Question

Stats/Probability help...

X and Y are random variables, and X|Y=y ~ Poisson(y), and Y~Gamma(k, theta). Show that the marginal distribution of X is Neg.Binomial(k, p) with pmf \[ f(x)={-k \choose x}p^k(p-1)^x, x=0,1,2,...,\] and \(p = 1/(1+\theta)\)

Answer 1

sorry, over my level, but check out MITs opencourseware for Poisson dist. it may help

Answer 2

Oh just to be clear we define our gamma pdf to be, for random variable X:
\[\large \frac{x^{k-1}e^{-x/\theta}}{\Gamma(k)\theta^k}\]

Answer 3

I have done quite a bit on Poisson distribution.. but I find this question a lot harder than usual

Answer 4

yea, I'm sorry. I'm on the number theory abstract algebra side, but I will recommend @ganeshie8 and maybe @e.mccormick to help.

Answer 5

ok thanks though.. I will check MIT if they have some useful stuff

Answer 6

they have the entire course on a graduate level too I think, but it may tke some sifting. Good luck!

Answer 7

I can't seem to find my notes on this, so I'll try from memory...
\[f_{X}(x)=\int_{-\infty}^{x}f_{X|Y}(x|y)f_Y(y)~dy\]
where
\[f_{X|Y}(x|y)=\begin{cases}\dfrac{e^{-y}y^x}{x!}&\text{for }x=0,1,2,...\\\\
0&\text{otherwise}
\end{cases}\]
and
\[f_Y(y)=\begin{cases}\dfrac{\theta^k}{\Gamma(k)}y^{k-1}\exp(-y\theta)&\text{for }y>0\\\\
0&\text{otherwise}\end{cases}\]
... is any of this right so far? Not so sure about that integral.

Answer 8

I put the integral as \[ \int_0^{\infty}\] as it was integrating with respect with dy
The gamma pdf was \[\large \frac{y^{k-1}e^{-y/\theta}}{\Gamma(k)\theta^k}, y>0, and ~0~otherwise\]

Answer 9

Oh I think I must be mixing up pdf with cdf... That's not a form of the gamma distribution I'm familiar with, but that's easy to get over. (The one I learned is written in terms of \(\alpha=k\) and \(\beta=1/\theta\))

Answer 10

Let's see... Google gives me the following formula for a marginal pdf:
\[f_X(x)=\int_{-\infty}^\infty f_{X,Y}(x,y)~dy\]
the limits of which, like you said, reduce to \((0,\infty)\) due to the gamma distribution's support.

So it looks like you need to find the joint distribution... Now that, I know there's a formula for that. From the definition for conditional distributions,
\[f_{X|Y}(x|y)=\frac{f_{X,Y}(x,y)}{f_Y(y)}~~\iff~~f_{X,Y}(x,y)=f_Y(y)f_{X|Y}(x|y)\]
I hope I'm not being misled here, but it looks like you have
\[\large f_{X,Y}(x,y)=\begin{cases}\dfrac{y^{k-1}e^{-y/\theta}}{\Gamma(k)\theta^k}\dfrac{e^{-y}y^x}{x!}&\text{for }x,y>0\\\\
0&\text{otherwise}\end{cases}\]
So you have
\[\large \begin{align*}f_X(x)&=\int_{-\infty}^\infty \dfrac{y^{k-1}e^{-y/\theta}}{\Gamma(k)\theta^k}\dfrac{e^{-y}y^x}{x!}~dy\\\\
&=\frac{1}{x!\Gamma(k)\theta^k}\int_{0}^\infty y^{k-1}e^{-y/\theta}e^{-y}y^x~dy\\\\
&=\frac{1}{x!\Gamma(k)\theta^k}\int_{0}^\infty e^{-y([1/\theta]+1)}y^{x+k-1}~dy
\end{align*}\]
That looks like it's not too bad to compute, but looks can be deceiving...

Answer 11

(you can continue typing.. but ah yes I have this so far which is a good sign! )

Answer 12

You can do some rewriting with the scalar:
\[\Gamma(k)=(k-1)!~~\iff~~\frac{1}{\Gamma(k)}=\frac{1}{(k-1)!}=\frac{k}{k!}\]
As for the integral, integration by parts seems manageable:
\[\large\begin{matrix}u=y^{x+k-1}&&&dv=e^{-y([1/\theta]+1)}~dy\\
du=(x+k-1)y^{x+k-2}~dy&&&v=-\frac{1}{\frac{1}{\theta}+1}e^{-y([1/\theta]+1)}\\
&&&v=-\frac{\theta}{1+\theta}e^{\cdots}
\end{matrix}\]
Keep on reducing the power on the \(u\) term. Notice that the \([uv]_0^\infty\) term will disappear:
\[\large-\frac{\theta(x+k-1)}{1+\theta}\left[y^{x+k-2}e^{-y([1/\theta]+1)}\right]_0^\infty\to(0-0)=0\]

