/*! 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 44 Suppose that a population consis... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that a population consists of a fixed number, say, \(m\), of genes in any generation. Each gene is one of two possible genetic types. If any generation has exactly \(i\) (of its \(m\) ) genes being type 1 , then the next generation will have \(j\) type 1 (and \(m-j\) type 2 ) genes with probability $$ \left(\begin{array}{c} m \\ j \end{array}\right)\left(\frac{i}{m}\right)^{j}\left(\frac{m-i}{m}\right)^{m-j}, \quad j=0,1, \ldots, m $$ Let \(X_{n}\) denote the number of type 1 genes in the \(n\) th generation, and assume that \(X_{0}=i\). (a) Find \(E\left[X_{n}\right]\). (b) What is the probability that eventually all the genes will be type \(1 ?\)

Short Answer

Expert verified
The short answer to this question is: (a) The expected value of the number of type 1 genes in the nth generation, given that the starting population has i type 1 genes, is: $$ E\left[X_{n}\right] = \sum_{j=0}^{m} j \cdot P_{ij} $$ (b) The probability that eventually all the genes will be type 1, given that the starting population has i type 1 genes, is given by B(i, m), where B is obtained by multiplying the fundamental matrix N (defined by N = (I - Q)^{-1}) with the matrix R (NR = B).

Step by step solution

01

Define the Markov chain

The given problem can be described by a Markov chain with states \(0, 1, 2, \dots, m\), where the nth state represents the number of type 1 genes in the nth generation. The transition probability matrix P is given by: $$ P_{ij}=\left(\begin{array}{c} m \\ j \end{array}\right)\left(\frac{i}{m}\right)^{j}\left(\frac{m-i}{m}\right)^{m-j}, \quad j=0,1, \ldots, m $$ for state i transitioning to state j.
02

Find Expected Value of \(X_n\)

To find the expected value of \(X_n\), we use the recursive relationship of the Markov chain given by: $$ E\left[X_{n}\right] = \sum_{j=0}^{m} j \cdot P_{ij} $$ Starting with \(X_0 = i\), we have: $$ E\left[X_{1}\right] = \sum_{j=0}^{m} j \cdot P_{ij} $$ Similarly, we can calculate the expected value for each generation \(X_n\) by plugging in the transition probabilities. The expected value of the number of type 1 genes in the nth generation, given that the starting population has i type 1 genes, is: $$ E\left[X_{n}\right] = \sum_{j=0}^{m} j \cdot P_{ij} $$ The expected number of type 1 genes in the nth generation does not depend on n, as we have a stationary Markov chain. Thus, we can find the expected value by finding the stationary distribution of the Markov chain.
03

Absorbing Markov chain

In order to find the probability that eventually all the genes will be type 1, we can treat this Markov chain as an absorbing one, in which the state with all type 1 genes (state m) is the absorbing state. We can write the matrix P as follows: $$ P = \begin{pmatrix} Q & R \\ 0 & 1 \\ \end{pmatrix} $$ where Q is an \((m-1) \times (m-1)\) matrix of transient states and R is an \((m-1) \times 1\) column matrix representing the probability of transitioning from a transient state to the absorbing state.
04

Finding Fundamental matrix

First, we need to find the fundamental matrix N for the absorbing Markov chain, which is given by: $$ N = (I - Q)^{-1} $$ where I is the identity matrix of size \((m-1)\).
05

Probabilities of absorbing

To find the probability matrix of absorbing for the stopping time of the chain, we need to multiply the fundamental matrix (N) with the matrix R: $$ B = NR $$ The element B(i,m) of the resulting matrix B will give the probability that eventually all the genes will be type 1, given that the starting population had i type 1 genes. Thus the probability that all genes will eventually be type 1 given that the starting population has i type 1 genes is given by B(i, m).

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Key Concepts

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

Transition Probability Matrix
In the context of Markov chains, the transition probability matrix plays a fundamental role in predicting the behavior of a stochastic process over time. This matrix, usually denoted as P, contains all the probabilities of moving from one state to another in one time step. Each entry Pij represents the probability of transitioning from state i to state j.

For the problem involving genetic types, the transition matrix is defined using binomial probabilities to account for the random selection of genes from one generation to the next. The matrix is constructed such that the sum of the probabilities in any given row is equal to 1, validating the concept that a transition from one state must result in some subsequent state. Thus, the Pij entries account for all possible outcomes of gene type distributions in the next generation.
Expected Value
The expected value in a Markov chain provides us with a way to forecast the average outcome over many iterations of the process. For our genetic example, the expected value of Xn, where Xn is the number of type 1 genes in the nth generation, can be calculated using the transition probabilities and the states' values. To add clarity to the existing solution, the formula for the expected value is given by:
\[E\left[X_{n}\right] = \sum_{j=0}^{m} j \cdot P_{ij}\]The equation sums up the product of each state's value (the number of type 1 genes) and the probability of reaching that state from the current state i. Even though we're dealing with a random process, the expected value remains constant over successive generations because the probabilities don't change—this is a property of stationary Markov chains, where the system's dynamics are stable over time.
Absorbing States
Within Markov chains, absorbing states are special because once entered, the process cannot leave them—there is no escape. The problem at hand sets the scene for an absorbing state when all genes become type 1, denoted as state m. This state is 'sticky' because once all genes are of type 1, no transitions to other states are possible, reflecting the real-world scenario where certain conditions become permanent.

