X-1: Reproduction

We shall now apply our knowledge of markov chains to an analysis of population growth. At each time epoch $t$, we let $X_t$ denote the number of individuals surviving at time $t$. Each time epoch will represent a generation of a species, so that at each time interval, offspring are generated, and the current population dies off. We now make the assumption that

1. Each member of the species has the same chance of producing offspring, given by a distribution over $\mathbf{N}$.
2. The population reproduces asexually and statistically independantly of one another.

These assumptions guarentee that $X_t$ is a markov chain on $\mathbf{N}$.

Now if $X_t = k$, and $Y_1, \dots, Y_k$ are independent random variables with the distribution described above, then \begin{equation} \mathbf{P}(X_{t+1} = k' | X_t = k) = \mathbf{P}\left(\sum_{i = 1}^k Y_i = k' \right) \end{equation} Let $\mu$ denote the mean amount of offspring a single individual will possess. Then by (1), \begin{equation} \mathbf{E}(X_{t+1}|X_t = k) = \mathbf{E}\left(\sum_{i = 1}^k Y_i \right) = \sum_{i = 1}^k \mathbf{E}(Y_i) = k\mu \end{equation} (2) gives us the relatively easy computation that \begin{equation} \mathbf{E}(X_{t+1}) = \mathbf{E}[\mathbf{E}(X_{t+1}|X_t)] = \sum_{k = 0}^\infty \mathbf{P}(X_t = k) k \mu = \mu \mathbf{E}(X_t) \end{equation} And therefore $\mathbf{E}(X_n) = \mu^n \mathbf{E}(X_0)$.

This already tells us the intuitive fact that

1. If $\mu < 1$, our expected population will tend to extinction.
2. If $\mu = 1$, our population will on average remain the same.
3. If $\mu > 1$, our expected population will become unbouned.

What's more, by Markov's inequality, if $\mu < 1$, then \[ \mathbf{P}(X_n = 0) = 1 - \mathbf{P}(X_n > 1/2) \geq 1 - 2\mathbf{E}(X_n) = 1 - 2 \mu^n \mathbf{E}(X_0) \to 1 \] Since $\sum_{n = 1}^\infty \mathbf{P}(X_n = 0) = \lim_{n \to \infty} \mathbf{P}(X_n) = 1$, the population will almost surely become extinct.

X-2 Reproduction Probabilities

Even if $\mu \geq 1$, there is a chance that the population will become extinct. The problem in this section will be in deriving this probability in terms of the reproduction probabilities. Let $a_n(k)$ denote the probability of extinction after $n$ generations given that we start with k individuals. Then, as we have derived above, the possibility of general extinction from $k$ individuals is $a(k) = \lim_{n \to \infty} a_n(k)$. Since all $k$ branches of the population act independantly, we have $a_n(k) = a_n(1)^k$, and it suffices to determine $a(1)$, which we shall denote $a$. Furthermore, for computation, denote the probability of $k$ offspring from a branch by $p_k$. If we look one generation ahead, then \begin{equation} a = \mathbf{P}(\text{extinction}|X_0 = 1) = \sum_{k = 0}^\infty \mathbf{P}(X_1 = k | X_0 = 1) \mathbf{P}(\text{extinction} | X_1 = k) = \sum_{k = 0}^\infty p_k a^k \end{equation} Thus $a = \varphi_1(a)$, where $\varphi_1$ is the generating function of $X_1$, assuming $X_0 = 1$. If we let $\varphi_n$ be the generating function of $X_n$, then \begin{align} \varphi_n(a) &= \sum_{k = 0}^\infty \mathbf{P}(X_n = k) a^k = \sum_{k = 0}^\infty \sum_{j = 0}^\infty \mathbf{P}(X_1 = j) \mathbf{P}(X_n = k | X_1 = j) a^k\\ &= \sum_{j = 0}^\infty p_j \sum_{k = 0}^\infty \mathbf{P}(X_n = k | X_1 = j) a^k = \sum_{j = 0}^\infty p_j \sum_{k = 0}^\infty \mathbf{P}(X_{n-1} = k | X_0 = j) a^k \end{align} Now $\mathbf{P}(X_{n-1}| X_0 = j)$ is distributed the same as the sum of $j$ independent random variables $Y_1, \dots, Y_j$, each distributed the same as $\mathbf{P}(X_{n-1} | X_0 = 1)$, so that each has the generating function $\varphi_{n-1}$, and so, in tandem with (6), \begin{align} \varphi_n(a) = \sum_{j = 0}^\infty p_j \mathbf{E}(a^{X_{n-1}}) = \sum_{j = 0}^\infty p_j \varphi_{n-1}(a)^j = \varphi(\varphi_{n-1}(a)) \end{align} Using (7), we can recursively determine $a_n(1) = \mathbf{P}(X_n = 0 | X_0 = 1) = \varphi_n(0)$ by calculating $\varphi_n(a)$