/*! 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 4 The Markov sequence \(\mathcal{A... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 4

The Markov sequence \(\mathcal{A}\) is realizable if and only if there exists a canonical triple realizing it.

Short Answer

Expert verified
To determine if the Markov sequence \(\mathcal{A} = (a_1, a_2, a_3, ... , a_n)\) is realizable, we need to create the corresponding canonical triple \((\sigma, \mu, \nu)\) and check whether this exists. The canonical triple consists of state space \(\sigma\), transition probabilities \(\mu_{ij}\), and initial distribution \(\nu_i\). The Markov sequence is realizable if it satisfies the Chapman-Kolmogorov equations and the initial distribution's probabilities sum to 1. If these conditions are met, the canonical triple exists, and the Markov sequence \(\mathcal{A}\) is realizable; otherwise, it is not realizable.

Step by step solution

01

Write down the given Markov sequence

We start by writing down the given Markov sequence \(\mathcal{A}\). Note: Since the exercise didn't specify the Markov sequence, we will use a general notation for the time being. If the exercise has provided specific values for the sequence, replace \(\mathcal{A}\) with the given sequence in the following steps. The given Markov sequence is: \[ \mathcal{A} = (a_1, a_2, a_3, ... , a_n) \]
02

Create the corresponding canonical triple

Next, we have to create the canonical triple \((\sigma, \mu, \nu)\) corresponding to the Markov sequence \(\mathcal{A}\). This triple will have: - State space \(\sigma = \{1, 2, ..., n\},\) - Transition probabilities \(\mu_{ij} = P(X_{t+1} = j \mid X_t = i)\), and - Initial distribution \(\nu_i = P(X_1 = i)\). We can organize all the transition probabilities into an \(n \times n\) matrix \(M\) with entries \(\mu_{ij}\) and denote the initial distribution as a row vector \(\nu = (\nu_1, \nu_2, ..., \nu_n)\).
03

Check if the canonical triple exists

Finally, we need to check whether the canonical triple exists based on the given Markov sequence. If the canonical triple exists, the Markov sequence will satisfy certain consistency conditions called the "Chapman-Kolmogorov equations". These equations relate the elements of the canonical triple in the following way: For every positive integers \(s\) and \(t\) and for every \(i\) and \(j\), \[ P(X_{s+t} = j \mid X_s = i) = \sum_{k=1}^n P(X_{s+t} = j \mid X_t = k)P(X_t = k \mid X_s = i) \Rightarrow M^{s+t}_{ij} = \sum_{k=1}^n M^{s}_{ik}M^{t}_{kj}. \] Additionally, the initial distribution's probabilities must sum to 1: \[ \sum_{i=1}^n \nu_i = 1. \] If the Markov sequence satisfies these two conditions, then the canonical triple exists, and the Markov sequence \(\mathcal{A}\) is realizable. Otherwise, it is not realizable.

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 91影视!

Key Concepts

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

Canonical Triple
To understand Markov sequence realizability, we must first understand the concept of a canonical triple. A canonical triple, denoted as \((\sigma, \mu, u)\), helps characterize a Markov sequence. It combines three components into one structure:

  • State Space (\(\sigma\)): This represents all possible states of the sequence. For an \(n\)-state Markov sequence, the state space is typically written as \(\{1, 2, ..., n\}\). These states are crucial as they outline the system or process modeled by the Markov chain.
  • Transition Probabilities (\(\mu\)): This is the probability \(P(X_{t+1} = j \mid X_t = i)\) that the system will transition from state \(i\) to state \(j\) at the next time step. These are usually arranged in a matrix format for clarity.
  • Initial Distribution (\(u\)): This row vector describes the starting probability distribution across the states. It specifies the likelihood of the system starting in each state. This is often one of the first steps in simulating a Markov process, as it dictates potential initial conditions.
Understanding these components is instrumental because a Markov sequence is only realizable if these three elements exist and satisfy certain conditions, such as those imposed by the Chapman-Kolmogorov equations.
Transition Probabilities
Transition probabilities are a core part of any Markov sequence and are essential for understanding how the process behaves over time. They describe the likelihood of moving from one state to another. For a Markov sequence with \(n\) states, these probabilities \( \mu_{ij} = P(X_{t+1} = j \mid X_t = i) \) are organized into an \(n \times n\) matrix \(M\).

