Showing tha a random variable is a martingale

[rickywaldron]

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

 

[DrDu]

Apparently E[X(n+1)|X(n)]=mX(n) as each individuum in the population has on the mean m offsprings.