a Galton-Watson riddle

riddleThe Riddler of this week has an extinction riddle which summarises as follows:

One observes a population of N individuals, each with a probability of 10⁻⁴ to kill the observer each day. From one day to the next, the population decreases by one individual with probability

K√N 10⁻⁴

What is the value of K that leaves the observer alive with probability ½?

Given the sequence of population sizes N,N¹,N²,…, the probability to remain alive is

(1-10^{-4})^{N+N^1+\ldots}

where the sum stops with the (sure) extinction of the population. Which is the moment generating function of the sum. At x=1-10⁻⁴. Hence the problem relates to a Galton-Watson extinction problem. However, given the nature of the extinction process I do not see a way to determine the distribution of the sum, except by simulation. Which returns K=27 for the specific value of N=9.

N=9
K=3*N
M=10^4
vals=rep(0,M)
targ=0
ite=1
while (abs(targ-.5)>.01){

for (t in 1:M){
  gen=vals[t]=N
  while (gen>0){
   gen=gen-(runif(1)<K*sqrt(gen)/10^4)
   vals[t]=vals[t]+gen}
  }
  targ=mean(exp(vals*log(.9999)))
print(c(as.integer(ite),K,targ))
  if (targ<.5){ K=K*ite/(1+ite)}else{
     K=K/(ite/(1+ite))}
  ite=ite+1}

The solution proposed on The Riddler is more elegant in that the fixed point equation is

\prod_{J=1}^9 \frac{K \cdot \sqrt{J}}{K \cdot \sqrt{J} + J}=\frac{1}{2}

with a solution around K=27.

2 Responses to “a Galton-Watson riddle”

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