TITLE: Probability of explosion in a Markov chain
QUESTION [1 upvotes]: I have the following problem: In a chain reaction a particle of a certain kind has probability 4/7 of hitting
a nucleus. If that happens then the particle disappears but 3 new particles of the same
kind are freed. With probability 3/7 the particle leaves without hitting a nucleus.
Determine the probability of explosion if the chain reaction starts with one particle.
My thoughts were, since this 0 is absorbing state and this is just a random walk on $\mathbb{N}$ with more probability going to right then left then the asked probability is just $p=1-h_1^{0}$ where h are the hitting probabilities.
Then I get from the jump matrix which is:
$$
P =
 \begin{pmatrix}
  1 & 0 & 0 &0&0\cdots & 0 \\
  \frac{3}{7} & 0 & 0&4/7&0\cdots & 0 \\
   0&3/7&0&0&\frac{4}{7}&\cdots &0\\
  \vdots  & \vdots  & \ddots & \ddots  \\
 \end{pmatrix}$$
So I get the following equations:
$$h_1=3/7+4/7*h_3$$
$$h_2=3/7*h_1+4/7*h_4$$
$$h_i=3/7*h_{i-1}+4/7*h_{i+2}$$, for $i \geq 1$
$$ $$
Well I'm having trouble solving this system since there are more variables than equation for each i. Is there any way to solve this system straight forward or do I need alternative perspective to solve this problem?
Regards, Raxel.

REPLY [1 votes]: The probability $p$ that the particle starts a chain reaction is the probability that it hits a nucleus and at least one of the resulting particles starts a chain reaction. Thus we have
$$
p=\frac47\left(1-(1-p)^3\right)\;,
$$
and thus either $p=0$ or
$$
p^2-3p+\frac54=0\;,
$$
with solutions $p=1/2$ and $p=5/2$. Since $5/2$ is not a valid probablity, the expected number of descendants is $12/7$ and the extinction probability of a branching process with expected number of descendants greater $1$ is less than $1$ (see Wikipedia), we have $p=1/2$.
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