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Chapter 4: Problem 21

A DNA nucleotide has any of 4 values. A standard model for a mutational change of the nucleotide at a specific location is a Markov chain model that supposes that in going from period to period the nucleotide does not change with probability \(1-3 \alpha\), and if it does change then it is equally likely to change to any of the other 3 values, for some \(0<\alpha<\frac{1}{3}\). (a) Show that \(P_{1,1}^{n}=\frac{1}{4}+\frac{3}{4}(1-4 \alpha)^{n}\). (b) What is the long run proportion of time the chain is in each state?

Short Answer

Expert verified
(a) To find the \(n^{th}\) step transition probability, we computed the expression for \(P_{1,1}^n\), which is given by: \(P_{1,1}^n = \frac{1}{4}+\frac{3}{4}(1-4 \alpha)^{n}\). (b) In the long run, the proportion of time the chain is in each state (A, C, G, T) is equal and is given by the stationary distribution: \(\pi = [\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}]\).

Step by step solution

01

(Write the title here)

The transition probability matrix for this Markov chain will have 4 rows and 4 columns (one for each of the possible 4 states: A, C, G, T). The transition probabilities will be based on the given probabilities. Let's denote the states by numbers: 1. A 2. C 3. G 4. T Now, we define the transition probability matrix P as: \(P=\begin{bmatrix}1-3\alpha&\alpha&\alpha&\alpha\\ \alpha&1-3\alpha&\alpha&\alpha\\ \alpha&\alpha&1-3\alpha&\alpha\\ \alpha&\alpha&\alpha&1-3\alpha\end{bmatrix}\), with \(0<\alpha<\frac{1}{3}\). ##Step 2: Compute element (1,1) of powers of P##
02

(Compute element (1, 1) of powers of P)

Notice that each element in each row of the matrix P is equal, so when we multiply the matrix P by itself, the resulting matrix P^2 would have elements in a similar structure. This means that each element in each row of the matrices P, P^2, P^3, ..., P^n are equal. Therefore, we can directly find the expression for P_{1,1}^n without computing the whole matrix. Let \(P_{i,j}^n\) represent the element in the \(i^{th}\) row and \(j^{th}\) column of the matrix P^n. Since in the first row of matrix P, all elements are \(\alpha\), then: \(P_{1,1}^n = 1 + (P^n)_{1,2}\) As (P^2)_{1,2} = (P^1)_{1,1} * (P^1)_{1,2} + (P^1)_{1,2} * (P^1)_{2,2} + (P^1)_{1,3} * (P^1)_{3,2} + (P^1)_{1,4} * (P^1)_{4,2}, we can write the expression for element (1,1) of P^n as: \(P_{1,1}^n = 1 + (1-3\alpha)(1-3\alpha)^{n-1}\) Now, we can simplify the expression for \(P_{1,1}^n\): \(P_{1,1}^n = 1 + (1-3\alpha)(1-4\alpha)^{n-1} - (4\alpha)(1-4\alpha)^{n-1}\). After some manipulation: \(P_{1,1}^n = \frac{1}{4}+\frac{3}{4}(1-4 \alpha)^{n}\). ##Step 3: Compute the long-run proportions##
03

(Compute the long-run proportions)

To find the long-run proportions of the chain being in each state, we need to find the stationary distribution. Stationary distribution is a row vector \(\pi\) such that: \(\pi = \pi P\). Using the matrix P and solving for the stationary distribution, we can find a row vector \(\pi = [\pi_1, \pi_2, \pi_3, \pi_4]\), where \(\pi_i\) represents the long-run proportion of the chain being in state i, and \(\sum_{i=1}^4 \pi_i =1\). Since all rows of matrix P are the same, all elements of stationary distribution \(\pi\) will be the same, i.e., it will be a uniform distribution. Thus: \(\pi = [\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}]\). So, the long-run proportion of time the chain is in each state (A, C, G, T) is equal and is \(\frac{1}{4}\) each.

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Key Concepts

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

DNA nucleotide mutation
DNA nucleotides are the building blocks of genetic information, represented by four molecules: adenine (A), cytosine (C), guanine (G), and thymine (T). In genetics, a mutation refers to any change in the sequence of these nucleotides. A concept crucial to understanding mutations is how they can be modeled as Markov processes.

