Members of an indefinitely large population are either immune to a given disease or are susceptible to it.
Let $X_n$ be the number of susceptible members in the population at time period $n$ and suppose $X_0=0$ and that in the absence of an epidemic $X_n =X_n+ 1$. Thus, in the absence of the disease, the number of susceptibles in the population increases in time, possibly owing to individuals losing their immunity, or to the introduction of new susceptible members to the population.
But in each period there is a constant but unknown probability $p$ of a pandemic disease. When the disease occurs all susceptibles are stricken. The disease is non-lethal and confers immunity, so that if $T$ is the first time of disease occurrence, then $X_T=0$
How can I compute the stationary distribution for $(X_n)$?
My attempt:
We have that
$$P[X_{n+1}=0| X_n=i]=p, \ P[X_{n+1}=i+1| X_n=i]=1-p $$, then we have the Markov Chain:
$$P = \begin{vmatrix} p & 1-p & 0 & 0 \cdots \\ p & 0 & 1-p & 0 \cdots \\ p & 0 & 0 & 1-p \cdots \\ p & 0 & 0 & 0 \cdots \\ \vdots & \vdots & \vdots & \vdots \end{vmatrix} $$
Then the probability equations are $$ \pi_j=\sum_{i=0}^\infty = \pi_iP_{ij}, \ \ \sum_{i=0}^\infty \pi_i=1$$
for $j=0$, we have $$ \pi_0 = \sum_{i=0}^\infty \pi_i p = p$$ for $j=1$, we have $$ \pi_1 = \sum_{i=0}^\infty \pi_i p = (1-p)\pi_0=(1-p)p$$
So,
$$\pi_j= \sum_{i=0}^\infty \pi_i P_{ij} = (1-p)\pi_{j-1}, \ j\geq 1 $$ Then, the stationary distribution is $\pi_i=p(1-p)^i, \ \forall i=0,1,\ldots$
Is this correct? I thought that I need to consider the case when there can be no epidemic, then my matrix $P$ is
$$P = \begin{vmatrix} 0 & 1 & 0 & 0 & \cdots \\ p & 1-p & 0 & 0 \cdots \\ p & 0 & 1-p & 0 \cdots \\ p & 0 & 0 & 1-p \cdots \\ p & 0 & 0 & 0 \cdots \\ \vdots & \vdots & \vdots & \vdots \end{vmatrix} $$
Could someone help me for this, pls.
Thanks for your time and help
The "new" matrix $P$ does not match with the problem described above, in my opinion. It suggests almost surely that $X_1 = 1$ under condition $X_0 = 0$ and further on $X_n \in \{0,1\}$ because $P_{12} = 0$. So the mass of a stationary distribution has to be contained in $\{0,1\}$.
The calculation would be $$\pi_0 = \sum_{i=0}^\infty \pi_i P_{i0} = p \sum_{i=1}^\infty \pi_i = p (1-\pi_0) \\ \Rightarrow \pi_0 = \frac p {1+p}$$ and $$\pi_1 = \sum_{i=0} \pi_i P_{ij} = \pi_0 P_{01} + \pi_1 P_{11} = \frac p {1+p} + \pi_1 (1-p) \\ \Rightarrow \pi_1 = \frac 1 {1+p}$$ thus $\pi_j = 0 $ for $j\geq 2$ and indeed: $$\pi_j = \sum_{i=0} \pi_i P_{ij} = \pi_j (1-p) \\ \Rightarrow \pi_j = 0$$