I'm looking at question 1 here : http://www.rss.org.uk/Images/PDF/pro-dev/Exam%20past%20papers/2014/rss-grad-diploma-module3-2014.pdf .

It's about a branching process, the probability generating function is: $G_n(s) = \frac{n - (n-5)s}{5 - n(s-1)}, n = 0,1,2..$

I've called the size of population in generation $n$ $X_n$. Population is extinct when $X_n = 0$.

The probability of extinction for a given generation (I think) is $G_n(0)$, as $G_n(s) = E[ s^{X_n} ] = \sum^\infty_{i=0} s^i P( X_n = i).$


$G_n(0) = P(X_n = 0)$.

$G_n(0) = \frac{n}{5+n}$

Part (v) of the question asks of the probability that extinction occurs in generation $T$ . I reasoned that $P( first \space extinct \space T=n) = P (X_n = 0) \Pi^{n-1}_{k=1} P(X_n > 0)$ i.e. Probability extinct when T = n x probability not extinct previously.

The question says this should be $P(T = n) = \frac{5}{(n+5)(n+4)}$

but I reasoned it was $\frac{5}{n+5} \cdot \frac{5^{n-1} 5!}{(n-1+5)!}$ based on $P(X_n > 0 ) = 1 - P(X_n = 0) = \frac{5}{5+n}$ and then multiplying these together for 1..$n-1$.

Could anyone suggest where I am going wrong? Thanks, Chris


1 Answer 1


This was tricky, but you explicitly go wrong because you assume $P(X_{n}=0,\cap_{i=1}^{n-1}X_{i}>0)=P(X_{n}=0)\prod_{i=1}^{n-1}P(X_{i}>0)$, that is you assume independence. But taking even the example of $T=2$ \begin{equation} P(T=2)=P(X_{2}=0,X_{1}>0)=P(X_{2}=0|X_{1}>0)P(X_{1}>0) \end{equation} we can reason that $P(X_{2}=0|X_{1}>0)\neq P(X_{2}=0)$ as $P(X_{2}=0)$ must take into account the events where all the individuals died out in a previous generation.

The correct approach is to note that \begin{equation} P(X_{n}=0)=P(\cup_{i=1}^{n}T=i) \\=P(T=n)+P(\cup_{i=1}^{n-1}T=i) \\=P(T=n)+P(X_{n-1}=0). \end{equation} The first equality is as we don't care when the population dies out when considering $X_{n}$. The second is due to inclusion-exclusion and that $P(T=n\cap_{i=1}^{n-1}T=i)=0$.

This gives \begin{equation} P(T=n)=P(X_{n}=0)-P(X_{n-1}=0)=\frac{n}{5+n}-\frac{n-1}{5+n-1} \end{equation} which gives the required result.

As I said tricky.

  • $\begingroup$ brilliant, thank you. I'm afraid I was fumbling around with this. Thank you for taking the time to give such a full answer. $\endgroup$
    – clumb
    Aug 31, 2015 at 20:14

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