Answer 13

In the end, reducing the powers \(x+k-1\) times will give you an integral like this one:
\[\large\left(-\frac{\theta}{1+\theta}\right)^{x+k-1}(x+k-1)!\int_0^\infty e^{-y([1/\theta]+1)}~dy\]
And let's bring the Gamma scalar back into the mix (I'll ignore the rewrite of \(\Gamma(k)\), it might not be necessary):
\[\large\frac{(x+k-1)!}{x!(k-1)!\theta^k}\left(-\frac{\theta}{1+\theta}\right)^{x+k-1}\int_0^\infty e^{-y([1/\theta]+1)}~dy\]
\[\large\frac{(x+k-1)!}{x!(k-1)!\theta^k}\left(-\frac{\theta}{1+\theta}\right)^{x+k}\left[e^{-y([1/\theta]+1)}\right]_0^\infty\]
\[\large\frac{(x+k-1)!}{x!(k-1)!\theta^k}\left(-\frac{\theta}{1+\theta}\right)^{x+k}\left[0-1\right]\]
\[\large(-1)^{x+k+1}\large\frac{(x+k-1)!}{x!(k-1)!(1+\theta)^k}\left(\frac{\theta}{1+\theta}\right)^x\]
Are you given that \(p=\dfrac{1}{1+\theta}\)? If so, you can substitution where you can,
\[\large(-1)^{x+k+1}\large\frac{(x+k-1)!}{x!(k-1)!}p^k\left(\left(\frac{1-p}{p}\right) p\right)^x\]
\[\large(-1)^{x+k+1}\large\frac{(x+k-1)!}{x!(k-1)!}p^k(1-p)^x\]
\[\large(-1)^{2x+k+1}\frac{(x+k-1)!}{x!(k-1)!}p^k(p-1)^x\]
I'm not sure what to make of the (-1) factor, but then again, I've never seen a binomial coefficient like \(\dbinom{-k}x\).

Answer 14

The Wikipedia page on the negative binomial distribution says
\[\frac{(k+x-1)!}{k!(x-1)!}=(-1)^k\frac{(-x)(-x-1)\cdots(-x-k+1)}{k!}=(-1)^k\binom{-x}k\]
Hmm... What do you think?

Answer 15

Slight correction on the integral.

I think it's helpful to see how to reduce the power using this general form:
\[\large\int_0^\infty x^ne^{-cx}~dx\]
Substitute \(u=cx\) so that \(x=\dfrac{u}{c}\) and \(dx=\dfrac{1}{c}du\).
\[\large\frac{1}{c}\int_0^\infty \left(\frac{u}{c}\right)^ne^{-u}~du\\
\large\frac{1}{c^{n+1}}\int_0^\infty u^ne^{-u}~du\]
Integrating by parts \(n\) times gives
\[\large\frac{n!}{c^{n+1}}\int_0^\infty e^{-u}~du=\frac{n!}{c^{n+1}}\]
So here we use \(u=y\), \(n=x-k+1\), and \(c=\dfrac{1}{\theta}+1=\dfrac{1+\theta}{\theta}\).
\[\large\begin{align*}&\frac{1}{x!\Gamma(k)\theta^k}\int_0^\infty e^{-y^{(1+\theta)/\theta}}y^{x+k-1}~dy\\\\
&\frac{1}{x!\Gamma(k)\theta^k}\int_0^\infty e^{-u}\left(\frac{\theta}{1+\theta}u\right)^{x+k-1}\left(\frac{\theta}{1+\theta}~du\right)\\\\
&\frac{1}{x!\Gamma(k)\theta^k}\left(\frac{\theta}{1+\theta}\right)^{x+k}\int_0^\infty e^{-u}u^{x+k-1}~du\\\\
&\frac{(x+k-1)!}{x!\Gamma(k)\theta^k}\left(\frac{\theta}{1+\theta}\right)^{x+k}\int_0^\infty e^{-u}~du\\\\
&\frac{(x+k-1)!}{x!(k-1)!\theta^k}\frac{\theta^x\theta^k}{(1+\theta)^x(1+\theta)^k}\\\\
&\frac{(x+k-1)!}{x!(k-1)!}\frac{\left(\dfrac{1-p}{p}\right)^x}{p^{-x}p^{-k}}\\\\
&\frac{(k+x-1)!}{x!(k-1)!}(1-p)^xp^k\\\\
&\frac{(k+x-1)(k+x-2)\cdots(k+1)(k)}{x!}(1-p)^xp^k\\\\
&(-1)^x\frac{(-k-x+1)(-k-x+2)\cdots(-k-1)(-k)}{x!}(1-p)^xp^k\\\\
&\frac{(-k-x+1)(-k-x+2)\cdots(-k-1)(-k)}{x!}(p-1)^xp^k\\\\
&\binom{-k}x(p-1)^xp^k
\end{align*}\]
as desired.

Answer 16

oh thank you so much ! That was incredibly helpful !