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A process moves on the integers \(1,2,3,4,\) and \(5 .\) It starts at 1 and, on each successive step, moves to an integer greater than its present position, moving with equal probability to each of the remaining larger integers. State five is an absorbing state. Find the expected number of steps to reach state five.

Short Answer

Expert verified
The expected number of steps to reach state five is 2.375.

Step by step solution

01

Define the States and Probabilities

The process starts at state 1 and can move to any of the larger integers. Given any state "i", the possible next moves are to states greater than "i" with equal probability. Absorbing state "5" means once the process reaches 5, it stops.
02

Set Up the Equations for Expected Steps

Let \( E(i) \) be the expected number of steps to reach state 5 from state \( i \). We know \( E(5) = 0 \) since it's the absorbing state. Our goal is to solve for \( E(1) \).
03

Establish Relationships for E(i)

For state 1: \( E(1) = 1 + \frac{1}{4}(E(2)) + \frac{1}{4}(E(3)) + \frac{1}{4}(E(4)) + \frac{1}{4}(E(5)) \).Similarly:- For state 2: \( E(2) = 1 + \frac{1}{3}(E(3)) + \frac{1}{3}(E(4)) + \frac{1}{3}(E(5)) \)- For state 3: \( E(3) = 1 + \frac{1}{2}(E(4)) + \frac{1}{2}(E(5)) \)- For state 4: \( E(4) = 1 + \frac{1}{1}(E(5)) \)
04

Solve from E(4) to E(1)

Start from the simplest equation to the hardest:- \( E(4) = 1 + E(5) = 1 + 0 = 1 \)- \( E(3) = 1 + \frac{1}{2}(1) + \frac{1}{2}(0) = 1.5 \)- \( E(2) = 1 + \frac{1}{3}(1.5) + \frac{1}{3}(1) = 2 \)- \( E(1) = 1 + \frac{1}{4}(2) + \frac{1}{4}(1.5) + \frac{1}{4}(1) + 0 = 2.375 \)

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Key Concepts

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

Absorbing State
In probability theory, an *absorbing state* is a particular state within a stochastic process where, once entered, the process will remain in that state indefinitely. Understanding absorbing states is crucial in studying *Markov Chains*, as they help determine the long-term behavior of processes.
  • If a Markov Chain includes one or more absorbing states, eventually the process ends up in one of these states.
  • Any state with a probability transition of 1 to itself can be considered absorbing.

In the given exercise, state 5 is an *absorbing state*. This means that once the process reaches this state, it terminates, and no further transitions occur. Recognizing this state is essential for calculating other properties, such as the expected number of steps required to reach it.
Expected Value
The concept of *expected value* is a fundamental component of probability theory. It represents the average outcome of a random variable over many trials, weighted by the likelihood of each possible outcome.
For solving problems involving Markov Chains, like in our exercise, the expected value signifies the average number of steps needed to transition between states. The expected number of steps to reach an absorbing state can be calculated using set equations for expected steps relative to each current state.
  • For each state "i", define the expected number of steps to reach the absorbing state (state 5 in our problem) as \( E(i) \).
  • Equations are developed based on the equal probability of moving from one state to possible subsequent states.

Ultimately, these equations allow us to solve sequentially from the simplest to the most complex to find the expected value for each starting state until reaching \( E(1) = 2.375 \). This value represents the expected number of steps needed from state 1 to reach the absorbing state 5.
Transition Probabilities
*Transition probabilities* are the probabilities of moving from one state to another in a stochastic process or Markov Chain. They are fundamental in describing how processes evolve over time within these models.
In the exercise, the transition probabilities determine the likelihood of the process moving from one integer state to a higher integer state.
  • The transition probability from one state "i" to a larger state determines the setup of the probabilities for all subsequent larger states.
  • Given our setup, for state 1, the probability of moving to any of the states 2, 3, 4, or the absorbing state 5, is 1/4.

Understanding these probabilities helps determine how the expected steps to reach the absorbing state are calculated, as it shows all potential paths and their respective chances of occurring.
Probability Theory
*Probability theory* is the branch of mathematics concerned with studying uncertainty and stochastic processes like Markov Chains. It provides the tools and frameworks necessary to analyze situations that involve random events and outcomes.
Markov Chains, as an essential application of probability theory, offer a structured way to predict changes over time. They can either be used for finite processes, like reaching an absorbing state, or for infinite processes to find long-term behaviors.
  • When applied to our exercise, probability theory helps us define and solve for expected values, factoring in all transition possibilities and their respective probabilities.
  • Each concept, such as absorbing states and transition probabilities, stems directly from principles laid by probability theory.

Mastering these fundamental principles allows for a deeper understanding of stochastic processes and the ability to apply such knowledge to real-life scenarios and various fields of study.

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Most popular questions from this chapter

Toss a fair die repeatedly. Let \(S_{n}\) denote the total of the outcomes through the \(n\) th toss. Show that there is a limiting value for the proportion of the first \(n\) values of \(S_{n}\) that are divisible by \(7,\) and compute the value for this limit. Hint: The desired limit is an equilibrium probability vector for an appropriate seven state Markov chain.

Let \(\mathbf{P}\) be a regular transition matrix and let \(\mathbf{w}\) be the unique non-zero fixed vector of \(\mathbf{P}\). Show that no entry of \(\mathbf{w}\) is 0 .

Assume that an experiment has \(m\) equally probable outcomes. Show that the expected number of independent trials before the first occurrence of \(k\) consecutive occurrences of one of these outcomes is \(\left(m^{k}-1\right) /(m-1) .\) Hint: Form an absorbing Markov chain with states \(1,2, \ldots, k\) with state \(i\) representing the length of the current run. The expected time until a run of \(k\) is 1 more than the expected time until absorption for the chain started in state \(1 .\) It has been found that, in the decimal expansion of pi, starting with the 24,658,601 st digit, there is a run of nine 7 's. What would your result say about the expected number of digits necessary to find such a run if the digits are produced randomly?

Prove that, if a 3 -by-3 transition matrix has the property that its column sums are 1 , then \((1 / 3,1 / 3,1 / 3)\) is a fixed probability vector. State a similar result for \(n\) -by- \(n\) transition matrices. Interpret these results for ergodic chains.

Consider the Markov chain with general \(2 \times 2\) transition matrix $$ \mathbf{P}=\left(\begin{array}{cc} 1-a & a \\ b & 1-b \end{array}\right) $$ (a) Under what conditions is \(\mathbf{P}\) absorbing? (b) Under what conditions is \(\mathbf{P}\) ergodic but not regular? (c) Under what conditions is \(\mathbf{P}\) regular?

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