Imagine an exotic particle that decays into any number of identical particles , with . So, for instance, when , the particle dissipates as energy; when , it's like nothing happens to the particle; when , the particle is effectively cloned with the help of the unlimited surrounding energy.
Now let be the number of a particle's offspring in the next step. Specifically, let be truncated Poisson variate with mean and possible values , which means where is between and and is an ordinary Poisson variate with mean
Let be the number of particles in the generation, with . What is the probability that this population of particles dies out eventually?
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First note that the number of particles at time n + 1 is a function of the number of particles at time n and some randomness, namely the sum of X n instances of K . This makes the process a Markov chain with states 0 through ∞ .
So one can in principle approximate the result by setting up the transition matrix of a finite Markov chain with some cut-off value. Let's say that value is 9 , the number of particles from where it's practically impossible for the population to die out since there are too many breeders. That makes the states 0 and 9 the two absorbing states in this 10-by-10 system with transition probabilities:
p 0 0 = 1 , p 9 9 = 1 p i j = P { k = 1 ∑ i K k = j } ; 0 < i < 9 , 0 ⩽ j < 9 p i 9 = P { k = 1 ∑ i K k ⩾ 9 } ; 0 < i < 9
Chance of absorption into state 0 from state 2 is 0 . 1 9 2 4 2 2 , which is correct to three decimal places.
One can calculate the answer more easily and precisely. Let us start with a single particle, so X 0 = 1 , and let's condition on the initial transition:
α = P { event D: population dies out ∣ X 0 = 1 } = j = 0 ∑ ∞ P { D ∣ X 1 = j , X 0 = 1 } p 1 j
Since each particle acts independently, the family, the dynasty if you will of the initial particle dies out eventually if all families of the first offspring die out eventually, so:
α = j = 0 ∑ 4 α j p 1 j = p 1 0 + α p 1 1 + α 2 p 1 2 + α 3 p 1 3 + α 4 p 1 4
Summing boundary is brought down to accommodate K . We are left with a polynomial with coefficients p 1 k = P { K = k } , and we are interested in the smallest positive root. Originally we have X 0 = 2 which means that not one, but two families should both die out eventually and that makes the final answer α 2 = 0 . 1 9 2 6 1 8 .