I am trying to understand the problem regarding the stock of a certain type of camera. The demand for the cameras in week $i$ has a Poisson distribution with expectation 1: $P(D_i = k) = e^{-1}/k!$ for $k = 0, 1, ...$. The maximum stock level is 3 cameras. The photographer orders 3 cameras only when the current stock is completely depleted and the cost of ordering cameras depends on the number of cameras.
I am struggling to determine the following:
a. State space $S$ and the matrix of transition probabilities for the Markov chain, which is represented by the number of cameras in stock at the end of each week.
b. Stationary distribution ${\pi_k : k \in S}$ and the long-term average stock level at the beginning of the week.
I would greatly appreciate any help or insights on this matter.
Thank you!
Conditioned on $\{X_0=0\}$, we have $\mathbb P(\{X_1=3\})=1$ as per the ordering policy.
Conditioned on $\{X_0=1\}$, we have $\mathbb P(X_1=1)=\mathbb P(D_0=0)=e^{-1}$ and $\mathbb P(X_1=0) = 1-\mathbb P(D_0=0)=1-e^{-1}$.
Continuing this logic, we find the transition matrix to be
$$ P=\left( \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 1-e^{-1} & e^{-1} & 0 & 0 \\ 1-2e^{-1} & e^{-1} & e^{-1} & 0 \\ 1-\frac52e^{-1} & \frac12e^{-1} & e^{-1} & e^{-1} \\ \end{array} \right). $$ Solving $\pi P=\pi$ and $\sum_{i=0}^3 \pi_i=1$ we find $$ \pi = \left( \begin{array}{cccc} \frac{2 (e-1)^3}{e (e (4 e-7)+7)-2} & \frac{e (1+e)}{e (e (4 e-7)+7)-2} & \frac{2 (e-1) e}{e (e (4 e-7)+7)-2} & \frac{2 (e-1)^2 e}{e (e (4 e-7)+7)-2} \\ \end{array} \right). $$