0
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ At the second moment ... 2 particles with probability $p(1-p)^2\color{red}{+(1-p)p}$. ? $\endgroup$ – Donald Splutterwit Aug 10 '17 at 15:23
  • $\begingroup$ @platty: I do not know how to solve this problem. I would appreciate if you suggest a reference. $\endgroup$ – Hasan Heydari Aug 10 '17 at 15:34
  • $\begingroup$ A reference: en.wikipedia.org/wiki/Branching_process $\endgroup$ – user940 Aug 10 '17 at 15:38
5
$\begingroup$

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 $E[x_{n+1}]=(1+p)^{n}$

$\endgroup$
  • 2
    $\begingroup$ 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 '17 at 15:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.