/*! 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 17 For the random walk of Example \... [FREE SOLUTION] | 91Ó°ÊÓ

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For the random walk of Example \(4.18\) use the strong law of large numbers to give another proof that the Markov chain is transient when \(p \neq \frac{1}{2}\). Hint: Note that the state at time \(n\) can be written as \(\sum_{i=1}^{n} Y_{i}\) where the \(Y_{i}\) s are independent and \(P\left\\{Y_{i}=1\right\\}=p=1-P\left\\{Y_{i}=-1\right\\}\). Argue that if \(p>\frac{1}{2}\), then, by the strong law of large numbers, \(\sum_{1}^{n} Y_{i} \rightarrow \infty\) as \(n \rightarrow \infty\) and hence the initial state 0 can be visited only finitely often, and hence must be transient. A similar argument holds when \(p<\frac{1}{2}\).

Short Answer

Expert verified
By applying the strong law of large numbers, we find that the sample average converges to the expected value: \(\frac{1}{n}\sum_{i=1}^{n}Y_i \xrightarrow[n\to\infty]{} E[Y_1] = 2p-1\). For \(p > \frac{1}{2}\), the sum \(\sum_{i=1}^{n}Y_i \xrightarrow[n\to\infty]{} \infty\), and for \(p < \frac{1}{2}\), the sum \(\sum_{i=1}^{n}Y_i \xrightarrow[n\to\infty]{} -\infty\). In both cases, the initial state 0 can be visited only finitely often, implying that the Markov chain is transient when \(p \neq \frac{1}{2}\).

Step by step solution

01

Understand given information

We are given a Markov chain with state at time n represented as \(\sum_{i=1}^{n} Y_{i}\) where \(Y_i\) are independent random variables with probabilities \(P\{Y_i=1\} = p\) and \(P\{Y_i=-1\} = 1-p\).
02

Apply the strong law of large numbers

By the strong law of large numbers, the sample average converges to the expected value, i.e., \[\frac{1}{n}\sum_{i=1}^{n}Y_i \xrightarrow[n\to\infty]{} E[Y_1]\] In our case, the expected value of \(Y_1\) can be computed as \[E[Y_1] = 1 \cdot p + (-1) \cdot (1-p) = 2p-1\]
03

Analyze the case when \(p > \frac{1}{2}\)

When \(p > \frac{1}{2}\), \(2p-1 > 0\) and \[\frac{1}{n}\sum_{i=1}^{n}Y_i \xrightarrow[n\to\infty]{} 2p-1 > 0\] Multiplying both sides by \(n\), we obtain \[\sum_{i=1}^{n}Y_i \xrightarrow[n\to\infty]{} \infty\] Thus, the initial state 0 can be visited only finitely often, and hence the Markov chain is transient in this case.
04

Analyze the case when \(p < \frac{1}{2}\)

When \(p < \frac{1}{2}\), \(2p-1 < 0\) and \[\frac{1}{n}\sum_{i=1}^{n}Y_i \xrightarrow[n\to\infty]{} 2p-1 < 0\] Multiplying both sides by \(n\), we obtain \[\sum_{i=1}^{n}Y_i \xrightarrow[n\to\infty]{} -\infty\] Again, the initial state 0 can be visited only finitely often, and hence the Markov chain is transient in this case as well. In conclusion, the strong law of large numbers helps us prove that the Markov chain is transient when \(p \neq \frac{1}{2}\).

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Key Concepts

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

Strong Law of Large Numbers
The strong law of large numbers (SLLN) is one of the foundation stones in the field of probability and statistics. It tells us about the stability of long-term averages obtained from independent random events. In essence, the SLLN states that if you have a sequence of independent and identically distributed random variables, such as the tosses of a fair coin, the average of their outcomes will almost surely converge to the expected value as the number of trials goes to infinity.

In formal terms, if \(Y_1, Y_2, \dots, Y_n\) are independent and identically distributed random variables with an expected value \(E[Y_1]\), the SLLN ensures that the sample average \(\frac{1}{n}\sum_{i=1}^{n}Y_i \rightarrow E[Y_1]\) almost surely as \(n \rightarrow \infty\). This concept is pivotal when analyzing random processes, such as the transient Markov chain discussed in the textbook problem, as it gives a solid basis for making inferences about their long-term behavior.

