**Transcription and Explanation for Educational Website****Text:**Let \( P_{ij} = P(X_{n+1} = j | X_n = i) \) where\(\{X_n, n = 0, 1, 2, 3, \ldots\}\) is a Markov chain. Show that for a fixed \( i, \sum_j P_{ij} = 1 \).**Explanation:**This excerpt deals with the concept of a Markov chain, which is a sequence of random variables with a specific property: the future state depends only on the current state and not on the sequence of events that preceded it. In this context:- \( X_n \) represents the state of the Markov chain at step \( n \).- \( P_{ij} \) represents the probability of transitioning from state \( i \) to state \( j \) in the next step.The statement requires us to demonstrate that for any fixed state \( i \), the sum of the transition probabilities to all possible subsequent states \( j \) is equal to 1. This reflects the fact that the process must move to some state in the next step, ensuring that probabilities are normalized. The summation across all \( j \) ensures that these probabilities cover all possible transitions from \( i \).This property is fundamental to the definition of stochastic processes like Markov chains, emphasizing that the probability distribution over the next state is complete and sums to 1 when starting from a given current state.

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Answered: LetPij = P(Xn+1 =jXn = i)where {Xn, n =…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Let X be a Markov chain on S, and let I: S" {0, 1}. Show that the distribution of Xn, Xn+1,..., conditional on {I (X₁,..., Xn) = 1} n {Xn = i}, is identical to the distribution of Xn, Xn+1,... conditional on {Xn = i}.
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LetPij = P(Xn+1 =jXn = i)where {Xn, n = Σ 01,2,3,...} is a Markov chain. Show that for a fixed i, j Pij = 1. =