/*! 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 23 A certain town never has two sun... [FREE SOLUTION] | 91Ó°ÊÓ

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A certain town never has two sunny days in a row. Each day is classified as being either sunny, cloudy (but dry), or rainy, If it is sunny one day, then it is equally likely to be either cloudy or rainy the next day. If it is rainy or cloudy one day, then there is one chance in two that it will be the same the next day, and if it changes then it is equally likely to be either of the other two possibilities. In the long run, what proportion of days are sunny? What proportion are cloudy?

Short Answer

Expert verified
In the long run, the proportions of days are as follows: \(30\%\) sunny days, \(40\%\) cloudy days, and \(30\%\) rainy days.

Step by step solution

01

Set up the Transition Matrix

First, we need to set up the transition matrix. Let S represent a sunny day, C represent a cloudy day, and R represent a rainy day. The transition matrix P is given as: \(P = \begin{bmatrix} P(S|S) & P(S|C) & P(S|R) \\ P(C|S) & P(C|C) & P(C|R) \\ P(R|S) & P(R|C) & P(R|R) \end{bmatrix}\) We are given the following probabilities: 1. \(P(S|S) = 0\) (there can never be two sunny days in a row) 2. \(P(C|S) = P(R|S) = \frac{1}{2}\) (it is equally likely to be either cloudy or rainy the next day) 3. \(P(C|C) = P(R|R) = \frac{1}{2}\) (there is a 50% chance of having the same weather the next day for cloudy and rainy days) 4. If the weather changes, it is equally likely to be either of the other two possibilities. With this information, we can fill the transition matrix: \(P = \begin{bmatrix} 0 & \frac{1}{4} & \frac{1}{4} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{4} \\ \frac{1}{2} & \frac{1}{4} & \frac{1}{2} \end{bmatrix}\)
02

Calculate the Steady State Probabilities

To find the long term probabilities, we need to find the steady state probabilities. Let \(\pi = [\pi_S, \pi_C, \pi_R]\) be the steady state vector. We have \(\pi P = \pi\) and \(\sum_{i} \pi_i = 1\). Multiplying the steady state vector with the transition matrix, we get: \(\begin{bmatrix} \pi_S, \pi_C, \pi_R \end{bmatrix} \begin{bmatrix} 0 & \frac{1}{4} & \frac{1}{4}\\ \frac{1}{2} & \frac{1}{2} & \frac{1}{4}\\ \frac{1}{2} & \frac{1}{4} & \frac{1}{2}\end{bmatrix} =\begin{bmatrix} \pi_S, \pi_C, \pi_R\end{bmatrix}\) This gives us the following set of equations: 1. \(\pi_S = \frac{1}{4}\pi_C + \frac{1}{4}\pi_R\) 2. \(\pi_C = \frac{1}{2}\pi_S + \frac{1}{2}\pi_C + \frac{1}{4}\pi_R\) 3. \(\pi_R = \frac{1}{2}\pi_S + \frac{1}{4}\pi_C + \frac{1}{2}\pi_R\) Additionally, we have the normalization condition: \(\pi_S + \pi_C + \pi_R = 1\).
03

Solve the Steady State Equations

Now, solve the equations to find the steady state probabilities: 1. From equation 1: \(\pi_S = \frac{1}{4}\pi_C + \frac{1}{4}\pi_R\) 2. Substituting the value of \(\pi_S\) into equation 2: \(\frac{1}{4}\pi_C + \frac{1}{4}\pi_R = \frac{1}{2}(\frac{1}{4}\pi_C + \frac{1}{4}\pi_R)+ \frac{1}{2}\pi_C + \frac{1}{4}\pi_R\) Solving this equation, we get: \(\pi_C = \frac{2}{5}\) 3. Now, substitute the value of \(\pi_C\) into normalization condition: \(\pi_S + \frac{2}{5} + \pi_R = 1\) And substitute the value of \(\pi_S\) into the equation: \(\frac{1}{4}\pi_C + \frac{1}{4}\pi_R + \frac{2}{5} + \pi_R = 1\) Solving this equation, we get: \(\pi_R = \frac{3}{10}\) Finally, substitute the value of \(\pi_C\) and \(\pi_R\) into equation 1: \(\pi_S = \frac{1}{4}\times \frac{2}{5} + \frac{1}{4}\times\frac{3}{10} = \frac{3}{10}\) Thus, the steady state probabilities are: \(\pi_S = \frac{3}{10}\) (sunny days) \(\pi_C = \frac{2}{5}\) (cloudy days) \(\pi_R = \frac{3}{10}\) (rainy days) In the long run, 30% of the days are sunny, 40% are cloudy, and 30% are rainy.

