/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 The transition matrix for a Mark... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The transition matrix for a Markov process is given by \(T=\begin{array}{l}\text { State } 1 \\ \text { State } 2\end{array}\left[\begin{array}{ll}\frac{1}{2} & \frac{3}{4} \\ \frac{1}{2} & \frac{1}{4}\end{array}\right]\) and the initial-state distribution vector is given by $$ X_{0}=\begin{array}{l} \text { State } 1 \\ \text { State } 2 \end{array} \quad\left[\begin{array}{l} \frac{1}{3} \\ \frac{2}{3} \end{array}\right] $$ Find \(T X_{0}\) and interpret your result with the aid of a tree diagram.

Short Answer

Expert verified
In summary, by multiplying the transition matrix \(T\) and the initial-state distribution vector \(X_0\), we find the probabilities for each state after one transition. The result is: \(T \times X_0 = \left[\begin{array}{l} \frac{7}{12}\\ \frac{5}{12} \end{array}\right]\) This means that after the first transition, there is a \(\frac{7}{12}\) probability of being in state 1 and a \(\frac{5}{12}\) probability of being in state 2. The tree diagram illustrates the transition probabilities from the initial states to the next states after one transition.

Step by step solution

01

Find the product T*X_0

To find the product of the transition matrix T and the initial-state distribution vector X_0, we simply perform matrix multiplication: $$ T \times X_0 = \left[\begin{array}{ll} \frac{1}{2} & \frac{3}{4} \\ \frac{1}{2} & \frac{1}{4} \end{array}\right] \times \left[\begin{array}{l} \frac{1}{3} \\ \frac{2}{3} \end{array}\right] $$
02

Perform the matrix multiplication

To multiply the matrix and the vector, we follow the procedure for matrix multiplication: 1. Multiply the first row of T by the elements of X_0, element-wise, and sum the results: \[ \frac{1}{2} \times \frac{1}{3} + \frac{3}{4} \times \frac{2}{3} = \frac{1}{6} + \frac{6}{12} \] 2. Multiply the second row of T by the elements of X_0, element-wise, and sum the results: \[ \frac{1}{2} \times \frac{1}{3} + \frac{1}{4} \times \frac{2}{3} = \frac{1}{6} + \frac{2}{12} \] Now, let's add the two results together:
03

Combine the results from step 2

Adding the results from step 2, we obtain our product vector: $$ T \times X_0 = \left[\begin{array}{l} \frac{1}{6} + \frac{6}{12}\\ \frac{1}{6} + \frac{2}{12} \end{array}\right] = \left[\begin{array}{l} \frac{7}{12}\\ \frac{5}{12} \end{array}\right] $$
04

Interpret the results with a tree diagram

The result vector represents the probabilities of being in each state after one transition from the initial states: - State 1: \(\frac{7}{12}\) probability - State 2: \(\frac{5}{12}\) probability Now, let's draw a tree diagram to visualize these probabilities: ``` Initial State ________________ | | ↙ ↘ 1/3 2/3 (State 1) (State 2) | | ↙ ↘ 7/12 5/12 (State 1) (State 2) ``` After the first transition, there is a total probability of \(\frac{7}{12}\) that the system will be in state 1 and a total probability of \(\frac{5}{12}\) that it will be in state 2.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix Multiplication
When we talk about matrix multiplication in the context of a Markov process, we're dealing with the mechanics of how one state can transition to another. Imagine that each row and column in the matrix represents different states, and the numbers inside capture the probabilities of moving from one state to another. In order to understand the next step, where the system will most likely be, we use matrix multiplication.

To multiply a matrix by a vector, as seen in the problem's step-by-step solution, we perform an element-wise multiplication of the rows of our matrix (the transition matrix) by the vector (the initial state distribution). Then, we sum up the results of these multiplications for each row. This process is conceptually similar to taking a weighted average, where the weights are the probabilities of transitioning from one state to another. It's important to always ensure that the number of columns in the matrix equals the number of rows in the vector to perform multiplication properly.

