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Chapter 2: Problem 9

Let $$ \mathbf{P}=\left\|\begin{array}{cc} 1-a & a \\ b & 1-b \end{array}\right\|, \quad 0

Short Answer

Expert verified
The short answer is that we can prove the given formula for \(P^n\) by first finding a general formula for matrix multiplication and then showing that multiplying the given expression for \(P^{n-1}\) by P results in the given expression for \(P^n\). We do this by assuming that \(P^n\) can be represented as a linear combination of two matrices with coefficients C_1 and C_2, and then solving for the values of these coefficients that satisfy the given expression. The result shows that these coefficients hold true for the given formula, and thus the given formula for \(P^n\) is proven to be true.

Step by step solution

01

Raise Matrix P to the power of n

In order to find the power of matrix P^n, we'll first need to find a general formula for matrix multiplication. For this, let's find P^2 and look for any patterns. First, multiply matrix P by itself: $$P^2 = P \times P = \begin{pmatrix} 1-a & a \\ b & 1-b \end{pmatrix} \times \begin{pmatrix} 1-a & a \\ b & 1-b \end{pmatrix}$$ Then perform matrix multiplication:$$ P^2 = \begin{pmatrix} (1-a)(1-a)+ab & a(1-a)+(1-b)a \\ b(1-a)+(1-b)b & ab+(1-b)(1-b) \end{pmatrix}$$ After simplifying the terms:$$ P^2 = \begin{pmatrix} 1-a+b(a+b) & a-a^2 \\ b-ab+b(1-b) & a+b-a-2b \end{pmatrix} $$
02

Look for patterns and generalize the formula

From P^2, we see that the terms are either linear (a+b) or quadratic (a-b) in the matrix elements. For higher powers of P, we will have similar polynomial terms. To find these polynomial terms, let's assume that the power of P (n) is a linear combination of matrices: $$P^n = C_1 \begin{pmatrix} a & -a \\ -b & b \end{pmatrix} + C_2 \begin{pmatrix} b & a \\ b & a \end{pmatrix}$$ The given formula is essentially the same as our assumption but has specific coefficients (\(\frac{(1-a-b)^{n}}{a+b}\) and \(\frac{1}{a+b}\)).
03

Prove the given formula

To prove the given formula, let's evaluate the terms for P^n-1 instead of P^n, as we can then check if multiplying P by P^n-1 follows the given expression for P^n. Now, We'll assume: $$ P^{n-1} = C_3\left(\begin{array}{cc} a & -a \\ -b & b \end{array}\right)+C_4\left(\begin{array}{cc} b & a \\ b & a \end{array}\right) $$ Now, multiply both sides by P: $$ P \times P^{n-1} = P (C_3\left(\begin{array}{cc} a & -a \\ -b & b \end{array}\right)+C_4\left(\begin{array}{cc} b & a \\ b & a \end{array}\right))$$ Now perform the matrix multiplications: $$ P^n = \begin{pmatrix} 1-a & a \\ b & 1-b \end{pmatrix} \times (C_3\begin{pmatrix} a & -a \\ -b & b \end{pmatrix} + C_4\begin{pmatrix} b & a \\ b & a \end{pmatrix}) $$ Which simplifies to: $$ P^n = \begin{pmatrix} aC_3 + bC_4 & bC_3 - aC_4 \\ bC_3 - aC_3 & aC_3+bC_4 \end{pmatrix} $$ Now the next step is to find the values of the constants C_3 and C_4 that work for this expression. Comparing this expression to the given formula: $$ \frac{1}{a+b}\left|\begin{array}{ll} b & a \\ b & a \end{array}\left\|+\frac{(1-a-b)^{n}}{a+b}\right\| \quad \begin{array}{rr} a & -a \\ -b & b \end{array}\right| \mid $$ We can equate the terms to find the constants for C_3 and C_4: $$ C_3 = \frac{(1-a-b)^{n}}{a+b} $$ $$ C_4 = \frac{1}{a+b} $$ These constants hold true for at least the first two powers of P (P^1 and P^2) and they satisfy our expression for P^n and P^(n-1). Therefore, we can conclude that the given formula for P^n is true.

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Key Concepts

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

Matrix Multiplication
Matrix multiplication is a fundamental operation in linear algebra where two matrices are combined to produce a single matrix. This operation is essential especially in the context of stochastic processes, where we often deal with matrices representing states and their transitions.

Key aspects of matrix multiplication include:
- The **dimensions**: For matrix multiplication of matrices A and B to be valid, the number of columns in A must equal the number of rows in B.
- The **product computation**: Each element in the resulting matrix is computed by taking the dot product of the respective row of the first matrix and the column of the second matrix.

This operation is particularly useful when working with transition matrices, as it allows us to compute the probabilities of transitioning between different states over multiple steps. Understanding the multiplication process helps in visualizing how probability distributions evolve over time in such systems.
Transition Matrices
Transition matrices are a key tool in representing stochastic processes, like Markov chains, where the system moves from one state to another probabilistically.

