As stated in the problem. Given a bacterium that at every time step $t$, either divides with a probability $p$ or dies off. What is the expected number of bacteria after $n$ time steps?
Does this use a binomial theorem? Markov Chains? I have no clue..
Experimental results show that with $p = 0.75$, here are the numbers per timestep $t$ (with $10^7$ simulations):
| t | population |
|----|------------|
| 1 | 1.500407 |
| 2 | 2.2507928 |
| 3 | 3.376627 |
| 4 | 5.0640936 |
| 5 | 7.5959494 |
| 6 | 11.3945926 |
| 7 | 17.0909234 |
| 8 | 25.6368554 |
| 9 | 38.4575074 |
| 10 | 57.6884722 |
| 11 | 86.5342376 |
| 12 | 129.802634 |
So the experimental results have shown that this follows a trend of $(2p)^i$ where $i$ is the timestep. I am not sure why that is though... any ideas?
If there are $n$ bacteria at time $t$, then on average $pn$ of them split and $(1-p)n$ die, giving an expected number of $2pn$ bacteria at time $t+1$.
If we start with one bacterium at time $t=0$, then by induction at time $t$ the expected number of bacteria is $(2p)^t$. For example, when $t=10$ and $p=0.75$, this gives us about $57.665$ bacteria in expectation.
More generally, the distribution of the bacteria follows a branching process. In particular, we can compute the distribution for any $t$ with generating functions: let $f(x) = 2px^2 + (1-p)$. Then the probability that there are $n$ bacteria at time $t$ is the coefficient of $x^n$ in the $t$-fold composition $$ \underbrace{f(f(f(\cdots(f(}_tx))\cdots))). $$ But deriving further properties of this distribution takes some work.