branching process $A_1(z)=\frac{1}{2-z}$

Question

If $$ A_1(z)=\frac{1}{2-z}$$

Show that $$ A_n(z)=\frac{n-(n-1)z}{n+1-nz}$$.

This equation come from a branching process where A(z) is the common offspring distribution. I tried writing out the first few terms but was unable to find a pattern. Can someone show me how to quickly identify the pattern?

Answer 1

Hint: Use induction on $n$. Hover below to see my solution, but you should try it first.

Note that $A_1(z) = \frac{1}{2 - z} = \frac{1 - (1-1)z}{1 + 1 - 1\cdot z}$ so the base case of $n = 1$ is taken care of. Assume the formula holds for $A_n(z)$. Then \begin{align*} A_{n+1}(z) = A_1(A_n(z)) &= \frac{1}{2-z} \\ &= \frac{1}{2 - \frac{n-(n-1)z}{n+1 - nz}} \\ &= \frac{n+1 - nz}{2n + 2 - 2nz -n + (n-1)z} \\ &= \frac{n+1 - nz}{n+2 - (n+1)z} \end{align*}