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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.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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

$\lim_{w\rightarrow\infty}\frac{1-e^{\lambda/w}}{\lambda/w}=1$

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