Statistics: geometric distribution proof problem

[TeenieBopper]

Statistics: geometric distribution "proof" problem

Homework Statement

If Y has a geometric distribution with success probability p, show that:

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

Homework Equations

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

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) = $\frac{p}{1-q^{2}}$

 

[Bacle]

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?

 

[LCKurtz]

Homework Statement

If Y has a geometric distribution with success probability p, show that:

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

Homework Equations

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

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) = $\frac{p}{1-q^{2}}$

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.