I am having problems with an exam question from a past paper, help would be appreciated:

Let $ X_n $ be the number of carriers of a family name in the $n$th generation and suppose $ X_0=a $. Suppose that the probability that a male individual is the progenitor of exactly $k$ boys is

$ p_k=(1-α)(1-p)p^{k-1} $, where $ k\ge1 $, $ p_0=α $

Find the probability generating function of $ X_1 $ and obtain the probability of ultimate extinction of the family name.

I believe this is a modified geometric distribution, and to find the pgf I have used

$ \pi(s)=\sum_{x\ge0}p_xs^x $

Therefore by letting $q=1-p$ I have found for this distribution:

$$ \pi(s)= p_0s^0 + \sum_{k\ge1}(1-α)(1-p)p^{k-1}s^k= α + (1-α)qs\sum_{k\ge1}(ps)^{k-1}, $$ that is, $$ \pi(s)= α + (1-α)qs\sum_{j\ge0}(ps)^{j} = α + (1-α)\frac{qs}{1-ps} $$

I am unsure if I have done this correctly this part and I am really stuck with where I go from this to find the probability of ultimate extinction.

Any help/hints/explanation for this question would be greatly appreciated!


The probability $z$ of extinction of a homogenous branching process starting from $1$ particle is the smallest nonnegative root of the equation $z=f(z)$, where $f$ denotes the generating function of the number of children of any individual. Here, $$f(s)=\alpha+(1-\alpha)(1-p)s/(1-ps),$$ hence $s=f(s)$ reads $ps^2-(\alpha+p)s+\alpha=0$, that is, $(s-1)(ps-\alpha)=0$, thus, $$z=\min\{\alpha/p,1\}.$$ The probability of extinction starting from $a$ particles is $z^a$.


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