Generating functions in the branching process.

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SUMMARY

The discussion focuses on generating functions in the context of branching processes, specifically for calculating the number of particles in the nth generation, denoted as Xn. The key relationship established is Fn+1(s) = E[sXn+1], which transitions to the expression ƩE[sXn+1|Xn=j] * P[Xn=j] through the law of total expectation. This derivation is crucial for linking the generating function to the distribution of family sizes, represented by Z, and understanding the behavior of Xn as a sum of independent random variables.

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stukbv
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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!
 
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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:

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