I am studying marcov chain.
The question is .
There are 5 people ( 4 diseased / 1 healthy)
Two people are selected randomly and assumed to interact. If one is diseased and the other is healthy, the probability of transmitting disease is 0.1. Otherwise, transmission dons not take place.
So I want to look at each case of one step transition, where Xn= number of people diseased.
For example, Probability is 0 from 4 diseased to 1 diseased (One step transition),2, and 3
So, I want to calculate probability of one step transition from 4 diseased to 4 diseased and 4 diseased to 5 diseased.
For the 4 to 4, there is probability 2/5 of choosing 1 healthy and 1 disease, and probability of not getting disease is(0.9) . so (0.9)(2/5)
For the 4 to 5, there is probability of 2/5 of choosing 1 healthy and 1 disease, and probability of getting disease is 0.1 . so (0.1)(2/5)
But, they do not sum up to 1, which is weird.
Can anyone explain what I did wrong
Your transition matrix looks like this: $$\mathcal P = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 - (0.4)(0.1) & (0.4)(0.1) & 0 & 0 & 0 \\ 0 & 0 & 1 - (0.6)(0.1) & (0.6)(0.1) & 0 & 0 \\ 0 & 0 & 0 & 1 - (0.6)(0.1) & (0.6)(0.1) & 0 \\ 0 & 0 & 0 & 0 & 1 - (0.4)(0.1) & (0.4)(0.1) \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$ This matrix is $6 \times 6$ because $X_t \in \{0, 1, \ldots, 5\}$. Note that if $X_t = n$, then $X_{t+1} \in \{X_t, X_t + 1\}$: the number of diseased people either stays the same or increases by one. Where do the $0.6$ and $0.4$ come from? Well, note that there are always $\binom{5}{2} = 10$ ways to choose two people at random without replacement from a group of $5$. Of these $10$ ways, if $1$ is diseased, then there are $4$ ways to choose the diseased person and one non-diseased person. Similarly, if $4$ are diseased, then there are $4$ ways to choose one diseased person and the non-diseased person.
If there are $2$ or $3$ diseased people, you also have a symmetry there: there are $6$ ways to choose one diseased person and one non-diseased person. And since the only possible choices for $X_{t+1} \mid X_t$ are $X_t$ and $X_t + 1$, the transition matrix can only have entries $p_{ii} + p_{i(i+1)} = 1$.
The flaw in your reasoning can be understood this way: let $C$ be the event of choosing one diseased and one non-diseased person. Let $D$ be the probability of transmitting disease. Then you are given $\Pr[D \mid C] = 0.1$ and $\Pr[\bar D \mid C] = 0.9$. You are not given $\Pr[D]$, the unconditional probability. So by the law of total probability, you must calculate $$\Pr[D] = \Pr[D \mid C]\Pr[C] + \Pr[D \mid \bar C]\Pr[\bar C].$$ But $\Pr[D \mid \bar C] = 0$, so $\Pr[D] = \Pr[D \mid C]\Pr[C]$ which is what you expect, but now $\Pr[\bar D] = 1 - \Pr[D]$.