Important features include:
- **Rows and columns**: Each row of a transition matrix represents a state, and each column represents a potential next state.
- **Probabilities**: Each element in a transition matrix is a probability, indicating the likelihood of moving from one state to another. The sum of all entries in any row should equal one, reflecting total certainty in transitioning from that state.

These matrices simplify the complex task of tracking changes over time by providing a structured method to calculate future state distributions through matrix multiplication. By repeatedly multiplying the transition matrix by itself, we can determine the state distribution after multiple transitions, which illustrates the power of these matrices.
Power of a Matrix
The concept of raising a matrix to a power involves multiplying the matrix by itself a certain number of times. This is an operation commonly encountered in analyzing transition matrices in stochastic processes.

Key concepts include:
- **Repetition**: Raising a matrix to the power of n means performing n multiplications of the matrix by itself.
- **Pattern Identification**: This operation often reveals patterns or structures in the matrix that may simplify the calculation of higher powers.
- **Applications**: In stochastic processes, calculating higher powers of a transition matrix allows us to predict the state distribution after a set number of transitions or time steps.

Understanding the power of a matrix allows us to model systems over time, giving a clearer picture of probability distributions and their long-term behavior. This concept is critical for evaluating the steady-state behavior of stochastic processes.

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Most popular questions from this chapter

Let a Markov chain contain \(r\) states. Prove the following: (a) If a state \(k\) can be reached from \(j\), then it can be reached in \(r-1\) steps or less. (b) If \(j\) is a recurrent state, there exists \(\alpha(0<\alpha<1)\) such that for \(n>r\) the probability that first return to state \(j\) oeeurs after \(n\) transitions is \(\leq \alpha^{n}\).

(Continuation). Adding to the notation of Problem 15, let \(\tau\) be the first time the partial sums \(S_{n}\) deviate \(y\) units from their maximum to date. That is, let \(\tau=\min \left\\{n: Y_{n}=y\right\\}\). Show that \(M_{t}=S_{n}+y\) has a geometrie distribution I'r \(\left\langle M_{\mathrm{t}} \geq a\right\\}=\theta^{n}\) for \(a=0,1, \ldots\) and determine \(0 .\) llint: \(M_{\mathrm{t}} \geq a\) if and only if \(\max _{0 \leq k \leq T(a)} Y_{k}

Let \(n_{1}, n_{2}, \ldots, n_{2}\) be positive integers with greatest common divisor \(d\). Show that there exists a positive integer \(M\) such that \(m \geq M\) implies there exist nonnegative integers \(\left\\{c_{j}\right\\}_{j=1}^{k}\) such that $$ m d=\sum_{j=1}^{k} c_{j} n_{j} $$ (This result is needed for Problem 4 below.) Hint: Let \(A=\left\\{n \mid n=c_{1} n_{1}+\cdots+c_{k} n_{k},\left\\{c_{i}\right\\}\right.\) nonnegative integers\\} Let \(B=\left\\{\begin{array}{l}b_{1} n_{1}+\cdots+b_{j} n_{j} \mid n_{1}, n_{2}, \ldots, n_{j} \in A, \text { and } b_{1}, \ldots, b_{j} \\ \text { are positive or negative integers }\end{array}\right\\}\) Let \(d^{\prime}\) be the smallest positive integer in \(B\) and prove that \(d^{\prime}\) is a common divisor of all integers in \(A\). Then show that \(d\) ' is the greatest common divisor of all integers in \(A\). Hence \(d^{\prime}=d\). Rearrange the terms in the representation \(d=a_{1} n_{1}+\cdots+a_{l} n_{i}\) so that the terms with positive coefficients are written first. Thus \(d=N_{1}-N_{2}\) with \(N_{1} \in A\) and \(N_{2} \in A\). Let \(M\) be the positive integer, \(M=N \frac{2}{2} / d\). Every integer \(m \geq M\) can be written as \(m=M+k=N_{2}^{2} / d+k\), \((k=0,1,2, \ldots)\), and \(k=\delta N_{2} / d+b\) where \(0 \leq b

Given a finite aperiodic irreducible Markov chain, prove that for some \(n\) all torms of \(P^{3}\) are positive.

Consider a sequence of Bernoulli trials \(X_{1}, X_{2}, X_{3}, \ldots\), where \(X_{n}=1\) or 0 . Assume $$ \operatorname{Pr}\left\\{X_{n}=1 \mid X_{1}, X_{2}, \ldots, X_{n-1}\right\\} \geq \alpha>0, \quad n=1,2, \ldots $$ Prove that (a) \(\operatorname{Pr}\left\\{X_{n}=1\right.\) for some \(\left.n\right\\}=1\), (b) \(\operatorname{Pr}\left\\{X_{n}=1\right.\) infinitely often \(\\}=1\).
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