# Galton-Watson process with geometric offspring distribution and finite expected depth

Consider the Galton-Watson tree with offspring distribution $$X$$ given by $$P(X=k) = (1-p)^kp$$. Let $$D$$ be the depth of the process. My question is how to calculate for which values of $$p$$ we have $$\mathbb{E}D<\infty$$.

Explanation of terms: a Galton-Watson tree consists of generations. For example the first generation consists of $$2$$ people (probability $$p(1-p)^2$$), the second one consists of the children of these two and their amount is also geometric($$p$$) distributed, and so on, so they continue multiplying in a geometric way.

The depth is defined by the largest generation which has $$>0$$ children. A result given in my lecture notes is that $$P(D\leq t) = G\circ ...\circ G(s)(0)$$ where there are $$t$$ times a $$G$$, and $$G(s)$$ is the pgf. So here it looks like a continued fraction.

The lecture's first question is to calculate $$P(D=t)$$, so I did $$P(D\leq t) - P(D. The next question is for which values $$\mathbb{E}D<\infty$$. I know that for $$p>1/2$$ the extinction probability is $$1$$, but unfortunaltely I calculated that for $$p=1/2$$, $$\mathbb{E}D=\infty$$, otherwise it was done.

Is there anyone who can help? Thanks in advance.

• Yeah I thought about that, but $\sum_{i=0}^{\infty} P(D>i)$ is too complicated too calculate for all $p$. For $p=1/2$ however, the sum is infinite. – Rocco van Vreumingen Mar 21 at 13:19
• I got an idea. It is for $p>1/2$, because then the average number of children will be $(1-p)/p<1$, so on average the number of children in the next generation decreases non-asymptotically. – Rocco van Vreumingen Mar 22 at 11:27
• Got it. The expected number of children in the $n$-th generation given $k$ in the first generation, $\mathbb{E}(Z_n|Z_1=k)$ say, is of the form $kc^{n-1}$ where $c=\mathbb{E}X = (1-p)/p<1$. So $P(D<n|Z_1=k)=P(Z_n=0|Z_1=k)\geq 1-kc^{n-1}$ and now we can derive the expectation $\mathbb{E}D$ – Rocco van Vreumingen Mar 22 at 15:54