So the question I am trying to solve is given below:
Find the extinction probability of the branching process generated by $ξ$ ∼ Bin$(2, p)$.
So I saw various approach has been used to solve this type of problem but I have few doubts. How I approached it is this way: Let the extinction probability obtained by the generating function $g(z)$ is given as $g(z)=\sum^{\infty}_{k=0}z^kP(ξ =k)$ which after substituting the values I get as follows:
$= 2(1-p)^2 + z2p(1-p)+z^2p^2$. After this, we have to equate it to zero to find the solution(That is what I am guessing). But I am not understanding how to proceed and I saw a lot of people are plotting graph which I do not understand properly. Moreover, I think the generating function I got is wrong. I saw the PGF for binomial distribution on the Wikipedia page, but I cannot implement it.
Please any help is appreciated. Thank you in advance.
You are asking multiple questions, please try to avoid this next time.
1.) The generating function of the binomial distribution ($\xi \sim Binom(n, p)$) $$g(z) = \sum_{k=0}^{\infty} z^k \mathbb{P}(\xi=k) = \sum_{k=0}^{n} \bigg(\begin{matrix} n \\ k \end{matrix}\bigg) (zp)^k (1-p)^{n-k} = (pz + (1-p))^n$$
2.) The branching process is defined as follows: $$X_0 = 1$$ $$X_{t+1} = \sum_{i=0}^{X_t} \xi_{i, t} ~\text{ , where }~ \xi_{i, t} \text{ is }\mathbb{N} \text{-valued i.i.d}$$ This means, that the generator function $$ g_{X_{t+1}}(z) = \sum_{k=0}^{\infty} z^k \mathbb{P}(X_{t+1} = k) = \mathbb{E}(z^{X_{t+1}}) = \mathbb{E}\big(z^{\sum_{i=0}^{X_t} \xi_{i, t}}\big) = \mathbb{E}\big[\mathbb{E}(z^{\xi} | X_t)^{X_t}\big] = g_{X_t}(g_{\xi}(z))$$ So $$ g_{X_{t}}(z) = (\underbrace{g_{\xi} \circ .. \circ g_{\xi}}_{t\text{-many}}) (z) \text{ , where $\circ$ is the composition operator}$$
What you want to know is the extinction probability, which is $\mathbb{P}(lim_{t \to \infty} X_t = 0) = ?$
First note that by the definition of the PGF $g_{X_{t}}(0) = \mathbb{P}(X_{t} = 0)$. Then using this identity
$$\mathbb{P}(lim_{t \to \infty} X_t = 0) = lim_{t \to \infty} \mathbb{P}(X_t = 0) = lim_{t \to \infty} (\underbrace{g_{\xi} \circ .. \circ g_{\xi}}_{t\text{-many}}) (0)$$
So basically you have an infinite composition at the end. The graph which you mentioned is presumably the plot of the PGF function of $\xi$ on the $[0, 1]$ interval. We know that the PGF function is convex, which means the plot will look like this.
Where the orange line is the PGF function, the black line is $y = x$. And $p_0 = \mathbb{P}(\xi = 0)$. The blue line shows iteratively how the infinite composition starting at zero behaves. If we have a fix point below 1 ($g_\xi(z) = z$, s.t. $z<1$), then $a_n = (\underbrace{g_\xi \circ ...\circ g_\xi}_{n\text{-many}})(0)$ converges to that fix point, which is the probability of extinction. If you don't have such fixed point, than it converges to $1$, so the population dies out in a finite amount of steps with probability 1.
You an apply this to your problem with binomially distributed $\xi$.
$$$$ Edit 1: Adding some additional steps to the generating function part.
Because of the law of total probability and independence of $\xi_{i,t}$, we have: $$\mathbb{E}\big(z^{\sum_{i=0}^{X_t} \xi_{i, t}}\big) = \sum_{n=0}^{\infty} \mathbb{E}\big(z^{\sum_{i=0}^{X_t} \xi_{i, t}} | X_t=n\big) \mathbb{P}(X_t=n)= \sum_{n=0}^{\infty} \mathbb{E}\big(z^{\sum_{i=0}^{n} \xi_{i, t}}\big) \mathbb{P}(X_t=n)$$ And because of $\xi_{i,t}$ being i.i.d, we have $$\mathbb{E}\big(z^{\sum_{i=0}^{n} \xi_{i, t}}\big) = \prod_{i=0}^{n}\mathbb{E}\big(z^{\xi_{i, t}}\big) = \mathbb{E}\big(z^\xi\big)^n$$ You put these together, and you get the result: $$\mathbb{E}\big(z^{X{t+1}}\big) = \sum_{n=0}^{\infty} \mathbb{E}\big(z^\xi\big)^n \mathbb{P}(X_t=n) = g_{X_t}(g_{\xi}(z))$$