Editing Branching process (section)

== Extinction problem for a Branching process==
The ultimate [[extinction probability]] is given by

:<math>\lim_{n \to \infty} \Pr(Z_n=0).</math>

For any nontrivial cases (trivial cases are ones in which the probability of having no offspring is zero for every member of the population - in such cases the probability of ultimate extinction is 0), the probability of ultimate extinction equals one if ''μ''&nbsp;≤&nbsp;1 and strictly less than one if ''μ''&nbsp;>&nbsp;1.

The process can be analyzed using the method of [[probability generating function]]. Let ''p''<sub>0</sub>,&nbsp;''p''<sub>1</sub>,&nbsp;''p''<sub>2</sub>,&nbsp;... be the probabilities of producing 0,&nbsp;1,&nbsp;2,&nbsp;... offspring by each individual in each generation. Let ''d''<sub>''m''</sub> be the extinction probability by the ''m''<sup>th</sup> generation, starting with a single individual (i.e. <math>Z_0=1</math>). Obviously, ''d''<sub>0</sub> = 0 and ''d''<sub>1</sub> = ''p''<sub>0</sub>. Since the probabilities for all paths that lead to 0 by the ''m''<sup>th</sup> generation must be added up, the extinction probability is nondecreasing in generations. That is,

:<math>0=d_0 \leq d_1\leq d_2 \leq \cdots \leq 1.</math>

Therefore, ''d''<sub>''m''</sub> converges to a limit ''d'', and ''d'' is the ultimate extinction probability. If there are ''j'' offspring in the first generation, then to die out by the mth generation, each of these lines must die out in ''m''&nbsp;−&nbsp;1 generations. Since they proceed independently, the probability is (''d''<sub>''m''−1</sub>) <sup>''j''</sup>. Thus,

:<math>d_m=p_0+p_1d_{m-1}+p_2(d_{m-1})^2+p_3(d_{m-1})^3+\cdots. \, </math>

The right-hand side of the equation is a probability generating function. Let ''h''(''z'') be the ordinary generating function for ''p''<sub>''i''</sub>:

:<math>h(z)=p_0+p_1z+p_2z^2+\cdots. \, </math>

Using the generating function, the previous equation becomes

:<math>d_m=h(d_{m-1}). \, </math>

Since ''d''<sub>''m''</sub> → ''d'', ''d'' can be found by solving

:<math>d=h(d). \, </math>

This is also equivalent to finding the intersection point(s) of lines ''y''&nbsp;=&nbsp;''z'' and ''y''&nbsp;=&nbsp;''h''(''z'') for&nbsp;''z''&nbsp;≥&nbsp;0. ''y'' = ''z'' is a straight line. ''y''&nbsp;=&nbsp;''h''(''z'') is an increasing (since <math>h'(z) = p_1 + 2 p_2 z + 3 p_3 z^2 + \cdots \geq 0</math>) and convex (since <math>h''(z) = 2 p_2 + 6 p_3 z + 12 p_4 z^2 + \cdots \geq 0</math>) function. There are at most two intersection points. Since (1,1) is always an intersect point for the two functions, there only exist three cases:[[File:Hgraph.png|thumb|Three cases of ''y'' = ''h''(''z'') intersect with ''y'' = ''z''.]]

Case 1 has another intersect point at ''z'' < 1 (see the red curve in the graph).

Case 2 has only one intersect point at ''z'' = 1.(See the green curve in the graph)

Case 3 has another intersect point at ''z'' > 1.(See the black curve in the graph)

In case 1, the ultimate extinction probability is strictly less than one. For case 2 and 3, the ultimate extinction probability equals to one.

By observing that ''h′''(1)&nbsp;=&nbsp;''p''<sub>1</sub>&nbsp;+&nbsp;2''p''<sub>2</sub>&nbsp;+&nbsp;3''p''<sub>3</sub>&nbsp;+&nbsp;...&nbsp;=&nbsp;''μ'' is exactly the expected number of offspring a parent could produce, it can be concluded that for a branching process with generating function ''h''(''z'') for the number of offspring of a given parent, if the mean number of offspring produced by a single parent is less than or equal to one, then the ultimate extinction probability is one. If the mean number of offspring produced by a single parent is greater than one, then the ultimate extinction probability is strictly less than one.