Chegg
Expert Q&A
1. In the market share analysis of Section 16.1, suppose that we are considering the Markov process associated with the shopping trips of one customer, but we do not know where the cu

Question

Chegg
Expert Q&A
1. In the market share analysis of Section 16.1, suppose that we are considering the Markov process associated with the shopping trips of one customer, but we do not know where the customer shopped during the last week. Thus, we might assume a 0.5 probability that the customer shopped at Murphy and a 0.5 probability that the customer shopped at Ashley at period 0; that is,
pi * 1(0) = 0.5
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and
pi * 1(0) = 0.5
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Given these initial state probabilities, develop a table similar to Table 16.2 showing the probability of each state in future periods. What do you observe about the long-run probabilities of each state?
2. Management of the New Fangled Softdrink Company believes that the probability of a customer purchasing Red Pop or the company major competition, Super Cola, is based on the customers most recent purchase. Suppose that the following transition probabilities are appropriate:
To
From Red Pop Super Cola
Red Pop 0.9 0.1 Super Cola 0.1 0.9
?
a. Show the two-period tree diagram for a customer who last purchased Re

VIPSolutions · Accepted Answer

Let’s break down your questions step by step for both problems.

---

## 1. Market Share Analysis with Markov Process

#### **Given:**
- Two stores: Murphy (M) and Ashley (A)
- Initial probabilities:  
  $ P_0(M) = 0.5 $, $ P_0(A) = 0.5 $

#### **Assume:**
- **Transition Matrix** (as in most textbook examples; let’s use the typical one):
  
  |        | To Murphy (M) | To Ashley (A) |
  |--------|:-------------:|:-------------:|
  | From M |     0.9       |     0.1       |
  | From A |     0.2       |     0.8       |

(If your transition matrix is different, you can substitute the values accordingly.)

#### **Step 1: Write the Transition Matrix**

Let’s denote the state vector as:
$$
\pi_n = \begin{bmatrix} P_n(M) \ P_n(A) \end{bmatrix}
$$

The transition matrix $ P $ is:
$$
P = \begin{bmatrix}
0.9 & 0.1 \
0.2 & 0.8
\end{bmatrix}
$$

#### **Step 2: Compute Probabilities for Each Period**
**Initial state ($ n=0 $):**
$$
\pi_0 = \begin{bmatrix} 0.5 \ 0.5 \end{bmatrix}
$$

**Period 1:**
$$
\pi_1 = P^T \cdot \pi_0
$$
$$
P^T = \begin{bmatrix}
0.9 & 0.2 \
0.1 & 0.8
\end{bmatrix}
$$
$$
\pi_1 = \begin{bmatrix}
0.9 & 0.2 \
0.1 & 0.8
\end{bmatrix}
\begin{bmatrix}
0.5 \
0.5
\end{bmatrix}
=
\begin{bmatrix}
0.9*0.5 + 0.2*0.5 \
0.1*0.5 + 0.8*0.5
\end{bmatrix}
=
\begin{bmatrix}
0.45 + 0.10 \
0.05 + 0.40
\end{bmatrix}
=
\begin{bmatrix}
0.55 \
0.45
\end{bmatrix}
$$

**Period 2:**
$$
\pi_2 = P^T \cdot \pi_1
$$
$$
=
\begin{bmatrix}
0.9 & 0.2 \
0.1 & 0.8
\end{bmatrix}
\begin{bmatrix}
0.55 \
0.45
\end{bmatrix}
=
\begin{bmatrix}
0.9*0.55 + 0.2*0.45 \
0.1*0.55 + 0.8*0.45
\end{bmatrix}
=
\begin{bmatrix}
0.495 + 0.09 \
0.055 + 0.36
\end{bmatrix}
=
\begin{bmatrix}
0.585 \
0.415
\end{bmatrix}
$$

**Period 3:**
$$
\pi_3 =
\begin{bmatrix}
0.9 & 0.2 \
0.1 & 0.8
\end{bmatrix}
\begin{bmatrix}
0.585 \
0.415
\end{bmatrix}
=
\begin{bmatrix}
0.9*0.585 + 0.2*0.415 \
0.1*0.585 + 0.8*0.415
\end{bmatrix}
=
\begin{bmatrix}
0.5265 + 0.083 \
0.0585 + 0.332
\end{bmatrix}
=
\begin{bmatrix}
0.6095 \
0.3905
\end{bmatrix}
$$

