Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Showing tha a random variable is a martingale

  1. May 22, 2013 #1
    I'm having a bit of a problem proving the second condition for a martingale, the discrete time branching process Z(n)=X(n)/m^n, where m is the mean number of offspring per individual and X(n) is the size of the nth generation.

    I have E[z(n)]=E[x(n)]/m^n=m^n/m^n (from definition E[X^n]=m^n) = 1
    which is less than infinity, so first condition passes

    Then I get lost with E[Z(n+1) given X(1),X(2)....X(n)]...any clues on how to show this is equal to Z(n)? Thanks
     
  2. jcsd
  3. May 23, 2013 #2

    DrDu

    User Avatar
    Science Advisor

    Apparently E[X(n+1)|X(n)]=mX(n) as each individuum in the population has on the mean m offsprings.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook