# Statistics: geometric distribution proof problem

1. Oct 4, 2011

### TeenieBopper

Statistics: geometric distribution "proof" problem

1. The problem statement, all variables and given/known data
If Y has a geometric distribution with success probability p, show that:

P(Y = an odd integer) = $\frac{p}{1-q^{2}}$

2. Relevant equations

p(y)=p(q)$^{2}$

3. The attempt at a solution

p(1)=pq^0
p(3)=pq^2
p(5)=pq^4
.
.
p(2k+1)=pq^2k

I also know the sum of a geometric series is basically $\frac{first-next}{1-ratio}$

Basically, I'm stuck on how to something set up so I can do some manipulation to eventually lead to P(Y = an odd integer) = $\frac{p}{1-q^{2}}$

2. Oct 4, 2011

### Bacle

Re: Statistics: geometric distribution "proof" problem

Would you explain your setup in more detail? Where are you drawing numbers from, the Reals, the Integers? How is the selection of numbers done?What is y? What is q? Sorry, I really don't understand what you're doing?

3. Oct 4, 2011

### LCKurtz

Re: Statistics: geometric distribution "proof" problem

You want the sum

p(1)+p(3)+p(5)+... = p + pq2 + pq4+...

This is an infinite geometric series. Do you know the formula for the sum of an infinite geometric series? All you need is the first term and the common ratio to figure it out.

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