The Galton Watson Process – Part II


By rewriting Eq.19 for generation n – 1 we obtain e_{n-1} = G_{n-1} \left( 0 \right). Then by using these two equations and Eq.18 we obtain the following important relation:

e_n = G_n \left( 0 \right) = G \left(G_{n-1} \left( 0 \right) \right) = G \left( e_{n-1} \right) \, \, \, \, \, \, \, \, (20)


Previously, we realized that G_1(s) = G(s) so using this in Eq. 20, we can explicitly write out the probability of extinction for the first couple of generations:

e_1 = G_1 \left( 0 \right) = G \left( 0 \right) \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,\, \, \, \, \, \, \, \, (21) \\ e_2 = G_2 \left( 0 \right) = G \left( e_1 \right) = G \left( G \left( 0 \right) \right)\, \, \, \, \, \, \, \, \, \,\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (22) \\ e_3 = G_3 \left( 0 \right) =G \left( e_2 \right) = G \left( G \left( G \left( 0 \right) \right) \right)\, \, \, \, \, \, \, \, \, \,\, \, \, \, \, (23)

This is the general form for the probability of extinction in terms of probability distribution functions. In order to make this more concrete we apply the Poisson distribution to the probability of having a certain number of children, i.e. G(s) = \exp{\left( \lambda (s - 1) \right)}. Therefore,

e_1 = G \left( 0 \right) = \exp{\left(- \lambda \right) } \\ e_2 = G \left( e_1 \right) = \exp{ \left( \lambda \left(\exp{ \left( -\lambda \right) } - 1\right) \right)} \\ e_3 = G \left( e_2 \right) = \exp{\left( \lambda \left( \exp{ \left( \lambda \left(\exp{ \left(-\lambda \right) } - 1\right) \right)} - 1 \right) \right)}

Since there are only two possibilities for a generation, either extinction or survival then we can say:

\mbox{Probability of survival} = 1 - \mbox{Probability of extinction}

The probability of survival for up to 1000 generations is then calculated with a spreadsheet software and plotted below for different expected values (i.e. \lambda) for the Poisson distribution:


Figure 2:The probability of survival for a given generation using different expectation values, \lambda, in the Poisson distribution.


All this hard work was to create this figure which will be used as the basis for comparison against the simulations in the next part. One interesting result is that if \lambda = 1, the probability of survival eventually drops to 0. Values of \lambda above 1 all have some finite probability that the lineage might continue on indefinitely. For values of \lambda less than 1, all will die after some finite number of generations.


This concludes the theory portion of the Galton Watson process. In Part 3 I will discuss how to build a Monte Carlo simulation of the Galton Watson process using Python, and compare the results to theory.

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