# Branching Process Extinction Probability

I'm doing a branching process problem and am not sure I did it correctly. Suppose $X_0 = 1$ and $p_0 = .5, p_1 = .1,$ and $p_3 = .4$ represent the probabilities that zero, one, and three individuals will be produced, respectively. There are two parts to the question:

a) Find $P(X_2 = 0 | X_1 > 0)$

b) Find $P(X_3 = 0 | X_2 > 0)$.

I imagine the solutions to these problems are very similar.

To solve part a), I first found the expected number of individuals produced by the zeroth generation to be $1.3$. Next, I found the extinction probability $a$ by solving the equation

$$a = .5 + .1a + .4a^3,$$

where $a$ came out to be roughly $0.725$. All I could think of doing was to solve for $$P(X_2 = 0 | X_1 = E[X_1]),$$ where $$E[X_1] = 1.3.$$ To solve this, I know that you can simply calculate $a^{1.3} = .658$. However, this doesn't feel like it's the right approach since it doesn't cover all possible values of $X_1$; it only takes into account the expected value at time $t = 1$. I figure one could correct this error by solving

$$P(X_2 = 0 | X_1 > 0) = \frac{P(X_2 = 0 \cap X_1 > 0)}{P(X_1 > 0)}$$

by Bayes' Theorem, but unfortunately I don't know how one would go about computing that. Is anything I have here at least remotely correct, or am I just throwing out jibber-jabber?

Side note: For part b), I raised the extinction probability to the expected number of individuals are time $t = 2$ $(E[X_2] = 2.6)$ and got $a^{2.6} = .433$. Again, I'm not so sure this is correct, but it was my best attempt.

Thanks!

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Given that $X_1>0$ there can be $1$ or $3$ children on generation $1$, so you could use conditioning on the number of children on generation $1$ to get $a)$. This would lead to \begin{align} P(X_2=0)&=P(X_2=0|X_1=0)P(X_1=0)+P(X_2=0|X_1=1)P(X_1=1)+P(X_2=0|X_1=3)P(X_1=3)\\[5pt] &= \frac{1}{2}+\frac{1}{2}\cdot\frac{1}{10}+\frac{1}{2^3}\cdot\frac{4}{10}\\[5pt] &=\frac{3}{5}. \end{align}

Now you can rewrite this as \begin{align*} P(X_2=0)&=P(X_2=0\,|\,X_1=0)P(X_1=0)+P(X_2=0\,|\,X_1>0)P(X_1>0)\\[5pt] \frac{3}{5}&=\frac{1}{2}+P(X_2=0\,|\,X_1>0)\frac{1}{2} \end{align*} which you can isolate to find $P(X_2=0\,|\,X_1>0)=\frac{1}{5}$.

You can do the same for $b)$, it only takes a few more cases.

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You've calculated $P(X_2=0, X_1>0)$ not $P(X_2=0\,|\, X_1>0)$. –  Byron Schmuland Sep 25 '12 at 13:44
You were right thanks! Fixed it. –  Jean-Sébastien Sep 25 '12 at 15:32
This is great, thanks! –  jrad Oct 10 '12 at 6:03

The function $\phi(t)={1\over 2}+{t\over 10}+{2 t^3\over 5}$ will prove very useful, since $\mathbb{P}(X_n=0)$ is equal to $\phi(\phi(\cdots(\phi(0))\cdots))$ where there are $n$ applications of the function $\phi$. Then you can easily calculate quantities like $$\mathbb{P}(X_2=0, X_1>0)=\mathbb{P}(X_2=0)-\mathbb{P}(X_1=0).$$ Can you take it from here?

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This all makes sense up until you write up your equation. I can see how to apply it, but I'm not seeing why $$P(X_2 = 0, X_1 > 0) = P(X_2 = 0) - P(X_1 = 0).$$ Could you explain that? –  jrad Oct 10 '12 at 6:01
Scratch that--A little consideration of my own helped me understand it. Thanks! –  jrad Oct 10 '12 at 6:02