Braching process and probabilities

[asset101]

Homework Statement

Consider the branching process with braching probabilities p0=1/2, p1=a, p2 = (([tex]\frac{3}{8}[/tex])-a)), p3 = 1/8 and pn = 0 otherwise, for some number a satisfying 0 [tex]\leq[/tex]a [tex]\leq[/tex]3/8

a) Find the probability generating function G(x).

b) Find the expected number of branches at a node.

c) Find those values of a for which G has two fixed points in the interval [0,1].

d) Find the probability of long-term survival when a = 0.

The Attempt at a Solution

(This attempt may be flawed and any correction would be appreciated)

The generating function is produced by the rule [tex]\sum(p_{n}x^{n})[/tex]

There for G(x) = (1/2)+(ax)+(([tex]\frac{3}{8}[/tex])-a))x[tex]^{2}[/tex]+[tex]\frac{x^{3}}{8}[/tex]

The remaining questions i am unsure with what to do any help would be appreciated.


Cheers

 

[mighty2000]

for sharing your problem with us! I can help you with the remaining questions:

b) To find the expected number of branches at a node, we simply need to take the derivative of the probability generating function G(x) at x=1. This is because the coefficient of x^n in G(x) gives the probability of n branches at a node, and taking the derivative at x=1 gives us the sum of all these probabilities, which is the expected number of branches. So, we have:

E(number of branches) = G'(1) = (a+(3/4)-a)/2 = 3/8

c) For G(x) to have two fixed points in the interval [0,1], it must have a point where G(x)=x and another point where G(x)=1. This means we need to solve the equations:

G(x) = x and G(x) = 1

Substituting G(x) = (1/2)+(ax)+((3/8)-a))x^{2}+(x^{3}/8) into the first equation, we get:

(1/2)+(ax)+((3/8)-a))x^{2}+(x^{3}/8) = x

Rearranging and simplifying, we get:

ax^{2}+((3/8)-a))x^{3} = 0

Since we want values of a for which G(x) has two fixed points, we can assume that x is not equal to 0. Therefore, we can divide both sides by x^{2} to get:

a+((3/8)-a))x = 0

This equation will have a solution for x=1 if a = 3/8 or a=1. So, the values of a for which G(x) has two fixed points in the interval [0,1] are a=3/8 and a=1.

d) To find the probability of long-term survival when a=0, we can use the formula for the extinction probability of a branching process, which is given by:

q = lim_{n \to \infty}G_{n}(0)

where G_{n}(0) is the nth derivative of G(x) evaluated at x=0. Since we already know the probability generating function G(x), we can find its nth derivative by using the general rule for the derivative of a