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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

A taxi driver provides service in two zones of a city. Fares picked up in zone \(A\) will have destinations in zone \(A\) with probability \(0.6\) or in zone \(B\) with probability \(0.4\). Fares picked up in zone \(B\) will have destinations in zone \(A\) with probability \(0.3\) or in zone \(B\) with probability \(0.7 .\) The driver's expected profit for a trip entirely in zone \(A\) is 6 ; for a trip entirely in zone \(B\) is \(8 ;\) and for a trip that involves both zones is 12 . Find the taxi driver's average profit per trip.

(a) Show that the limiting probabilities of the reversed Markov chain are the same as for the forward chain by showing that they satisfy the equations $$ \pi_{j}=\sum_{i} \pi_{i} Q_{i j} $$ (b) Give an intuitive explanation for the result of part (a).

Consider a Markov chain in steady state. Say that a \(k\) length run of zeroes ends at time \(m\) if $$ X_{m-k-1} \neq 0, \quad X_{m-k}=X_{m-k+1}=\ldots=X_{m-1}=0, X_{m} \neq 0 $$ Show that the probability of this event is \(\pi_{0}\left(P_{0,0}\right)^{k-1}\left(1-P_{0,0}\right)^{2}\), where \(\pi_{0}\) is the limiting probability of state 0 .

In a Markov decision problem, another criterion often used, different than the expected average return per unit time, is that of the expected discounted return. In this criterion we choose a number \(\alpha, 0<\alpha<1\), and try to choose a policy so as to maximize \(E\left[\sum_{i=0}^{\infty} \alpha^{i} R\left(X_{i}, a_{i}\right)\right]\) (that is, rewards at time \(n\) are discounted at rate \(\left.\alpha^{n}\right)\) Suppose that the initial state is chosen according to the probabilities \(b_{i} .\) That is, $$ P\left\\{X_{0}=i\right\\}=b_{i}, \quad i=1, \ldots, n $$ For a given policy \(\beta\) let \(y_{j a}\) denote the expected discounted time that the process is in state \(j\) and action \(a\) is chosen. That is, $$ y_{j a}=E_{\beta}\left[\sum_{n=0}^{\infty} \alpha^{n} I_{\left[X_{n}=j, a_{n}=a\right\\}}\right] $$ where for any event \(A\) the indicator variable \(I_{A}\) is defined by $$ I_{A}=\left\\{\begin{array}{ll} 1, & \text { if } A \text { occurs } \\ 0, & \text { otherwise } \end{array}\right. $$ (a) Show that $$ \sum_{a} y_{j a}=E\left[\sum_{n=0}^{\infty} \alpha^{n} I_{\left\\{X_{n}=j\right\\}}\right] $$ or, in other words, \(\sum_{a} y_{j a}\) is the expected discounted time in state \(j\) under \(\beta\). (b) Show that $$ \begin{aligned} \sum_{j} \sum_{a} y_{j a} &=\frac{1}{1-\alpha}, \\ \sum_{a} y_{j a} &=b_{j}+\alpha \sum_{i} \sum_{a} y_{i a} P_{i j}(a) \end{aligned} $$ Hint: For the second equation, use the identity $$ I_{\left\\{X_{n+1}=j\right\\}}=\sum_{i} \sum_{a} I_{\left\\{X_{n-i}, a_{n-a}\right\rangle} I_{\left\\{X_{n+1}=j\right\\}} $$ Take expectations of the preceding to obtain $$ E\left[I_{\left.X_{n+1}=j\right\\}}\right]=\sum_{i} \sum_{a} E\left[I_{\left\\{X_{n-i}, a_{n-a}\right\\}}\right] P_{i j}(a) $$ (c) Let \(\left\\{y_{j a}\right\\}\) be a set of numbers satisfying $$ \begin{aligned} \sum_{j} \sum_{a} y_{j a} &=\frac{1}{1-\alpha}, \\ \sum_{a} y_{j a} &=b_{j}+\alpha \sum_{i} \sum_{a} y_{i a} P_{i j}(a) \end{aligned} $$ Argue that \(y_{j a}\) can be interpreted as the expected discounted time that the process is in state \(j\) and action \(a\) is chosen when the initial state is chosen according to the probabilities \(b_{j}\) and the policy \(\beta\), given by $$ \beta_{i}(a)=\frac{y_{i a}}{\sum_{a} y_{i a}} $$ is employed. Hint: Derive a set of equations for the expected discounted times when policy \(\beta\) is used and show that they are equivalent to Equation \((4.38) .\) (d) Argue that an optimal policy with respect to the expected discounted return criterion can be obtained by first solving the linear program $$ \begin{array}{ll} \operatorname{maximize} & \sum_{j} \sum_{a} y_{j a} R(j, a), \\ \text { such that } & \sum_{j} \sum_{a} y_{j a}=\frac{1}{1-\alpha}, \\ \sum_{a} y_{j a}=b_{j}+\alpha \sum_{i} \sum_{a} y_{i a} P_{i j}(a) \\ y_{j a} \geqslant 0, \quad \text { all } j, a \end{array} $$ and then defining the policy \(\beta^{*}\) by $$ \beta_{i}^{*}(a)=\frac{y_{i a}^{*}}{\sum_{a} y_{i a}^{*}} $$ where the \(y_{j a}^{*}\) are the solutions of the linear program.

Let \(\pi_{i}\) denote the long-run proportion of time a given irreducible Markov chain is in state \(i\). (a) Explain why \(\pi_{i}\) is also the proportion of transitions that are into state \(i\) as well as being the proportion of transitions that are from state \(i\). (b) \(\pi_{i} P_{i j}\) represents the proportion of transitions that satisfy what property? (c) \(\sum_{i} \pi_{i} P_{i j}\) represent the proportion of transitions that satisfy what property? (d) Using the preceding explain why $$ \pi_{j}=\sum_{i} \pi_{i} P_{i j} $$

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