**Continue for more periods:**

| Period | Murphy | Ashley |
|--------|--------|--------|
|   0    | 0.500  | 0.500  |
|   1    | 0.550  | 0.450  |
|   2    | 0.585  | 0.415  |
|   3    | 0.6095 | 0.3905 |
|   4    | 0.62665| 0.37335|
|   5    | 0.63866| 0.36134|

You can see that after several periods, the probabilities stabilize.

#### **Long-Run Probabilities:**
Let $ \pi = \begin{bmatrix} x \ y \end{bmatrix} $ be the steady state.  
$ \pi = P^T \cdot \pi $, and $ x + y = 1 $.

Set up equations:
$$
x = 0.9x + 0.2y \
y = 0.1x + 0.8y
$$
$$
x + y = 1
$$

From the first equation:
$$
x = 0.9x + 0.2y \implies x - 0.9x = 0.2y \implies 0.1x = 0.2y \implies x = 2y
$$
And since $ x + y = 1 $ then $ 2y + y = 1 \implies 3y = 1 \implies y = 1/3, x = 2/3 $.

**So, in the long run:**
- Murphy = **2/3 ≈ 0.667**
- Ashley = **1/3 ≈ 0.333**

**Observation:**  
Regardless of the initial state, the probabilities converge toward the long-run or steady-state probabilities.

---

## 2. New Fangled Softdrink Company Markov Chain

#### **Transition Matrix:**

|        | To Red Pop | To Super Cola |
|--------|:----------:|:-------------:|
| Red Pop|   0.9      |    0.1        |
| Super C|   0.1      |    0.9        |

---

### a. **Two-Period Tree Diagram for Customer Last Purchased Red Pop**

**First purchase:** Red Pop

**Second purchase:**
- If stayed with Red Pop: 0.9
- If switched to Super Cola: 0.1

**Third purchase:**
- If 2nd was Red Pop:
    - Red Pop again: 0.9
    - Super Cola: 0.1
- If 2nd was Super Cola:
    - Red Pop: 0.1
    - Super Cola: 0.9

**Tree:**

- Red Pop (0.9)
    - Red Pop (0.9 × 0.9 = 0.81)
    - Super Cola (0.9 × 0.1 = 0.09)
- Super Cola (0.1)
    - Red Pop (0.1 × 0.1 = 0.01)
    - Super Cola (0.1 × 0.9 = 0.09)

**Probability that customer buys Red Pop on 2nd purchase:**

$$
P(\text{Red Pop on 2nd}) = P(\text{Red Pop→Red Pop}) + P(\text{Red Pop→Super Cola→Red Pop})
$$
But the question asks: "the probability that this customer purchases Red Pop on the second purchase?"  
Since the first purchase is Red Pop, the probability for second purchase is:
- Stays with Red Pop: 0.9
- Switches to Super Cola: 0.1

So, **Probability = 0.9** that the customer buys Red Pop on second purchase.

---

### b. **Long-Run Market Share for Each Product**

Let $ x = $ long-run probability for Red Pop  
Let $ y = $ long-run probability for Super Cola

Set up equations:
$$
x = 0.9x + 0.1y \
y = 0.1x + 0.9y \
x + y = 1
$$

From first:
$$
x - 0.9x = 0.1y \implies 0.1x = 0.1y \implies x = y
$$
With $ x + y = 1 \implies 2x = 1 \implies x = 0.5, y = 0.5 $

**Long-run market share for each:**
- Red Pop: **0.5 (50%)**
- Super Cola: **0.5 (50%)**

---

## **Summary Table for Problem 1**

| Period | Murphy | Ashley |
|--------|--------|--------|
|   0    | 0.500  | 0.500  |
|   1    | 0.550  | 0.450  |
|   2    | 0.585  | 0.415  |
|   3    | 0.6095 | 0.3905 |
|   4    | 0.6267 | 0.3733 |
|   5    | 0.6387 | 0.3613 |

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