Stochastic Process - Creating a Probability Transition Matrix

[cse63146]

Homework Statement

The total population size is N = 5, of which some are diseased and the rest are healthy. During any single period of time, two people are selected at random from the population and assumed to interact. The selection is such that an encounter between any pair of individuals is independent of any other pair. If one of the persons is diseased and the other isn't, the with probability 0.1 the disease is transmitted to the healthy person. Otherwise, no disease is transmitted. Let denote the number of diseased people in the population at the end of the nth period. Find the transition probability matrix.

Homework Equations

The Attempt at a Solution

I did the TPM for two people from the population:

H ~ Healthy Person D ~ Diseased person

$$\begin{bmatrix}0.9 & 0.1\\ 0 & 1\end{bmatrix}$$

The columns are (H,D) and the rows are (H,D)'

But I'm not sure how to account for the N=5 population

 

[mighty2000]

size and the transition from one period to the next. Can someone help me with this?

Thank you for your question. In order to account for the population size and the transition from one period to the next, we need to consider the probability of each possible outcome for the two selected individuals in each period.

Let's denote the number of diseased individuals in the population at the beginning of the nth period as D(n) and the number of healthy individuals as H(n). We can then create a transition probability matrix that takes into account the population size and the transition from one period to the next as follows:

\begin{bmatrix}P(D(n+1)=0|D(n)=0, H(n)=0) & P(D(n+1)=1|D(n)=0, H(n)=0) & P(D(n+1)=2|D(n)=0, H(n)=0) \\ P(D(n+1)=0|D(n)=1, H(n)=0) & P(D(n+1)=1|D(n)=1, H(n)=0) & P(D(n+1)=2|D(n)=1, H(n)=0) \\ P(D(n+1)=0|D(n)=2, H(n)=0) & P(D(n+1)=1|D(n)=2, H(n)=0) & P(D(n+1)=2|D(n)=2, H(n)=0) \\ P(D(n+1)=0|D(n)=0, H(n)=1) & P(D(n+1)=1|D(n)=0, H(n)=1) & P(D(n+1)=2|D(n)=0, H(n)=1) \\ P(D(n+1)=0|D(n)=1, H(n)=1) & P(D(n+1)=1|D(n)=1, H(n)=1) & P(D(n+1)=2|D(n)=1, H(n)=1) \\ P(D(n+1)=0|D(n)=2, H(n)=1) & P(D(n+1)=1|D(n)=2, H(n)=1) & P(D(n+1)=2|D(n)=2, H(n)=1) \end{bmatrix}

Note that the rows and columns represent the possible combinations of diseased and healthy individuals at the beginning and