1
$\begingroup$

This is a homework question, I feel like I'm doing it right, but I can't seem to get the answer to match up. I have a uniform RV from 2 to 4, and an exponential with mean 4, so $X \sim \text{UNI}(2,4)$, $Y \sim \text{EXP}(4)$.

I'm looking for the density of $U=\dfrac{Y}{X}$. So $f(x)=\dfrac{1}{2}$, $f(y)=\dfrac{e^{-y/4}}{4}$, and $f(x,y)=\dfrac{e^{-y/4}}{8}$.

Now if I substitute $Y=Ux$, and multiply in a Jacobian ($\dfrac{\text{d}}{\text{d}u}ux=x$, this might be a mistake, I'm not completely clear on this), I can get $f(x,u)=\dfrac{xe^{-xu/4}}{8}$. The marginal density $f(u)$ is found then by integrating out the $x$, which I can do by turning that joint density into a gamma function, where I need to get a $\dfrac{16}{u^2}$ on the bottom, so I end up with $\dfrac{2}{u^2}$ outside of the gamma integral, and after the gamma integrates to 1, f(u)=$\dfrac{2}{u^2}$. Anyone see a mistake here?

$\endgroup$
  • 2
    $\begingroup$ Well, your method of proceeding is poor, though, regrettably, it is often taught as a recipe that can be followed without thinking too much about what's going on, and so I will not attempt to read through your work to see if there is a mistake. But the final answer should be testable. Since $Y \in (0,\infty)$, $Y/X$ takes on values in $(0,\infty)$ also. Is $2u^{-2}, u > 0$ a valid pdf? $\endgroup$ – Dilip Sarwate Apr 19 '14 at 22:32
  • $\begingroup$ Thanks. I'm kind of swimming in new material, and it's easy to overlook the obvious, like does it integrate to 1? $\endgroup$ – user143719 Apr 20 '14 at 2:40
2
$\begingroup$

Your answer is clearly wrong since the density does not integrate to $1$ across the positive reals, as Dilip Sarwate.

The way I would approach this would be to take your $f(x,y)=\dfrac{e^{-y/4}}{8}$ and look at $$\Pr\left(\frac{Y}{X} \le u\right)=\int_{x=2}^{4} \int_{y=0}^{ux} \dfrac{e^{-y/4}}{8} dy\, dy= \int_{x=2}^{4}\left(\frac12- \dfrac{e^{-ux/4}}{2}\right) dx $$ $$=1-\frac{2}{u}\left({e}^{-\frac{u}{2}}-{e}^{-{u}}\right),$$ which has the correct limits as $u \to 0$ and $u \to \infty$, and then take the derivative for the density.

$\endgroup$
  • $\begingroup$ Your method is how I would have done it. I'm led to believe that the assumptions to use the Jacobian method aren't satisfied, and my guess is that $U$ is not one-to-one with respect to $X$ or $Y$. Am I correct? $\endgroup$ – Clarinetist Apr 19 '14 at 22:50
  • 2
    $\begingroup$ @Clarinetist: I try not to use Jacobians or even their 1-dimensional equivalent, since I often find CDFs much easier to work with. $U$ is monotonic with respect to $X$ and with respect to $Y$ $\endgroup$ – Henry Apr 19 '14 at 22:57
1
$\begingroup$

You have assumed that $X$ and $Y$ are independent random variables and so I suppose that the actual problem you are solving is stating this somewhere, or it is taken from a section titled Independent Random Variables in some book.

For a (continuous) random variable $Y$ with density $f_Y(y)$, the density of $Z = aY$ is $$f_Z(z) = \frac{1}{|a|}f_Y\left(\frac za\right).$$ Thus, for $x > 0$, given that $X = x$, the conditional density of $U =\frac YX = \frac Yx$ is $$f_{U\mid X}(u \mid X=x) = xf_Y(xu) = \frac{x e^{-xu/4}}{4}.$$ Now find the unconditional density of $U$ using $$f_U(u) = \int_{-\infty}^\infty f_{U\mid X}(u \mid X=x)f_X(x)\,\mathrm dx = \int_2^4 \frac{x e^{-xu/4}}{4}\times\frac 12 \,\mathrm dx.$$

$\endgroup$
  • $\begingroup$ This integral gives the same result as the derivative of my expression, as it should. $\endgroup$ – Henry Apr 20 '14 at 8:53
  • $\begingroup$ @Henry Thanks for the confirmation. I wrote a different answer from yours (which I did upvote) to point out to the OP the implicit assumption of independence and also to give the OP a different approach to the problem. I am very much of the same view that you expressed in your comment to Clarinetist: that it is easier to work with the CDF instead of Jacobians. $\endgroup$ – Dilip Sarwate Apr 20 '14 at 11:54

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.