Find the marginal and conditional densities without explicitly having the joint density?

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.

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

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 $1$.