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Key Concepts

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

Transition Matrix
A transition matrix is a powerful tool used to represent a Markov chain, especially when modeling systems where each state transitions to one or more other states with certain probabilities. In the context of weather modeling, it helps describe how the weather moves from one state to another, such as from sunny to cloudy to rainy and vice versa.
The transition matrix is often expressed as a square matrix, where each element \( p_{ij} \) represents the probability of transitioning from state \( i \) to state \( j \). For example, in a 3-state weather model (sunny, cloudy, rainy), the transition matrix is: \[P = \begin{bmatrix}P(S|S) & P(S|C) & P(S|R) \P(C|S) & P(C|C) & P(C|R) \P(R|S) & P(R|C) & P(R|R)\end{bmatrix}\] In this matrix, the rows represent the current state and the columns represent the next state. The transition probabilities must satisfy two conditions: each probability is between 0 and 1, and the probabilities in each row must sum to 1.
For example, for a town's weather model stated in the exercise, with specific rules such as no consecutive sunny days and equal probabilities of a change to cloudy or rainy days, the transition matrix becomes: \[P = \begin{bmatrix}0 & \frac{1}{2} & \frac{1}{2} \\frac{1}{2} & \frac{1}{2} & 0 \\frac{1}{2} & 0 & \frac{1}{2}\end{bmatrix}\]Here, each element's value reflects the logic given, ensuring a logical representation of weather behavior over days.
Steady State Probabilities
Steady state probabilities in a Markov chain refer to a situation where the probabilities of being in each state remain constant over time. This occurs when the system reaches equilibrium, meaning that the distribution of states doesn't change despite ongoing transitions. For weather modeling, steady state probabilities help predict the long-term expected frequency of each type of weather.
To find steady state probabilities, we solve the equation \( \pi P = \pi \), alongside the condition \( \sum \pi_i = 1 \). Here, \( \pi \) is the steady state vector \([\pi_S, \pi_C, \pi_R ]\), representing the probabilities of sunny, cloudy, and rainy days respectively. Using the transition matrix from our weather example, this results in a system of linear equations.
While solving, it is often necessary to employ algebra to simplify these equations. For instance, simplifying the equation results, and using normalization condition \( \pi_S + \pi_C + \pi_R = 1 \), we derive:- \( \pi_S = \frac{1}{4}\pi_C + \frac{1}{4}\pi_R \)- \( \pi_C = \frac{1}{2}\pi_S + \frac{1}{2}\pi_C + \frac{1}{4}\pi_R \)- \( \pi_R = \frac{1}{2}\pi_S + \frac{1}{4}\pi_C + \frac{1}{2}\pi_R \) By solving these, we find \( \pi_S = 0.3 \), \( \pi_C = 0.4 \), and \( \pi_R = 0.3 \), indicating long-term proportions of sunny, cloudy, and rainy days.
Weather Modeling
Weather modeling using Markov chains is a practical application that utilizes probabilities to predict future weather conditions based on present conditions and their potential changes. By understanding the transition matrix and solving for steady state probabilities, models can simulate and analyze the probability distribution of different weather states over time.
Unlike deterministic models which give precise predictions, Markov models account for randomness and are stochastic. They provide insights into the likelihood of different types of weather occurring, thus assisting in planning and decision-making processes.
To create an effective model, it's crucial to incorporate realistic rules and constraints, like those observed in the town's weather scenario where no two sunny days occur consecutively. Such rules shape the transition probabilities, offering more reliable predictions.
  • A town's specific weather transition rules can be used to construct a transition matrix.
  • Solving the system of equations associated with the transition matrix yields steady state probabilities.
  • Markov chains help estimate the long-term expected distribution of weather types, aiding financial, agricultural, and daily planning.
This way, weather modeling informs decisions by capturing the essence of weather pattern probabilities.

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Most popular questions from this chapter

Consider a Markov chain with states \(0,1,2,3,4\). Suppose \(P_{0,4}=1\); and suppose that when the chain is in state \(l, i>0\), the next state is equally likely to be any of the states \(0,1, \ldots, i-1\). Find the limiting probabilities of this Markov chain.

Consider a population of individuals each of whom possesses two genes which can be either type \(A\) or type \(a\). Suppose that in outward appearance type \(A\) is dominant and type \(a\) is recessive. That is, an individual will only have the outward characteristics of the recessive gene if its pair is aa.) Suppose that the population has stabilized, and the percentages of individuals having respective gene pairs \(A A, a a\), and \(A a\) are \(p, q\), and \(r .\) Call an individual dominant or recessive depending on the outward characteristics it exhibits. Let \(S_{11}\) denote the probability that an offspring of two dominant parents will be recessive; and let \(S_{10}\) denote the probability that the offspring of one dominant and one recessive parent will be recessive. Compute \(S_{11}\) and \(S_{10}\) to show that \(S_{11}=S_{10}^{2}\). The quantities \(S_{10}\) and \(S_{11}\) are known in the genetics literature as Snyder's ratios.)

Let \(P^{(1)}\) and \(P^{(2)}\) denote transition probability matrices for ergodic Markov chains having the same state space. Let \(\pi^{1}\) and \(\pi^{2}\) denote the stationary (limiting) probability vectors for the two chains. Consider a proeess defined as follows: (i) \(X_{0}=1 .\) A coin is then flipped and if it comes up heads, then the remaining states \(X_{1}, \ldots\) are obtained from the transition probability matrix \(P^{(1)}\) and if tails from the matrix \(P^{(2)} .\) Is \(\left\\{X_{n}, n \geq 0\right\\}\) a Markov chain? If \(p=P\) coin comes up heads], what is \(\lim _{n \rightarrow \infty} P\left(X_{n}=f\right] ?\) (ii) \(X_{0}=1\). At each stage the coin is flipped and if it comes up heads. then the next state is chosen according to \(P^{(1)}\) and if tails comes up, then it is chosen according to \(P^{(2)}\). In this case do the successive states constitute a Markov chain? If so, determine the transition probabilities. Show by a counterexample that the limiting probabilities are not the same as in part (i).

Three white and three black balls are distributed in two urns in such a way that each contains three balls. We say that the system is in state \(l_{3}\) \(i=0,1,2,3\), if the first urn contains \(i\) white balls. At each step, we draw one ball from each urn and place the ball drawn from the first urn into the second, and conversely with the ball from the second urn. Let \(X_{n}\) denote the state of the system after the \(n\) th step. Explain why \(\left[X_{n}, n=0,1,2, \ldots\right]\) is a Markov chain and calculate its transition probability matrix.

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