1
$\begingroup$

Take a random variable $Y$. You're given $f_Y(x)=xe^{-x}$ when $x>0$. Given $Y$, a new random variable, $M$, is uniform over the interval $(0,Y)$.

I need to calculate the marginal density for $Y$, the conditional density for $M$ given $Y$, and I must show that $M$ and $(M-Y)$ are independent, then proceed to find their joint density function.

I am confused when it comes to finding such functions without explicitly having the joint density function. I'm accustomed to calculating a marginal density as:

$\int_{-\infty}^{\infty}f_{X,Y}(x,y)dxdy$, for instance.

The conditional density that I know is:

$f_{(X|Y)}(x|y)=\frac{f_{X,Y}(x,y)}{f_{Y}(y)}$.

I'm struggling to get started without the joint density. Any hints are appreciated.

$\endgroup$
  • $\begingroup$ The conditional density of $M$ is $1/y$. They tell you it is uniform $(0,Y)$. $\endgroup$ – Em. Feb 2 '16 at 23:25
  • $\begingroup$ Oops - I mean the joint density! Edited to fix that. $\endgroup$ – Taylor Feb 2 '16 at 23:26
2
$\begingroup$

It appears that your end goal is to find the unconditional distribution of $M$, \begin{align*} f_M(m) = \int_m^{\infty} f_{M,Y}(m,y)\,dy=\int_m^{\infty} f_{M|Y}(m|y)f_Y(y)\,dy = \int_m^\infty \frac{1}{y}\cdot ye^{-y}\,dy = e^{-m} \end{align*}

This is unconditional distribution of $M$; it is an exponential distribution with mean $1$.

$\endgroup$
  • $\begingroup$ That's a helpful computation for me to see; thank you. From that, though, I'm still not quite sure how to compute the other quantities. For instance, how do I find the joint density from this? Given the joint density, I would know how to find the conditional and marginal densities, but I'm not sure about this first step. $\endgroup$ – Taylor Feb 3 '16 at 0:34

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.