Question

A2. A newspaper publisher uses one roll of newsprint every day. A local supplier delivers a random number of rolls each evening, where the number of rolls has a binomial distribution with parameters 3 and 3, i.e. P(delivers x rolls)-C)G) r 0, 1,2,3 , x-0.1, 2.3 Deliveries are independent over days. If the newspaper has no rolls of newsprint in stock at the start of a day, it must obtain some from an emergency supplier to cover the day's requirement. Also the newspaper's warehouse only has room to hold three rolls; if on any day the delivery is too large to be accommodated then the excess is taken back by the supplier Writing Xn for the number of rolls of newsprint in the warehouse at the start of day n, show that (Xn) is a Markov chain and obtain the transition matrix. (e

Answer 1

The number of rolls the publisher has are 0, 1, 2, 3 at the start of the day. Let use define states as 0, 1, 2, 3 denoting the number of rolls of newsprint in the warehouse at the start of the day.

P(Xn+1 = j | Xn = i) is the probability that number of rolls of newsprint in the warehouse at the start of the day is j, given number of rolls of newsprint in the warehouse at the start of the previous day is i. One roll is used by the warehouse. Then

j = i - 1 + x where x is the number of rolls delivered and follows binomial distribution.

Since, number of deliveries are independent over days, j depends only on i (previous state).

Thus, the next transition depends only on the previous state and hence {Xn} is a Markov chain.

$P(x = 0) = \binom{3}{0}\left ( \frac{1}{2} \right )^3 = \frac{1}{8}$

$P(x = 1) = \binom{3}{1}\left ( \frac{1}{2} \right )^3 = \frac{3}{8}$

$P(x = 2) = \binom{3}{2}\left ( \frac{1}{2} \right )^3 = \frac{3}{8}$

$P(x = 3) = \binom{3}{3}\left ( \frac{1}{2} \right )^3 = \frac{1}{8}$

The transition from state 0 to state 0 is possible for x = 0, 1. So, the transition from state 0 to state 0 is (1/8) + (3/8) = 1/2

The transition from state 0 to state 1 is possible for x = 2 with probability 3/8

The transition from state 0 to state 2 is possible for x = 3 with probability 1/8

The transition from state 1 to state 0 is possible for x = 0 with probability 1/8

The transition from state 1 to state 1 is possible for x = 1 with probability 3/8

The transition from state 1 to state 2 is possible for x = 2 with probability 3/8

The transition from state 1 to state 3 is possible for x = 3 with probability 1/8

The transition from state 2 to state 1 is possible for x = 0 with probability 1/8

The transition from state 2 to state 2 is possible for x = 1 with probability 3/8

The transition from state 2 to state 3 is possible for x = 2, 3 with probability (3/8 + 1/8) = 1/2

The transition from state 3 to state 2 is possible for x = 0 with probability 1/8

The transition from state 3 to state 3 is possible for x = 1, 2, 3 with probability (3/8 + 3/8 + 1/8) = 7/8

The transition matrix is,

$P = \begin{bmatrix} 1/2&3/8 &1/8 &0 \\ 1/8& 3/8& 3/8& 1/8\\ 0&1/8 &3/8 &1/2 \\ 0 & 0 & 1/8 & 7/8 \end{bmatrix}$