Applying this law gives us an intuitive understanding of why, when a probability \(p \eq \frac{1}{2}\), the Markov chain's state will drift away from the starting point and not return, indicating transience.
Probability
Probability is the measure of the likelihood that an event will occur. It quantifies uncertainty and is represented as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. In the context of the transient Markov chain problem, probability is used to predict the behavior of the chain based on the likelihood of individual steps. Here, \(P\{Y_i=1\} = p\) quantifies the chance of moving to a higher state, while \(P\{Y_i=-1\} = 1-p\) represents the chance of moving to a lower state.

The concept of probability is crucial for understanding the outcomes of random processes like random walks and for applying the strong law of large numbers. By acknowledging the different probabilities of moving up or down in the Markov chain, we can infer the overall direction and behavior of the Markov process under various circumstances.
Random Walk
A random walk is a mathematical formalization of a path consisting of a series of random steps. It is widely used across disciplines such as economics, physics, and computer science to model seemingly random yet statistically predictable phenomena. In the context of our problem, a random walk describes the sequence of states in the transient Markov chain.

The one-dimensional random walk can be visualized as a person walking along a straight line who takes steps either forward or backward at each time period. If the probability of taking a step forward (\(p\)) is not equal to the probability of taking a step backward (\(1-p\)), the walk is said to be biased. This unequal probability influences the chain's overall tendency to drift in one direction, which is the central idea when establishing transience in the chain. The strong law of large numbers, when applied to a biased random walk, supports the conclusion that such a walk will result in the walker moving infinitely far away from the starting point.
Expected Value
The expected value, also known as the mean or the first moment, is a fundamental concept in probability theory that represents the average outcome of a random variable if the experiment is repeated a large number of times. It is calculated by summing all possible values of the random variable weighted by their respective probabilities.

In the random walk exercise, the expected value \(E[Y_1] = 1 \cdot p + (-1) \cdot (1-p) = 2p-1\) plays a crucial role. It sums up the average result of one step in the process and determines the direction in which the walk is biased. When \(p > \frac{1}{2}\), the expected value is positive, indicating that the random walk will, on average, move towards positive infinity. Conversely, when \(p < \frac{1}{2}\), the expected value is negative, suggesting a move towards negative infinity. Understanding this concept helps students grasp the idea that, depending on the value of \(p\), the random walk will diverge from the origin, thus rendering the Markov chain transient.

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Most popular questions from this chapter

Specify the classes of the following Markov chains, and determine whether they are transient or recurrent: $$\mathbf{P}_{1}=\left\|\begin{array}{lll} 0 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 0 \end{array} \mid, \quad \mathbf{P}_{2}=\right\| \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array} \|$$ $$\mathbf{P}_{3}=\left\|\begin{array}{|ccccc|} \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ \frac{1}{4} & \frac{1}{2} & \frac{1}{4} & 0 & 0 \\ \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \end{array} \mid, \quad \mathbf{P}_{4}=\right\| \begin{array}{ccccc} \frac{1}{4} & \frac{3}{4} & 0 & 0 & 0 \\ \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{1}{3} & \frac{2}{3} & 0 \\ 1 & 0 & 0 & 0 & 0 \end{array} \|$$

Consider three urns, one colored red, one white, and one blue. The red urn contains 1 red and 4 blue balls; the white urn contains 3 white balls, 2 red balls, and 2 blue balls; the blue urn contains 4 white balls, 3 red balls, and 2 blue balls. At the initial stage, a ball is randomly selected from the red urn and then returned to that urn. At every subsequent stage, a ball is randomly selected from the urn whose color is the same as that of the ball previously selected and is then returned to that urn. In the long run, what proportion of the selected balls are red? What proportion are white? What proportion are blue?

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?

It follows from Theorem \(4.2\) that for a time reversible Markov chain $$ P_{i j} P_{j k} P_{k i}=P_{i k} P_{k j} P_{j i}, \quad \text { for all } i, j, k $$ It turns out that if the state space is finite and \(P_{i j}>0\) for all \(i, j\), then the preceding is also a sufficient condition for time reversibility. (That is, in this case, we need only check Equation \((4.26)\) for paths from \(i\) to \(i\) that have only two intermediate states.) Prove this. Hint: Fix \(i\) and show that the equations $$ \pi_{j} P_{j k}=\pi_{k} P_{k j} $$ are satisfied by \(\pi_{j}=c P_{i j} / P_{j i}\), where \(c\) is chosen so that \(\sum_{j} \pi_{j}=1\)

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 \(i, i>0\), 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.

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