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So I'm looking to calculate the probability that a bacterial population dies out with the following conditions:

  • Initial population is 1
  • At each iteration the bacteria can die, do nothing, divide into 2 or divide into three. So each bacteria can change into either 0, 1, 2, 3 bacteria.
  • Each event has a 25% likelihood.

So I know from simulation that the likelihood of the population dying out is ~36.3%. However, I'm looking for a method to calculate this analytically.

Some MATLAB code for the simulation:

ntries = 100000; mnum = 10000; res = ones(ntries, 1);   for i = 1:ntries     while res(i) > 0 && res(i) < mnum         res(i) = sum(randi(4,res(i), 1)-1);     end end mean(res==0) 

Thanks,

Will

  • 0
    @joriki: Your reading abilities are impressive... Well done.2011-05-02

2 Answers 2

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Call $q_n$ the probability that the initial bacteria dies out at or before generation $n$ and condition by the size $i$ of the first generation. The event that every bacteria of this generation dies out at or before generation $n+1$ has probability $q_n^i$ and $i=0$, $1$, $2$ or $3$ with probability $1/4$, hence $q_{n+1}=g(q_n)$ with $ g(q)=q^0/4+q^1/4+q^2/4+q^3/4. $ Since $q_0=0$, $q_n=g\circ g\circ\cdots\circ g(0)$ ($g$ composed $n$ times). One can check, for instance by considering the graph of $g$ on the interval $(0,1)$, that $q_n$ converges to the smallest nonnegative root $q$ of the equation $q=g(q)$.

Now $g(q)=q$ if and only if $0=q^3+q^2-3q+1=(q-1)(q^2+2q-1)$, hence $q=1$ or $(q+1)^2=2$, that is $q=-1\pm\sqrt{2}$. Finally the population dies out with probability $q=\sqrt2-1\approx41\%$.

See this chapter of lecture notes by Steve Lalley, section 1.2.

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    @joriki: Yep. :-) Thanks.2011-04-28
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I find the probability to be $\sqrt{2}-1\approx 41.4\%$. If we let $p(n)$ be the probability that a population of $n$ dies out, we have $p(0)=1, p(n)=(p(n-1)+p(n)+p(n+1)+p(n+2))/4$. I first just did a numeric relaxation in Excel and recognized $p(1), p(2)=3-\sqrt{2}, p(3)=5\sqrt{2}-7.$ This was aided by the inverse symbolic calculator. The recurrence becomes $p(n+2)=-p(n+1)+3p(n)-p(n-1)$, whose characteristic polynomial has roots $1$, $\sqrt{2}-1$, and $-1-\sqrt{2}$. Finally we can check that $\sqrt{2}-1=(1+\sqrt{2}-1+3-2\sqrt{2}+5\sqrt{2}-7)/4$