Joint Distribution, Geometric Distribution

Question

Let X and Y be IID Geom(p) (independent and identically distributed geometric probability) and $N=X+Y$. Find the joint PMF (probability mass function) of $X,Y,N$.

I know the solution to this but I could not come up with it myself, and even after looking the solution up im not sure my reasoning is right.....

So the solution is $$f(x,y,n)=f(N=x+y |X=x, Y=y) =1$$ therefore $$f(x,y,n)=f(x,y)= pq^x *pq^y = p^2 * q^n$$

My reasoning for this is that if I Conidtion on x, y... they are true so if x and y are true than $n = x+y$ has to be true with $p=1$.

I know my explanation may sound stupid, i just started with probability after a i haven’t done any math for years.

I've also tried to reason mathematically but didn’t come very far... $$ f_x(x)=pq^{x-1}\\ f_y(y)=pq^{y-1} $$ $f_n(n=X+Y)$ should be neg. Binomial $Nbin(2,p)$? $\binom{n+1}{1}p^2q^n$

That is not very conclusive, and in particular, why is there a $q^x$ and not $x-1$? There is so much i don’t get and unfortunately my books don’t help, So many thanks for taking the time and reading this

Answer 1

Suppose that $X,Y \sim Geometric(p) $

The joint pdf of two random variables is

$$f_{X,Y}(x,y) = f_{X}(x) f_{Y}(y) $$

Now the probability mass function the geometric random variable is given by

$$ p(k) = P(X=k) = (1-p)^{k-1} p $$

then we have

$$ f_{X,Y}(x,y) = ((1-p)^{x-1}p)(1-p)^{y-1}p $$ $$ f_{X,Y}(x,y) = (1-p)^{x+y-2}p^{2} $$

Then we have

$$N= X+Y \sim NegBin(2,p) $$

The probability mass function for the negative binomial given by

$$ X\sim NegBin(r;p) $$

$$f(k;r,p) = Pr(X=k) = \binom{k+r-1}{k} p^{k}(1-p)^{r} $$ In the geometric distribution we had our mass function $$ Pr(X=k) = (1-p)^{k-1}p $$ then our number of failures are $$x+y-2 $$ $$ N = \binom{x+y-2+2-1}{x+y-2}p^{x+y-2}(1-p)^{2} $$ $$ N = \binom{x+y-1}{x+y-2}p^{x+y-2}(1-p)^{2} $$

Then we have

$$f_{N,XY}(n,xy) = f_{N}(n)f_{X,Y}(x,y) = \binom{x+y-1}{x+y-2} p^{x+y-2}(1-p)^{2} (1-p)^{x+y-2}p^{2} $$

Does this work?

Answer 2

1) Why is that $q^{x}$ but not $q^{x-1}$?

You need to look at the definition of the book that you use. Some books define $X\sim Geom(p)$ as the total number of steps include the success. While some books define as the number of failures only (do not count the success). In this solution, it seems your book uses the later definition. Thus the total steps will be x failures. Therefore:

$f_{X}=pq^{x}$ (x failures + 1 success)

$f_{Y}=pq^{y}$ (y failures + 1 success)

2)

You mistaken the probability of joint distribution of $(X,Y)$ with the distribution of N, the total steps.

The joint distribution of $(X,Y)$ tells the probability of one specific event $(X,Y)$ that satisfies $X + Y = N$:

$f_{X,Y}(x,y,n) = P(X=x,Y=n-x) = f_{X}(x)f_{Y}(n-x) $

$f_{X,Y}(x,y,n) = pq^{x}pq^{n-x} = p^{2}q^{n}$

While the distribution of N tells the probability of all events $(X,Y)$ that satisfy $X+Y=N$:

$f_{N}(n) = P(X+Y=N) = \sum_{x=0}^{n}P(X=x)P(Y=n-x) = \sum_{x:0}^{n} pq^{x}pq^{n-x} = (n+1)p^{2}q^{n}$

3) Another solution is to use Moment Generating Function. You can look it up.