In a textbook titled, Essentials of Stochastic Process by the author, Durrett.
I am unable to understand the relation between the rows and the columns in which the author have meant for it to be.
The author suggested first a 1,3 inventory policy and in which case he replied on the potential of sales and probability associated with each sales to construct the transition matrix. In my understanding, the rows represents the number of units at the start of the day and the columns represents the number of units at the end of the day. From this, starting at 3 units and ending the day with 0 units, there is a sale of 3 units and that sale has an associated probability of 0.1 — 0.2, 0.4 and 0.3 for 2, 1 and 0 units of sale, respectively.
In moving up the row index and looking at 2, we begin with 2 units at the start of the day and ending the day with 0 units implies a sale of 2 units. That has a probability of 0.2 but p_{2,0} = 0.3 in the table.
This implies either a misunderstanding on my part or a typological error.
Can someone clarify?
The transition matrix should be interpreted as
$$P(x,y) = \mathbf P\left( \text{end next day with $y$ units, having ended today with $x$ units}\right).$$
Note that this is different to the interpretation you gave of "ending the day with $y$, having started that day with $x$", because of the fact that we re-stock in the morning.
The 2,3 strategy In this case it does not matter how many units there were the night before, because we always restock to have $3$ units the next morning. This is why the rows of this matrix are constant. Note that in your original interpretation it would be the case that $P(x,y) = 0$ for all $x \neq 3$, since we know that we start the morning with $3$ units.
The 1,3 strategy If in this case $x = 0,1,3$ then as above this means that we would start the next morning with $3$ units. Hence these rows in the matrix are the same as before.
Now in the case that we end the previous evening with $x = 2$ then we do not restock. So we have to look at what this means for our likely sales.
The important assumption that Durrett is implicitly making is: the number of units we have to sell, does not impact the purchasers desire to buy. I.e. having fewer units does not mean that fewer people will try to buy one.
So, given that we have $x = 2$ units the evening before (and hence at the start of the next day), the probability that we end the day with $2$ units is the probability of sellin no units, which is the same as if we had three units. i.e $P(2,2) = 0.3$.
The probability that we end the day with $1$ unit is the probability that we sell $1$ unit (of $2$) and so $P(2,1) = 0.4$.
Finally the probability that we end the day with $0$ units is the probability that we sell $2$. This is the same as the probability that two or more people want to buy a unit: it is not the probability that exactly two people want to buy a unit. As such it is the same as the probability that if we had 3 units we would sell two of them, plus the probability that we would have sold three. In this case, if we had only two units to sell somebody walks away without one. Hence $P(2,0) = 0.2 + 0.1 = 0.3$.
Finally it is hopefully clear that $P(2,3) = 0$ since we only have two units.
Combining these, we obtain the matrix proposed by Durrett.