  • \(\textbf{Row-Stochastic Matrix:}\) A key property of the transition matrix is that each of its rows must sum to 1. This ensures that all probabilities are accounted for when the system transitions from one state to the next.
  • Ensuring Validity: To make sure transition probabilities are valid, they must be non-negative and adhere to the laws of probability.
Understanding the transition matrix allows us to predict future states of the Markov process given its current state. It is this predictability that makes Markov models so powerful in fields like economics, genetics, and even artificial intelligence. Transition probabilities can be considered the "engine" of the Markov sequence, determining how it evolves and ensuring consistent probabilistic behavior aligning with the initial distribution and, eventually, with stationary distributions if they exist.
Chapman-Kolmogorov Equations
The Chapman-Kolmogorov equations are a fundamental aspect of Markov processes. They provide the conditions necessary for the composite transitions in a Markov sequence to hold true, relating short-term dynamics to longer-term transitions. This helps verify if a canonical triple for a Markov sequence is valid.

  • Equation Definition: The Chapman-Kolmogorov equations state that for any integers \(s\) and \(t\), and for every pair of states \(i\) and \(j\), the probability of transitioning from state \(i\) to state \(j\) over \(s+t\) time steps is equal to the sum of probabilities over intervening states \(k\). Mathematically: \[ P(X_{s+t} = j \mid X_s = i) = \sum_{k=1}^n P(X_{s+t} = j \mid X_t = k)P(X_t = k \mid X_s = i) \]
  • Connecting Transitions: This equation ensures that the composite transitions between states over non-consecutive times can be broken down into products of simpler one-step transitions.
These equations are indispensable for ensuring the validity of the state transitions in a Markov sequence. They ascertain that the initial and transition conditions set by the canonical triple remain consistent over any number of steps. By evaluating these, we can definitively state if the sequence is realizable within the boundaries defined by the canonical triple.

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

If two controllable triples are similar, then the similarity must be given by the formulas obtained in the proof of Theorem 27, so in particular similarities between canonical systems are unique. This is because if \(T\) is as in Definition \(6.5 .8\), then necessarily $$ T^{-1} A^{i} B=\widetilde{A}^{i} \widetilde{B} $$ for all \(i\), so it must hold that \(T^{-1} \mathbf{R}=\widetilde{\mathbf{R}}\), and therefore $$ T=\mathbf{R} \widetilde{\mathbf{R}}^{\\#} $$ In particular, this means that the only similarity between a canonical system and itself is the identity. In the terminology of group theory, this says that the action of \(G L(n)\) on triples is free.

The topological system \(\Sigma\) is locally observable about the state \(x^{0} \in X\) if there exists some neighborhood \(V\) of \(x^{0}\) such that every state in \(V\) different from \(x^{0}\) is distinguishable from \(x^{0}\).

There is a continuous-time version of the above result. Consider continuous- time polynomial systems $$ \dot{x}=f(x, u), \quad y=h(x) $$ where \(\mathcal{X}, \mathcal{U}, \mathcal{y}, h\) are as above and \(f\) is polynomial. We assume completeness. Show that, again, there exists then a finite set of controls \(\omega_{1}, \ldots, \omega_{k}\) such that, for any pair of states \(x\) and \(z\), these states are distinguishable if and only if one of the controls \(\omega_{i}\) distinguishes them. (Hint: Use analytic controls and argue in terms of the corresponding derivatives \(y(0), \dot{y}(0), \ldots\) that there must exist for each such control a polynomial \(\Delta_{\omega}\) so that \(x\) and \(z\) are indistinguishable by \(\omega\) if and only if \(\Delta_{\omega}(x, z)=0\). The Hilbert Basis Theorem is needed for that. Now apply the Basis Theorem again.)

Let \(\Sigma\) be a complete time-invariant system with outputs and card \(y \geq 2\). Consider the system $$ \tilde{\Sigma}:=(\mathcal{T}, \tilde{\mathcal{X}}, \mathcal{U}, \tilde{\phi}) $$ defined as follows. Its state space \(\widetilde{\mathcal{X}}\) is the set of all maps \(x \rightarrow y\), and for each \(\omega \in \mathcal{U}^{[\sigma, \tau)}\) and each state \(\alpha: X \rightarrow y\) of \(\widetilde{X}\) $$ \widetilde{\phi}(\tau, \sigma, \alpha, \omega) $$ is the map $$ x \mapsto \alpha(\phi(\tau, \sigma, x, \widetilde{\omega})) $$ where \(\widetilde{\omega}\) is the time-reversed control \(\widetilde{\omega}(t):=\omega(\sigma+\tau-t) .\) Note that the output map \(h\) can be seen as an element of \(\widetilde{X}\). Prove that if \(\widetilde{\Sigma}\) is reachable from \(h\), i.e., $$ \mathcal{R}(h)=\widetilde{X} $$ then \(\Sigma\) is observable.

If \(m=1\) or \(p=1\) and \(W_{\mathcal{A}}=P / q\) with \(q\) monic of degree equal to the rank of \(\mathcal{A}\), then \(q\) is the (common) characteristic polynomial of the canonical realizations of \(\mathcal{A}\).
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
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.