# Finding the Joint PMF of X and X+Y

I have a question for class that says: Let X and Y be i.i.d. Geom(p), and N= X+Y. Find the joint PMF of X and N.

EDIT: The entire question, as someone requested: Let X and Y be Geom(p), and N=X+Y.

a) Find the joint PMF of X, Y, and N.

b) Find the joint PMF of X and N.

c) Find the conditional PMF of X given N=n.

Here is my thinking:

In a previous part of the problem I found that the joint PMF for X, Y, and N was $$p^2q^n$$ (where q is 1-p), and in theory I should be able to find the joint PMF of X and N by summing that PMF over all possible values for Y. But I don't know how to do that with a geometric distribution.

So instead I tried this approach: $$P(X=x,N=n) = P(X=x|N=n)P(N=n)$$

I found in my text that a negative binomial distribution can be represented as a sum of i.i.d. geometrics, so N should have a negative binomial distribution.

The PMF for a negative binomial is $$(^{n+r-1}_{r-1})p^rq^n$$, where r is the number of successes. I believe in this problem r=2 (because we have two variables). That would mean the PMF for this distribution is:

$$(^{n+1}_{n})p^2q^n$$; I believe the binomial coefficient works out to 1, so: $$P(N=n) = p^2q^n$$

I also know that: $$P(X=x|N=n) = \frac{P(X=x, N=n)}{P(N=n)}$$

However, I get stuck there. This feels circuitous because if I had the numerator, I would have the answer to my problem.

1. Does the work I have so far look correct?
2. Is this the approach to take? (If not, what is?)
3. Where do I go next?
• Could you please post the actual question? And also the second time you rearranged Bayes' rule you made a mistake with the denominator – Meeta Jo Nov 6 '18 at 22:30
• Ok, I have added the entire text of the question and fixed the mistake (I hope). – JStorm Nov 6 '18 at 22:48

## 1 Answer

You are overthinking.

The joint PMF of $$X,Y$$ is $$\mathbb{P}((X,Y)=(x,y))=p^2q^{x+y}\mathbf{1}_{(x,y)\in\mathbb{Z}^2, x\geq 0, y\geq 0}$$ (assuming geometric distribution count the number of failures, not the number of Bernoulli trials) so that gives $$\mathbb{P}(X=x,N=n)=\begin{cases} p^2q^n & x=0,1,2,\dots,n\\ 0 & \text{otherwise}. \end{cases}$$