I have tested this result. My result as well as that of andre are both correct. We need to user the L'hopital's law as follows $\lim_{w\rightarrow\infty}\frac{1-e^{\lambda/w}}{\lambda/w}=1$ $\endgroup$ – DouglasKeuk Jul 21 '14 at 22:59

I have tested this result. My result as well as that of andre are both correct. We need to user the L'hopital's law as follows



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.