In a given population, each individual has a number of offspring $Y$ with discrete Uniform distribution on $\{0, 1, \ldots, N \}$, for some fixed $N \ge 1$, i.e. $P(Y = k) =\frac{1}{N+1}$ for all $k = 0, 1, \ldots, N$. Let $\pi_N$ denote the probability of eventual extinction of this population when it starts from one individual.
a. Write down an equation solved by $\pi_N$.
b. For each fixed $N$, give the number of solutions that this equation has in the interval $[0, 1]$. Give a short explanation. If there is more than one solutions, indicate which one is $\pi_N$.
For part a, is it something along the lines of $\ \pi_N = G(\pi_N)$? I'm kind of lost.
If the first individual has $k\ne 0$ offspring then the eventual extinction probability is $\pi_0^k$. So $\pi_k$ satisfies
$\pi_0=\frac{1}{N+1}+\sum_{k=1}^N\frac{\pi_0^k}{N+1}$.
This simplifies to $(N+1)\pi_0=\sum_{k=0}^N \pi_0^k$.
$\pi_0=1$ is one of the solutions.
$f(\pi_0)=\left(\sum_{k=0}^N\pi_0^k\right)-(N+1)\pi_0$ factorises as $f(\pi_0)=(\pi_0-1)(\pi_0^{N-1}+2\pi_0^{N-2}+3\pi_0^{N-3}+...+(N-1)\pi_0-1)$.
Let $g(\pi_0)=\pi_0^{N-1}+2\pi_0^{N-2}+3\pi_0^{N-3}+...+(N-1)\pi_0-1$. When $N=1$, $g$ has no roots. When $N=2$, $g$ has one root at $\pi_0=1$.
Now consider the case $N>2$. $g$ has at least one root in $[0,1]$ since $g(0)<0$ and $g(1)>0$. In fact, $g$ cannot have more than one root in $[0,1]$ since $g'(\pi_0)>0$ for all $\pi_0>0$.
So there is one solution when $N=1,2$ and there are two solutions when $N>2$.