The joint density for N and X is given by...

Question

anonymous · Answer

$$fNX(nx) = \frac{ 2x^n e^-3x }{ n! }$$ x>0 ; n=0,1,2,3.... 

a. Determine the marginal and conditional distributions.
b.  Are N and X independent?

Any help on how to approach this?

perl · Answer

I think if f(N)*F(X) = F(N*X)

kirbykirby · Answer

Notice that $N$ is discrete due to its support (integers) and $X$ is continuous since its support is a continuous interval $(0, \infty)$. So when finding the marginals, you should be careful when you are integrating or summing. 

Recall the marginal distribution for 2 distributions $X, Y$ which are both continuous. To find the marginal of $Y$, you compute:
$$ f_Y(y)=\int_{-\infty}^{\infty}f_{X,Y}(x,y) \,dx$$ And the bounds on the integral come from the support of $X$, because you're integrating with respect to $dx$. 

Thus: $f_N(n)$
$$f_N(n)=\int_{-\infty}^{\infty}f_{N,X}(n,x)\, dx $$
Here, you do an integral because the $X$ r.v. is continuous, and that's what you're integrating out, and the bounds are from the support of $X$.

$$\begin{align} f_N(n)&=\int_{0}^{\infty}f_{N,X}(n,x)\, dx\&=\int_0^{\infty}\frac{2x^ne^{-3x}}{n!}\, dx\&=\frac{2}{n!}\int_{0}^{\infty}x^ne^{-3x}\, dx  \end{align}$$
Notice how this looks a lot like the Gamma function, but you have a $-3x$ as the exponent for $e$ insead of just $-x$, so try this substitution:
$u = 3x \implies \dfrac{u}{3}=x$
$du = 3\, dx \implies \dfrac{du}{3}=dx$
Also, the bounds conveniently won't change. 
$$\begin{align} f_N(n)&=\frac{2}{n!}\int_{0}^{\infty} \left(\frac{u}{3}\right)^ne^{-u}\, \frac{du}{3}\&=\frac{2}{n!}\frac{1}{3^n}\frac{1}{3}\int_{0}^{\infty}u^ne^{-u}\, du \ \&=\frac{2}{3^{n+1}n!}\int_{0}^{\infty}u^{(n+1)-1}e^{-u} \, du  \&= \frac{2}{3^{n+1}n!}\Gamma(n+1)\&= \frac{2}{3^{n+1}}\end{align}$$
I used $\Gamma(n+1)=n!$ since the support of $n$ is discrete. 
---------
$$f_X(x)=\sum_{n=0}^{\infty}f_{N,X}(n,x) $$
Now we do a sum because we are summing out the r.v of $N$ and it is discrete, and the summation is over the support of $N$.

$$\begin{align} f_X(x)&=\sum_{n=0}^{\infty}\frac{2x^ne^{-3x}}{n!} 
\&=2e^{-3x}\sum_{n=0}^{\infty}\frac{x^n}{n!} \&=2e^{-3x}e^{x}\&= 2e^{-2x} \end{align} $$ using the fact that $\large  \sum\limits_{n=0}^{\infty}\frac{x^n}{n!} $ is the power series for $e^x$
------------
Finally, the conditional distributions should be easy to find as you just do a division.
$$f_{X|N}(x|n)=\frac{f_{N,X}(n,x)}{f_N(n)}  \
~ \
f_{N|X}(n|x)=\frac{f_{N,X}(n,x)}{f_X(x) }$$

kirbykirby · Answer

I kind of assumed you meant, too, that you had $e^{-3x}$ and not $e^{-3}x$ since the integral wouldn't converge otherwise.

anonymous · Answer

Wow great explanation.  I believe my professor did something like this but your explanation is much clearer.  I got intimidated with the complicated looking joint density and completely missed that one part was discrete and the other continuos.  You're amazing thanks for the help once again.

And you are right it is e^-3x sorry about that.

kirbykirby · Answer

yw :)