Identifying absorbing states is crucial for understanding the long-term behavior of the system. In such states, the transition probability to itself is 1, and to all other states is 0. This is significant because it allows us to predict ultimate outcomes and their probabilities, which—when calculated—can be of immense value in various fields such as genetics, finance, and even sports.
Stationary Distribution
The stationary distribution of a Markov chain, a key concept within probability theory, is a distribution of probabilities that remains unchanged as the process evolves over time. In other words, it represents an equilibrium state of the stochastic process.

For a Markov chain to have a stationary distribution, it must be regular, meaning it is possible to go from any state to any other state in a finite number of steps. This stationary distribution can be found by solving a system of linear equations that come from the transition probability matrix. When a Markov chain has a stationary distribution, the expected number of genes of each type remains stable across generations—even as the actual number fluctuates randomly from generation to generation—reflecting the process's underlying stability.
Fundamental Matrix
The fundamental matrix, often denoted by N, is a concept tightly linked with absorbing Markov chains. It tells us about the expected number of times the chain will visit a transient state before being absorbed. In our example involving genes, the fundamental matrix can help understand how many generations, on average, it will take before all genes are of type 1 under different starting conditions.

To find the fundamental matrix, we subtract the transition matrix of transient states, denoted by Q, from an identity matrix of the same dimension, taking the inverse of the result:
\[N = (I - Q)^{-1}\]N is then used to determine many important quantities, such as the mean first passage times—the average time to be absorbed starting from a transient state—and different long-term behavior probabilities. It can be especially insightful when evaluating processes that have clear end conditions or 'goals,' providing a measure of the expected time to reach such goals.

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Most popular questions from this chapter

Consider a process \(\left\\{X_{H}, n=0,1, \ldots\right\\}\) which takes on the values 0,1 , or 2 . Suppose $$ \begin{aligned} &P\left\\{X_{n+1}=j \mid X_{n}=i, X_{n-1}=i_{n-1}, \ldots, X_{0}=i_{0}\right\\} \\ &\quad=\left\\{\begin{array}{ll} P_{i j}^{\mathrm{I}}, & \text { when } n \text { is even } \\ P_{i j}^{\mathrm{II}}, & \text { when } n \text { is odd } \end{array}\right. \end{aligned} $$ where \(\sum_{j=0}^{2} P_{i j}^{\mathrm{I}}=\sum_{j=0}^{2} P_{i j}^{\mathrm{II}}=1, i=0,1,2 .\) Is \(\left\\{X_{n}, n \geqslant 0\right\\}\) a Markov chain? If not, then show how, by enlarging the state space, we may transform it into a Markov chain.

Let \(\pi_{i}\) denote the long-run proportion of time a given 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 $$ \pi_{j}=\sum_{i} \pi_{i} P_{i j} $$

For a branching process, calculate \(\pi_{0}\) when (a) \(P_{0}=\frac{1}{47} \cdot P_{2}=\frac{3}{4}\) (b) \(P_{0}=\frac{1}{4}, P_{1}=\frac{1}{2}, P_{2}=\frac{1}{4}\) (c) \(P_{0}=\frac{1}{6}, P_{1}=\frac{1}{2}, P_{3}=\frac{1}{3}\)

A group of \(n\) processors is arranged in an ordered list. When a job arrives, the first processor in line attempts it; if it is unsuccessful, then the next in line tries it; if it too is unsuccessful, then the next in line tries it, and so on. When the job is successfully processed or after all processors have been unsuccessful, the job leaves the system. At this point we are allowed to reorder the processors, and a new job appears. Suppose that we use the one- closer reordering rule, which moves the processor that was successful one closer to the front of the line by interchanging its position with the one in front of it. If all processors were unsuccessful (or if the processor in the first position was successful), then the ordering remains the same. Suppose that each time processor \(i\) attempts a job then, independently of anything else, it is successful with probability \(p_{i}\). (a) Define an appropriate Markov chain to analyze this model. (b) Show that this Markov chain is time reversible. (c) Find the long-run probabilities.

Suppose that coin 1 has probability \(0.7\) of coming up heads, and coin 2 has probability \(0.6\) of coming up heads. If the coin flipped today comes up heads, then we select coin 1 to flip tomorrow, and if it comes up tails, then we select coin 2 to flip tomorrow. If the coin initially flipped is equally likely to be coin 1 or coin 2 , then what is the probability that the coin flipped on the third day after the initial flip is coin \(1 ?\)

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