In this Python problem we want to simulate the branching pro

In this Python problem we want to simulate the branching process X_n, n Greaterthanorequalto 0 with p_0 = 1/5 = 9_1, p_2 = 1/2, p_3 = 1/10. First note that X_n = sigma_k = 1^X_n - 1 xi_n, k, where xi_n, k represents the number of offspring of k-th individual in the n - the generation. Here {xi_n, k} are I.I.D. random variables with probability distribution P(xi_n, k = 0) = 1/5 = P(xi_n, k = 1), P(xi_n, k= 2) = 1/2, P(xi _ n, k = 3) = 1/. Generate the branching process X_n with the initial distribution X_0 = 1 until extinction. Simulate this branching process {X_n, n Greaterthanorequalto 1} 10000 times, and estimate the probability of extinction.

Solution

Define qn as the probability of ultimate extinction at the nth generation, i.e., qn         =              P ( X n = 0)         

                                                                                                                                                           

                                                                n ( s ) = P ( X n = k ) sk                                                            

                                                                k                                                                                             

                                                                = P ( X n =                                                                       

                                                                                0) + P ( X n = k ) sk                                        

                                                                                                k =1                                                       

                                                                = qn                                                                                  

                                                                                + P ( X n = k ) sk                                                             

                                                                                k =1                                                                       

                                                qn     = n (0)                                                                                               

                                                q1           = 1 (0) = (0) =           p0                                                          

Now,                                                     n ( s )                = (n1 ( s ))                                                

                                                                qn     = (qn1 )                                          (5.5)      

Now,                                     p0 = P(an object generates no similar objects)                                                  

So if p0 = 0          q1 = 0 , i.e., extin If p0 = 1 q1 = 1, q2 = 1, ... , i.e. the probability of no offspring is one and the extinction is certain to

occur right after the 0th generation.

Consider the case when 0 < p0 < 1. Now                ( s ) =                                                            sk            is a strictly increasing function of s so,                                              

                                p          k                                                                             

                                k                                                                                                                                                                             

                                                                                                                                                                                                               

                q   =                    q             =          (              p             0 )           > ( 0 ) = q                                        

                2                              (              1 )                                                                           1                                             

                                                                                                q        2                              >              q                                            

                                                                                                                                                                1                                             

Now, assume that qn   > qn1 . So                                                                                                                                                                                                           

                qn +1 = ( qn ) > ( qn1 ) = qn                                           

                                                                qn +1 >           qn                                                          

Thus by mathematical induction q1 <    q2 < q3 < ...         and this monotonically increasing sequence of {qn} is  

bounded above by 1, so there must exist an upper limit for the sequence, i.e., lim qn   = q (say); 0                       q 1 ,  

                                                                                                                                                                n                                   

which is the probability of ultimate extinction, and from (5.5), we find that q satisfies the equation      q             = ( q ) ,               

i.e., q is a root of the equation                                                                                                                                                                                                 

                s = ( s )                                                                                                                                                             (5.6)      

Now we show that q is the smallest positive root of the equation (5.6).

Let s0 be any other arbitrary positive root of (5.6). Then

q1 = ( 0 )         < (s0 ) = s0 .

Assume that qm   < s0   , then qm+1 = ( qm ) < (s0 ) = s0 . Hence, by induction

qn   < s0                n

q = lim            qn   s0

n    ction can never occur since the probability of no offspring is zero.         

Now, consider the graph of y = (s ) in 0 s 1.

It starts with the point (0,p0) and ends at the point (1,1). As P(s) is increasing function of s, so the curve is a convex curve, lying entirely in the first quadrant and intersecting the line y = s in at most two points. So (5.6) can have at most two roots one of which is unity.