Nuclear chain reaction, Poisson offspring

Question

Each neutron in a reactor triggers a random number of next-generation neutrons, distributed Poisson with mean $\lambda = 1.5$ , independently. The chain starts from one neutron.

What is the probability the chain reaction eventually dies out?

Accepted Answer

Let q be the extinction probability from one neutron. The offspring generating function for a Poisson(\lambda) count is g(s) = E\!\left[s^{	ext{offspring}}ight] = e^{\lambda(s - 1)}. Extinction is the smallest solution in [0, 1] of q = g(q): q = e^{1.5\,(q - 1)}. Solve by iteration. There is no closed form, but the map q \mapsto e^{1.5(q-1)} converges to the smallest fixed point when started from q0 = 0: q1 = e^{-1.5} = 0.223,\quad q2 = 0.312,\quad q3 = 0.356,\quad q4 = 0.381,\quad \dots The iterates climb and settle at q \approx 0.417. So the chain reaction fizzles out with probability ab…

Nuclear chain reaction, Poisson offspring

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