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Beginning with a population of $n_0$ individuals, let each individual have a probability $p$ to survive until it replicates into two independent and identical individuals, where $p<\frac12$.

It follows that the population goes extinct in the long-run with probability 1, and the expected number of replications before extinction is $n_0p/(1-2p)$.

The population must reach some maximum size $N$ before going extinct, where $N\geq1$. What is the expected value of this maximum size $N$?

Alex
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    What have you tried? I'm very interested in this problem. – Unit Apr 11 '19 at 13:02
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    @Unit, I think that the probability that there are $x$ replications before extinction is $C_x p^x(1-p)^{x+1}$, where $C_x$ is $x$th Catalan number. Then, given that there are $x$ replications, the expectation (and cumulative mass function) for the maximum size $N$ seems to be given by this excellent paper, which considers a different but seemingly equivalent problem: http://www.cs.tau.ac.il/~nachumd/papers/CatalanHeight.pdf – Alex Apr 11 '19 at 14:14
  • I should add that my comment applies only to the case where $n_0=1$. – Alex Apr 11 '19 at 14:18

1 Answers1

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I have written this $\texttt{R}$ code to compute a point estimate for the mean and variance of $N$ given fixed values for $p$ and $n_0$:

rm(list=ls())

N <- 10000
maxpop <- numeric(N)
p <- 0.3
n_0 <- 2

for(n in 1:N) {
  pop <- n_0
  maxpop[n] <- n_0
  while(pop > 0) {
    if(runif(1)>p) {
      pop <- pop - 1
    }
    else {
      pop <- pop + 1
      maxpop[n] <- max(maxpop[n], pop+1)
    }
  }
}

print(mean(maxpop))
print(var(maxpop))

Perhaps it can be useful.

Math1000
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  • Thanks for this useful approach! Using this simulation method for the case $n_0=1$, I have found that $N\approx 1+p$ for small $p\approx 0$, and $N\approx-\ln(1-2p)$ for $p\approx \tfrac12$. – Alex Sep 30 '24 at 22:20