The result of this matrix-vector multiplication gives us a new vector, which illustrates the probabilities of being in each state after the transition.
Initial-State Distribution Vector
The initial-state distribution vector is a fundamental concept in understanding Markov processes. It represents the starting point of our system; in other words, it's a snapshot of where we are before any transitions occur. This vector is composed of probabilities that add up to 1, signifying that the system must be in one of the possible states.

In our sample exercise, the vector \(X_{0}\) contained the probabilities for the system being in state 1 or state 2 initially. These values were \(\frac{1}{3}\) and \(\frac{2}{3}\), respectively. When we multiply our transition matrix by the initial-state distribution vector, we effectively simulate a step forward in time, seeing how those initial probabilities spread and change due to the transition rules defined in the matrix. This kind of operation enables us to predict the behavior of the system over time, which is incredibly useful in various fields such as finance, epidemiology, and computer science.
Probability Tree Diagram
A probability tree diagram is a visual representation that helps us map out the possible outcomes of a process, like our Markov process. It's a tool that lays out all possible paths that can be taken from a given starting point, along with the probabilities associated with each path.

In the solution to the exercise, a simple tree diagram was used to illustrate the one-step transition probabilities. It began with the two initial states and branched out to show the resulting probabilities for each subsequent state. With each 'branch,' the probabilities along the paths were multiplied together, providing the cumulative probability of reaching that end state from the starting point.

The use of a tree diagram can be incredibly informative in more complex Markov processes involving multiple steps, where it would show a spreading network of possible paths. It is an excellent way to conceptually understand the process, allowing us to see the probabilities at a glance and make predictions about future states of the system.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Within a large metropolitan area, \(20 \%\) of the commuters currently use the public transportation system, whereas the remaining \(80 \%\) commute via automobile. The city has recently revitalized and expanded its public transportation system. It is expected that 6 mo from now \(30 \%\) of those who are now commuting to work via automobile will switch to public transportation, and \(70 \%\) will continue to commute via automobile. At the same time, it is expected that \(20 \%\) of those now using public transportation will commute via automobile, and \(80 \%\) will continue to use public transportation. In the long run, what percentage of the commuters will be using public transportation?

The payoff matrix for a game is $$ \left[\begin{array}{rrr} 4 & -3 & 3 \\ -4 & 2 & 1 \\ 3 & -5 & 2 \end{array}\right] $$ a. Find the expected payoff to the row player if the row player \(R\) uses the maximin pure strategy and the column player \(C\) uses the minimax pure strategy. b. Find the expected payoff to the row player if \(R\) uses the maximin strategy \(40 \%\) of the time and chooses each of the other two rows \(30 \%\) of the time, while \(C\) uses the minimax strategy \(50 \%\) of the time and chooses each of the other columns \(25 \%\) of the time. c. Which of these pairs of strategies is most advantageous to the row player?

The management of Acrosonic is faced with the problem of deciding whether to expand the production of its line of electrostatic loudspeaker systems. It has been estimated that an expansion will result in an annual profit of \(\$ 200,000\) for Acrosonic if the general economic climate is good. On the other hand, an expansion during a period of economic recession will cut its annual profit to \(\$ 120,000 .\) As an alternative, Acrosonic may hold the production of its electrostatic loudspeaker systems at the current level and expand its line of conventional loudspeaker systems. In this event, the company will make a profit of \(\$ 50,000\) in an expanding economy (because many potential customers will be expected to buy electrostatic loudspeaker systems from other competitors) and a profit of \(\$ 150,000\) in a recessionary economy. a. Construct the payoff matrix for this game. Hint: The row player is the management of the company and the column player is the economy. b. Should management recommend expanding the company's line of electrostatic loudspeaker systems?

Determine whether the matrix is an absorbing stochastic matrix. \(\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\)

Find \(X_{2}\) (the probability distribution of the system after two observations) for the distribution vector \(X_{0}\) and the transition matrix \(T\). \(X_{0}=\left[\begin{array}{l}.6 \\ .4\end{array}\right], T=\left[\begin{array}{ll}.4 & .8 \\ .6 & .2\end{array}\right]\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.