(This is probably an easy question, but I don't have much experience in probability. I'm looking for inspiration/the right viewpoint.)
This is a toy example for a more general process.
We define it as follows, where at each time $t\in \mathbb N_0$ we have a certain number $A_t\ $ of adults and $C_t\ $ of children:
Setup
We start with $A_0 = 1$ and $C_0 = 0$, say.
Fixing $p\in (0,1)$; between times $t$ and $t+1$,
- each child becomes an adult, and
- each adult independantly becomes either
- 2 adults (with probability $p$), or
- 4 children (with probability $1-p$).
I.e., inductively, given $A_t$ and $C_t$, let $Y_t \sim \operatorname{Bin}(A_t,p)$. Then
$$A_{t+1} = 2Y_t + C_t,\qquad C_{t+1} = 4(A_t - Y_t).$$
The task
Show that
$$ 2^{-t} A_t \to \frac 1{2-p}$$
almost-surely as $t\to \infty$.
My error-ridden attempt to show this is below. All comments are welcome:
- Does it work?
- Is there an easier way, e.g., a martingale argument?
- Does the method generalise to the setting where adults can also turn into 8 grandchildren (who turn into children), etc.?
Sketchy proof attempt
We can inductively check that, for all $t$, we have
$$ 2^{-t}A_t + 2^{-t-1}C_t = 1, \tag1$$
so we can forget about $C_t$ and write $$ (A_{t+1}|A_t) \sim 2\operatorname{Bin}(A_t,p) + 2^{t+1} - 2A_t.$$ Using (1) at the end, it seems that the conditional variance satisfies $$\operatorname{Var}(2^{-t-1}A_{t+1}\mid A_t) = 4^{-t-1}A_t p (1-p) \leq 2^{-t}.$$ Since the bound is independent of $A_t$, this suggests $\operatorname{Var}(2^{-t-1}A_{t+1}) \leq 2^{-t}$, and via Chebyshev/Borel-Cantelli this gives that $A_t - \mathbb E (A_t) \to 0$ almost surely. We also get from the above formula: $$\mathbb E\left(\frac{A_{t+1}}{2^{t+1}} \;\bigg|\; A_t\right) = 1 - (1-p) \frac{A_t}{2^t} $$ which I hope means, for $f(x) := 1 - (1-p)x$, that $$ \mathbb E\left(\frac{A_t}{2^t}\right) = f^{\circ n}(1) \to \frac 1 {2-p},$$ as required.