0
$\begingroup$

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 $G_n(s) = E[ s^{X_n} ] = \sum^\infty_{i=0} s^i P( X_n = i).$

so

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

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

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

$\endgroup$

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.