Case 2 of Markov chain example: $P(D_n \ge i - 1) = 1 - P(D_n \le i - 2) =: p_{i5}$

Question

I am presented with the following Markov chain example:

A trader sells large and expensive machines.

$X_n$ is the number of machines in stock at the start of week $n$.

$D_n$ is the number of machines demanded by customers during week $n$.

Assume that $D_n \sim \text{Poi}(3)$, the $D_n$ are independent, and that $D_n$ and $X_n$ are independent for each $n$.

There are two stipulations:

1. Inventory control: If $0$ or $1$ machines are left in stock by the end of a week, then machines are ordered and delivered to raise the stock to $5$ by the start of the next week. If $2$ or more machines are in stock at the end of the week, then no orders are placed.

2. Lost business: If $D_n > X_n$, then the unsatisfied demands are lost.

We seek to show that $X_n$ is a Markov chain.

$(X_n - D_n)^+$ is the number of machines in stock at the end of week $n$.

$$X_{n + 1} \begin{cases} X_n - D_n & \text{if} X_n - D_n \ge 2 \\ 5 & \text{if} X_n - D_n \le 1 \end{cases}$$

$S = \{ 2, 3, 4, 5 \}$

Denote the history of the process up to time $n$ by $H_n = \{ X_0, X_1, \dots, X_n \}$.

Given that $X_n = i$, the independence assumptions ensure that $X_{n + 1}$ is conditionally independent of $H_{n - 1}$.

Case 1: $j = 2, 3,$ or $4$, and $i = j, \dots, 5$:

$$\begin{align} P(X_{n + 1} = j \vert X_n = i, H_{n - 1}) &= P(X_n - D_n = j \vert X_n = i, H_{n - 1}) \\ &= P(i - D_n = j \vert X_n = i, H_{n - 1}) \\ &= P(i - D_n = j) \\ &= P(D_n = i - j) \\ &= e^{-3} \dfrac{3^{i - j}}{(i - j)!} \end{align}$$

Case 2: $i = 2, 3,$ or $4$, and $j = 5$:

$$\begin{align} P(X_{n + 1} = 5 \vert X_n = i, H_{n - 1}) &= P(X_n - D_n \le 1 \vert X_n = i, H_{n - 1}) \\ &= P(i - D_n \le 1 \vert X_n = i, H_{n - 1}) \\ &= P(i - D_n \le 1) \\ &= P(D_n \ge i - 1) \\ &= 1 - P(D_n \le i - 2) \\ &=: p_{i5} \end{align}$$

Case 3: $i = 5, j = 5$:

$$p_{55} = P(D_n = 0) + P(D_n \ge 4) = P(D_n = 0) + 1 - P(D_n \le 3).$$

In case 2, the author has that $P(D_n \ge i - 1) = 1 - P(D_n \le i - 2) =: p_{i5}$. I don't understand how they concluded this. I would greatly appreciate it if people would please take the time to clarify this.

Answer 1

If you are talking about the first equality:

$P(D_n \ge i - 1) = 1 - P(D_n \le i - 2)$

$D_n$ is Poisson distributed, so it's discrete. Therefore the probability of the RV being greater than or equal to some value is simply just $1$ minus it's CDF of the discrete value just before it (i.e. $i-2$).

Perhaps it is easier to see if you just move the $1$ to it's own side:

$P(D_n \ge i - 1) + P(D_n \le i - 2) = 1$

The sum of probabilities for all possible values is simply $1$.

If you are talking about the definition, well that is just how $p_{i5}$ is defined, from the first probability in the very first line.