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Suppose a particle in each time step, with probability $p$ turns into two particles and with probability $1-p$ remains without any change.


For example, if there exist only one particle in the time step $0$, then in the $1$st time step with probability $p$, we have $2$ particles and with probability $1-p$ we have $1$ particles. In the $2$th time step, we have $4$ particles with probability $p^3$, $3$ particles with probability $2p^2(1-p)$, $2$ particles with probability $p(1-p)^2 + p(p-1)$ and $1$ particle with probability $(1-p)^2$.


Let there exist only one particle in the time step $0$. What is the expectation of the number of particles after $n$ time steps?

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    $\begingroup$ At the second moment ... 2 particles with probability $p(1-p)^2\color{red}{+(1-p)p}$. ? $\endgroup$ Aug 10, 2017 at 15:23
  • $\begingroup$ @platty: I do not know how to solve this problem. I would appreciate if you suggest a reference. $\endgroup$ Aug 10, 2017 at 15:34
  • $\begingroup$ A reference: en.wikipedia.org/wiki/Branching_process $\endgroup$
    – user940
    Aug 10, 2017 at 15:38

1 Answer 1

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Suppose you have $x_{n}$ particles after $n$ timesteps. Provided a particle turning into two or not is independent of any other particle doing so, then the expected number of new particles is $x_{n}p$ by linearity of expectation. So you can work out $E[x_{n+1}]$ as a function of $x_{n}$.

Then provided a particle turning into two or not is independent of past events, then you can use this expression for $E[x_{n+1}]$ as a recurrence relation to give you the closed-form expectation, again using linearity of expectation.

Recurrence relation:

$E[x_{n+1}] = x_{n} + x_{n}p = x_{n}(1+p)$, and since $E[E[x_{n+1}]]=E[x_{n+1}]$, we have $E[x_{n+1}]=E[E[x_{n+1}]]=E[x_{n}(1+p)]=(1+p)E[x_{n}]$ from linearity of expectation and since $p$ is constant

Solution

$E[x_{n+1}] = E[x_{0}](1+p)^{n}$; in your case $x_{0}=1$ so $E[x_{n+1}]=(1+p)^{n}$

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    $\begingroup$ Strictly speaking you should use some conditional expectation notation here, i.e. $E[x_{n+1}]=E[E[x_{n+1} \mid x_n]]=E[x_n(1+p)]$. $\endgroup$
    – Ian
    Aug 10, 2017 at 15:42

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