next up previous
Next: Analyzing the Population Changes Up: Due in Recitation on Previous: A Population Problem

Markov Chain Models

Suppose a population consists of k disjoint sub-populations labeled by the integers $1 \ldots k$. Let pi be the fraction of the population in sub-population i. Form the vector $\vec{S} \in R^k$ called the state vector of the system:

\begin{displaymath}
\vec{S} = \left(
\begin{array}{l}
p_1 \\
p_2 \\
\cdots \\
p_k
\end{array}\right)
\end{displaymath}

The sum $\Sigma_{i=1}^k p_i$ of the population fractions is assumed to be 1.

The state vector $\vec{S}$ will depend on time t. In a discrete model, we might measure the population fractions pi at times $\{t_1,t_2, \ldots t_n, \ldots \}$. Use $\vec{S}[t_n]$ and pi[tn] respectively to denote values at the end of the n'th time period.

In a Markov chain model one assumes that there are transition probabilities Tij representing the proportion of the population in sub-population j during time period n which enter sub-population i for time period n+1. These transition probabilities are independent of time. Let T be the k x k matrix whose i,j entry is Tij.

We also assume that the entire population in time period n+1 comes by this transition process from the population during time period n. Thus

\begin{displaymath}
p_i(t_{n+1}) = T_{i1} p_1(t_n) + T_{i2} p_2(t_n) + \cdots T_{ik} p_k(t_n) +
\end{displaymath}

This is just the formula for the i'th entry of a matrix product of T and the n'th state vector $\vec{S}[t_n]$. Thus in matrix form, $\vec{S}[t_{n+1}] = T \circ \vec{S}[t_{n}]$.

More generally, we can keep track of the change in population over r time periods by computing the r'th power of T. Specifically $\vec{S}[t_{n+r}] = T^r \circ \vec{S}[t_{n}]$.

The sum of the entries in each column (e.g. the j'th) of the transition matrix T should be 1 because the entries of that column reflect all the possible destinations at time tn+1 for the people in region j at time tn.


next up previous
Next: Analyzing the Population Changes Up: Due in Recitation on Previous: A Population Problem
root
2002-08-21