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Here's the question

Let X and Y be independent random variables, where X is uniformly distributed over (2,4) and Y is exponentially distributed with mean 3. Find the density of U=X/Y

Here's what I got: (3/2)e^(-u/3)

I tried doing the PDF of both of them and divide them subbing in U for Y and it seems that's not the way to do it. I asked my professor for help and he told me that "I was on the right track and to keep thinking about it like that." However, I am at a complete loss as to what exactly I'm suppose to do.

Can anyone help me out with this one?

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    $\begingroup$ Since $X/Y$ roams over the interval $(0,\infty)$, the density cannot be the one you got, for that one does not integrate to $1$. $\endgroup$ Apr 18, 2016 at 22:59
  • $\begingroup$ So then how would I go about doing it? $\endgroup$
    – Nimbus
    Apr 18, 2016 at 23:05
  • $\begingroup$ I will write a partial answer. $\endgroup$ Apr 18, 2016 at 23:09

3 Answers 3

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Let $U=X/Y$ Then since $X$ and $Y$ are strictly positive.

$$\begin{align}F_U(u)~=~&\Pr(U\leq u) \\[1ex] ~=~& \mathsf P(Y\geq uX) \\[1ex] =~& \int_2^4 \mathsf P(Y\geq X/u\mid X=x)~f_X(x)\operatorname d x \\[1ex] =~& \tfrac 1 2\int_2^4 \mathsf P(Y\geq x/u)\operatorname d x\quad\mathbf 1_{u\in(0;\infty)} \\[1ex] =~& \tfrac 1 2\int_2^4 \mathsf e^{-x/3u}\operatorname d x\quad\mathbf 1_{u\in(0;\infty)} \\[3ex] f_U(u) ~=~& \dfrac{\operatorname d~F_U(u)}{\operatorname d~u\qquad}\end{align}$$

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  • $\begingroup$ I have a question as to what would be the best way to do it. Should I take the derivative of that final integral and apply the fundamental theorem of calculus? $\endgroup$
    – Nimbus
    Apr 19, 2016 at 0:51
  • $\begingroup$ I ended up getting Integral(2 to 4) of te^(-t/(3u))dt/(6u²) and then when I integrated this I got the right answer without even having to differentiate at all. $\endgroup$
    – Nimbus
    Apr 20, 2016 at 3:20
  • $\begingroup$ @Nimbus. Yes, that is possible because: $$\dfrac{\partial ~~}{\partial u}\int_2^4 1-F_X(t/u)\operatorname d t = \int_2^4\frac{f_X(t/u)}{2u^2}\operatorname d t$$ $\endgroup$ Apr 20, 2016 at 3:51
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Let $W=X/Y$. We find the cumulative distribution function $F_W(w)$ of $W$, that is, $\Pr(W\le w)$. Then one can differentiate to find the density.

The joint density function of $X$ and $Y$ is $\frac{1}{6}e^{-y/3}$ on the strip $2\le x\le 4$, $0\lt y\lt\infty$.

The probability that $W\le w$ is the probability that $X/Y\le w$, or equivalently the probability that $Y\ge \frac{X}{w}$.

Draw the line $y=\frac{x}{w}$. We want the probability that $(X,Y)$ lands in the part $K_w$ of our strip that is above the line $y=\frac{x}{w}$. This is $$\iint_{K_w} \frac{1}{6}e^{-y/3}\,dx\,dy.$$ To evaluate the double integral, it may be useful to make a sketch of the region $K_w$.

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  • $\begingroup$ would i want y to go from 0 to x and x from 2 to 4 or am I thinking about this wrong? $\endgroup$
    – Nimbus
    Apr 18, 2016 at 23:27
  • $\begingroup$ We are integrating over the region above $y=x/w$. But if you like, you can integrate over the part of our strip that is below the line. That may be geometrically more comfortable for you. Then whatever answer $g(w)$ we get, the cdf of $W$ is $1-g(w)$. If you choose to integrate over the finite region, it is $y$ going $0$ to $x/w$ (not $x$), and then $x$ going from $2$ to $4$. $\endgroup$ Apr 18, 2016 at 23:31
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    $\begingroup$ The cdf $F_W(w)$ that we are computing is a function of $w$, and the density function that we get after differentiating is a function of $w$. In your computation of the double integral over $K_w$, or, if you prefer, first over the part below $K_w$, treat $w$ as a constant. $\endgroup$ Apr 19, 2016 at 0:21
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    $\begingroup$ It is the usual substitution process, so if $u=x/(3w)$ then $dx=3w\,du$. Probability theory can be integration-heavy! $\endgroup$ Apr 19, 2016 at 0:50
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    $\begingroup$ P.S: If you know how to differentiate under the integral sign (but you may not), then you can bypass the second integration. $\endgroup$ Apr 19, 2016 at 0:53
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f(x) = 1/2 for 2≤x≤4 f(x) = 0 elsewhere (uniform distribution on [2,4])

g(y) = e^(-y/3)/3 for y≥0 g(y) = 0 elsewhere (exponential distribution with mean of 3)

The density function of u, h(u) is then calculated this way:

h(u) = Integral(-∞ to +∞)f(ux)g(x)|x|dx

g(x) = 0 if x<0 so our lower bound is 0. x will always be positive so |x| = x on our interval and:

h(u) = Integral(0 to +∞)f(ux)xe^(-x/3)dx/3

let us change the variable to t = ux dx = dt/u and: h(u) = Integral(0 to +∞)f(t)te^(-t/(3u))dt/(3u²) f(t) = 1/2 on [2,4] and 0 elsewere, so: h(u) = Integral(2 to 4)te^(-t/(3u))dt/(6u²) and when I plugged this into wolframalpha because I did NOT feel like actually integrating this by hand I got:

(e^(-4/(3 u))(-3 u+e^(2/(3 u))(3 u+2)-4))/(2 u) which was the correct answer.

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