Q: A simplified model for the spread of a contagion in a small population of size 4 is as follows. At each discrete time unit, two individuals in the population are chosen uniformly at random to meet. If one of these persons is healthy and the other has the contagion, then with probability 1/4 the healthy person becomes sick. Otherwise the system stays the same.
Let $X_n$ be the number of healthy people at step $n$. Specify the transition probabilities of $X_n$
What I've done... I know the probability that in the next step there is one less healthy person (so Xn+1= Xn -1 given Xn) = (1/6)Xn(4-Xn)*(1/4)
I know the 1/6 comes from there being six combos of 2 from the four people (4 choose 2), each with equal likelihood of being chosen. Then I know the 1/4 is the prob that if a pair has one healthy and one sick person, there is a 1/4 chance of the healthy person being sick.
I don't fully understand the Xn(4-Xn) part. Is this the probability that pair of one healthy and one sick are chosen?
Then, from there, can you help me set up the transition probabilities? Thanks!
The $X_n(4-X_n)$ is the number of pairs with one healthy and one sick (it is the number of ways to choose one healthy from $X_n$ and one sick from $4-X_n$). Since each pair has probability $1/6$, the probability of choosing such a pair is just the number of pairs times $1/6$.
Now the only possibilities for $X_{n+1}$ are that it is $X_n-1$ or $X_n$, so $\Pr(X_{n+1}=X_n)=1-\frac16X_n(4-X_n)\frac14$, and all other transition probabilities are $0$. Now you just need to work out the probabilities these expressions give for $X_n=0,1,2,3,4$.