Generating functions in the branching process.

  • Thread starter Thread starter stukbv
  • Start date Start date
  • Tags Tags
    Functions Process
stukbv
Messages
112
Reaction score
0

Homework Statement



I am told that I have particles which each yield a random number of offspring of known distribution independently from each other and from the past generations.

Xn is the number of particles in the nth generation
The distribution of a typical family size is Z and so Xn is the sum of appropriate Zi's

I need a generating function of the number Xn in the nth generation.


2. The attempt at a solution

I know that Fn+1(s) = E [sXn+1]
from the definition of generating functions and how to derive them.

But my lecturer then goes on to say that = ƩE[sXn+1|Xn=j] * P[Xn=j ]

Summed over j.

How does he get from one to the other? If I can make this link then I can go on to show what I need to!

Thank you!
 
Physics news on Phys.org
stukbv said:

Homework Statement



I am told that I have particles which each yield a random number of offspring of known distribution independently from each other and from the past generations.

Xn is the number of particles in the nth generation
The distribution of a typical family size is Z and so Xn is the sum of appropriate Zi's

I need a generating function of the number Xn in the nth generation.2. The attempt at a solution

I know that Fn+1(s) = E [sXn+1]
from the definition of generating functions and how to derive them.

But my lecturer then goes on to say that = ƩE[sXn+1|Xn=j] * P[Xn=j ]

Summed over j.

How does he get from one to the other? If I can make this link then I can go on to show what I need to!

Thank you!

It's just a standard result in Probability. Suppose \{A_k \} is a partition of the sample space \Omega, meaning that the A_k are disjoint and their union is Ω. Then, for any discrete random variable B we have \Pr \{B=j\} = \sum_k \Pr\{B=j|A_k\} \Pr \{A_k\}. Thus, for any f >= 0 we have
E f(B) = \sum_j f(j) \Pr\{B=j \} = \sum_k \Pr\{A_k\} \sum_j f(j) \Pr\{B=j|A_k\}<br /> =\sum_k E[f(B)|A_k] \Pr\{A_k\},
where I have swapped the order of summation, which is OK for a positive function.

RGV
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top