Branching process - probability that the branching process survives forever with 4 individuals

Question

Consider a branching process where the number of offspring of an individual is a binomial random variable with parameters $(2, p),$ with $p \in (0, 1).$ Initially there is one individual.

(a) Calculate the probability, as a function of $p,$ that the branching process survives forever.

(b) Suppose we observe that the initial individual has $2$ offspring, and each of these then has $2$ offspring of its own. What is the probability that the branching process survives forever?

I have calculated for (a), the answer is $1-\min\left\{1,\left(\frac{p-1}{p}\right)^2\right\}$.

However, I don't know how to solve for (b). My instinct tells me it might be $4 \cdot \left(1-\min\left\{1,\left(\frac{p-1}{p}\right)^2\right\}\right)$. Am I correct?

score 1 · Accepted Answer · answered Jul 05 '19 at 05:12

1

Let's call the probability of the branching process surviving forever $a$.

There is a $(1-p)^2$ probability of zero offspring, $2p(1-p)$ probability of one offspring, and a $p^2$ probability of two offspring.

Then, assuming there is one offspring, the probability of the branching process continuing forever is $a$. If there is two offspring, both branches must not die off, so the probability is $1-(1-a)^2$. Therefore $$a = 0*(1-p)^2 + a*(2p(1-p)) + p^2 * (1-(1-a)^2) = 2pa-p^2*a^2$$Simplifying and setting equal to $0$, $$p^2*a^2+(1-2p)a = 0 \rightarrow a=(2p-1)/p^2$$

Therefore, the probability of the branching process surviving forever is $$a = \frac{2p-1}{p^2}$$. Since this is less than $0$ if $p < 0.5$, $a$ is instead $$\max(0, \frac{2p-1}{p^2})$$ which is the same formula you arrived at.

If there are four branches, then all must not fail. $$b=1-(1-a)^4$$

answered Jul 05 '19 at 05:12

Varun Vejalla

8,873

Thank you for your reply. How is $P(\text{both branches must not die off}) = 1-(1-a)^2$? Wouldn't it be $2a$ because now there are two offsprings and the chance of either/both of the two surviving is doubled? – underdisplayname Jul 05 '19 at 19:11
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Let's say $a > 0.5$. This would definitely not be true. The reason for $1-(1-a)^{2}$ is because the probability of a branch dying out is $1-a$. The probability of two branches dying out is $(1-a)^{2}$. The probability of that not happening is $1-(1-a)^{2}$ – Varun Vejalla Jul 05 '19 at 19:14
I see. So the answer is $b = 1-(1-a)^4$, where $a = max(0,\frac{2p-1}{p^2})$? – underdisplayname Jul 05 '19 at 19:25
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Yeah, that is correct. – Varun Vejalla Jul 05 '19 at 19:26
Thank you. That is a really clever way to get the value of $a$. I have some more branching process problems that I am struggling with. Do you mind if I send you the link to those questions later when I ask them on MathStackExchange? – underdisplayname Jul 05 '19 at 19:30
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Sure that works – Varun Vejalla Jul 05 '19 at 19:34
Let us continue this discussion in chat. – underdisplayname Jul 05 '19 at 20:08

Branching process - probability that the branching process survives forever with 4 individuals

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