I am solving some questions in Introduction to Probability Models by Sheldon Ross as part of my preparation for my upcoming exams. I am stuck on this problem number 27 from Chapter 4 in the third edition of the book.
Each individual in a population of size $N$ is, in each period, either active or inactive. If an individual is active in a period then, independent of all else, that individual will be active in the next period with probability $\alpha$. Similarly, if an individual is inactive in a period then, independent of all else, that individual will be inactive in the next period with probability $\beta$. Let $X_n$ denote the number of individuals that are active in period n.
(a) Argue that $X_n$, $n\geq 0$ is a Markov chain. (b) Find $E[X_n|X_0 = i]$. (c) Derive an expression for its transition probabilities. (d) Find the long-run proportion of time that exactly j people are active
My attempt
So for (a) I think my argument is good enough. I am able to write transition probabilities that only depend on the current state of the markov chain. So it does not matter how many were active in the last few periods. If we know how many are active in the current period then we can write the probability for how many will be active in the next period solely as a function of those constants alpha and beta. So it is a Markov Chain.
For part b, I am not able to think of how to find any recurrence relation. The probabilities themselves are complicated and I can't simply go by definition. I just need help with b)
c) Please verify if my transition probabilities are right. So if $k$ of the currently active $i$ individuals remain active, we need $j-k$ more individuals to turn active from the $N-i$ previously inactive crowd. Or if $j<i$, we need a minimum of $i-j$ of the $i$ active crowd to turn inactive and if more turn inactive we need the difference $(j-(i-k))$ to come from the inactive $N-i$.
So if $j>i$,
$$P_{ij} = \sum_{k=0}^i {i\choose k} \alpha^k {N-i \choose j-k}(1-\beta)^{j-k}$$
or if $j<i$,
$$P_{ij} = \sum_{k=(i-j)}^i {i\choose k} (1-\alpha)^k {N-i \choose j-(i-k)}\beta^{(j-(i-k))}$$
I did not get a chance to really try d) yet. Any hints for that will help too, I can't imagine actually building the $P$ matrix with the above probabilties and solving the linear equation $\pi P = \pi$ is what the question expects from us.
The simple way to find the equilibrium is that the same average number must become active each time as become inactive. If $j$ people are active, we expect $\alpha j$ to become inactive. If $j$ people are active, $N-j$ are inactive and we expect $(N-j)\beta$ to become active. Equate those and solve for $j$.