1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Statistics: geometric distribution proof problem

  1. Oct 4, 2011 #1
    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) = [itex]\frac{p}{1-q^{2}}[/itex]


    2. Relevant equations

    p(y)=p(q)[itex]^{2}[/itex]

    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 [itex]\frac{first-next}{1-ratio}[/itex]

    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) = [itex]\frac{p}{1-q^{2}}[/itex]
     
  2. jcsd
  3. Oct 4, 2011 #2
    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?
     
  4. Oct 4, 2011 #3

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook