Bacteria Breakup

Suppose that $\left(X_{n}\right)_{n \geq 0}$ represents the popuLation size of a colony of bacteria where $X_{0}=1 .$ Over each time step each bacterium (independently of all others in the population) can only do one of three things; it either dies out, remains the as an individual or divides in to two with probabilities $a, b$ and $c$ respectively.
(i) Find an expression for $G(s)$ and hence find the extinction probability for the process. Hint: Your solution will have 2 answers which are associated with conditions, that you should state explicitly, depending on the unknown offspring distribution constants.
(ii) Herein for the rest of this question suppose that $a=1 / 4, b=1 / 4$ and $c=1 / 2 .$ Determine the probability of extinction.
(iii) Explain in one sentence why the maximum population at time $n$ is given by $2^{n}$.
(iv) Derive the function $f(n)$ and constant $d$ for the following expression:
$$
\mathbb{P}_{1}\left(X_{n}=2^{n}\right)=(d)^{f(n)}
$$

More Branching Processes. The aim of this question is to slightly modify the definition of the branching process from the lectures and check how this affects the results of the final two theorems in Chapter $3 .$

Let $\left(X_{n}\right)_{n \geq 0}$ represent the number of individuals in a population at time $n$. At each generation each individual (independently of all others) gives birth to a random number of offspring that forms the next generation but now the offspring distribution is determined by whether the generation index is odd or even! In fact, the offspring distribution alternates between consecutive generations.
To put this mathematically, with $X_{0}=1,$ then
$$
X_{n+1}=\sum_{k=1}^{X_{n}} Z_{k}^{n}
$$
where all $Z_{k}^{n}$ are independent and
$$
Z_{k}^{n} \sim\left\{\begin{array}{l}
Z_{1} \text { if } n \text { is odd } \\
Z_{2} \text { if } n \text { is even }
\end{array}\right.
$$
where $Z_{1}, Z_{2}$ are RVs on $\mathbb{N}$ such that $\mathbb{E}\left(Z_{1}\right)=\mu_{1}<\infty$ and $\mathbb{E}\left(Z_{2}\right)=\mu_{2}<\infty$. Also, assume that $0<\mathbb{P}\left(Z_{i}=0\right)<1$ for $i=1,2$

(i) Define $F_{n}(s):=\mathbb{E}_{1}\left[s^{X_{n}}\right]$ and $G_{i}(s):=\mathbb{E}\left[s^{Z_{i}}\right] .$ Using the same methodology as in lectures (or otherwise) show that for $0 \leq s<1$
$$
F_{2 n}(s)=G_{2}\left(G_{1}\left(F_{2 n-2}(s)\right)\right)
$$
(ii) By either differentiating wrt $s$ (using chain rule), OR carefully using Tower property, derive an expression of
$$
\mathbb{E}_{1}\left(X_{2 n}\right)
$$
(iii) Adjust a proof in lectures to prove that the probability of extinction of this process is given by $\alpha$ which is the minimal non-neg solution to:
$$
\alpha=G_{2}\left(G_{1}(\alpha)\right)
$$
(iv) Again by replicating and adjusting a proof from the lectures; show that if $\mu_{1} \mu_{2} \leq 1$ then the process becomes extinct with probability 1 whereas if $\mu_{1} \mu_{2}>1$ then extinction is not certain.
(v) Suppose that $Z_{1} \sim \operatorname{Po}(5)$ and $Z_{2} \sim \operatorname{Po}\left(\mu_{2}\right) .$ Give a range of values for $\mu_{2}$ that guarantee extinction for the population.

To infinity and beyond: This question is all about showing that an asymmetric RW will “drift” off in the direction of the bias, This is done without needing SLL.N or even WLLN. It simply uses careful application of the Gambler’s ruin result from lectures.

Consider a Simple Random Walk, $\left(X_{n}\right)_{n \geq 0},$ on $Z$ where the dynamics are given by the following:
$$
(P)_{i j}=\left\{\begin{array}{ll}
p & \text { if } j=i+1 \\
q=(1-p) & \text { if } j=i-1 \\
0 & \text { otherwise. }
\end{array}\right.
$$
Now suppose that $1>p>q>0 .$ Hence, we know (from lectures) that the process $\left(X_{m}\right)_{n} \geq 0$ is transient for all states $i \in \mathbb{Z}$
(i) For $m \in I,$ let $H^{m}:=\inf \left\{n \geq 0: X_{n}=m\right\}$. Suppose that $m<0$. Then show that for $m<i \in I$
$$
h_{i}:=\mathbb{P}_{i}\left(H^{m}<\infty\right)=a^{f(i, m)}
$$
where $a$ is a constant and $f(i, m)$ is a function of $i$ and $m$ that you should display clearly at the end of your solution. Hint: Just adapt the Gambler’s ruin proof from lectures.

(ii) Define the event:
$$
E:=\left\{\omega \in \Omega: \exists N=N(\omega) \in \mathbb{N} \forall n \geq N, \quad X_{n}(\omega)>-N\right\}
$$
So $E$ is the event that the chain has a finite minimal point for any given “trajectory”. We want to show that under the law $\mathbb{P}$ o this event has probability $1 .$ Take a look at how we wrote the extinction event in a branehing process as a union of increasing events. The aim now is to do something similar for $E$.

On the event that $X_{0}=0,$ write an identity for the event $E$ in terms of a countable union of events involving the random variables $H^{-m}$ for $m \in \mathbb{N}$. Hint: $E$ could be read as “eventually we can find some $N \in \mathbb{N}$ such that the chain does not hit $-N$ (and therefore any state below that)”.
(iii) Justify that
$$
\mathbb{P}_{0}(E)=1
$$
(iv) Using the fact that we know 0 is a transient state and the state space is irreducible, show that for any $K>0$ the Markov chain eventually leaves the interval $[-K, K]$ If done correctly this should take no more than 5 concise statements to prove this and so long unclear answers will be lose credit. Hint: One approach is to assume there is no final visit to the set and prove a contradiction.
(v) Combine the results from (iii) and (iv) to show that for any fixed $K \in \mathbb{N}$ the Markov Chain will eventually exceed and then remain above the value $K$