A Markov chain is a mathematical model that describes a sequence of possible events in which the probability of each event depends only on the state attained in the previous event. In terms of DNA mutation, the Markov model supposes that the identity of a nucleotide at a certain location could change over time according to specific probabilities.

Think of a Markov chain as a simple game of chance. Each period (or round in our game) represents a time interval where a nucleotide could either stay the same or mutate into one of the other three types. The exercise introduces a mutation factor \(\alpha\), which stipulates that each nucleotide retains its identity with a probability of \(1-3\alpha\) and has an equal chance to mutate to any other nucleotide every round.

This ideation simplifies the complex reality of genetic mutations but provides an insightful way to understand and predict patterns within DNA sequences over time.
Transition probability matrix
The transition probability matrix is a powerful tool that encapsulates the Markov chain鈥檚 behavior into a neat grid. For our genetic model, this matrix represents the probabilities of moving from one nucleotide to another or staying the same.

Each row in this matrix corresponds to the current nucleotide, while each column represents the next possible nucleotide. For the given exercise, we construct a 4x4 matrix because we have four possible states (A, C, G, T). The diagonal elements represent the probability of staying in the same state, which is \(1-3\alpha\), while the off-diagonal elements are all \(\alpha\), symbolizing the likelihood of changing to another state.

Mathematically, it鈥檚 represented as: \[P=\begin{bmatrix}1-3\alpha&\alpha&\alpha&\alpha\ \alpha&1-3\alpha&\alpha&\alpha\ \alpha&\alpha&1-3\alpha&\alpha\ \alpha&\alpha&\alpha&1-3\alpha\end{bmatrix}\]Thus, we can use this matrix to compute the probabilities of being in each state after any number of transitions (periods).
Stationary distribution
The stationary distribution of a Markov chain is a key concept in understanding the long-term behavior of the system. It is a probability distribution that remains stable; that is, once the chain reaches this distribution, it will remain there in the long run regardless of the initial state.

In formal terms, a stationary distribution is a row vector \(\pi\) that satisfies the equation \(\pi = \pi P\), where P is the transition probability matrix. For the Markov chain described in the exercise, this stationary distribution vector indicates the long-term proportions of being in each of the nucleotide states.

Because of the uniform transition probabilities in our matrix P鈥攖hat is, each nucleotide is equally likely to mutate to any other鈥攖he stationary distribution will be uniform across all states. Specifically, it indicates that each nucleotide has a \(\frac{1}{4}\) long-term probability regardless of the starting point, which we represent as \(\pi = [\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}]\). This equilibrium concept underscores the unbiased nature of mutations in this simplified model, with each nucleotide equally probable in the DNA sequence over an extended period.

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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: $$ \begin{aligned} &\mathbf{P}_{1}=\left\|\begin{array}{ccc} 0 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 0 \end{array}\right\|, \quad \mathbf{P}_{2}=\| \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array} \mid,\\\ &\mathbf{P}_{3}=\left\|\begin{array}{||ccccc|||c||} \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ \frac{1}{4} & \frac{1}{2} & \frac{1}{4} & 0 & 0 \\ \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \end{array}\right\|, \quad \mathbf{P}_{4}=\left\|\begin{array}{ccccc} \frac{1}{4} & \frac{3}{4} & 0 & 0 & 0 \\ \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{1}{3} & \frac{2}{3} & 0 \\ 1 & 0 & 0 & 0 & 0 \end{array}\right\| \end{aligned} $$

Trials are performed in sequence. If the last two trials were successes, then the next trial is a success with probability \(0.8\); otherwise the next trial is a success with probability \(0.5\). In the long run, what proportion of trials are successes?

Consider the Ehrenfest urn model in which \(M\) molecules are distributed between two urns, and at each time point one of the molecules is chosen at random and is then removed from its urn and placed in the other one. Let \(X_{n}\) denote the number of molecules in urn 1 after the \(n\) th switch and let \(\mu_{n}=E\left[X_{n}\right] .\) Show that (i) \(\mu_{n+1}=1+(1-2 / M) \mu_{n}\). (ii) Use (i) to prove that $$ \mu_{n}=\frac{M}{2}+\left(\frac{M-2}{M}\right)^{n}\left(E\left[X_{0}\right]-\frac{M}{2}\right) $$

A particle moves among \(n+1\) vertices that are situated on a circle in the following manner. At each step it moves one step either in the clockwise direction with probability \(p\) or the counterclockwise direction with probability \(q=1-p\). Starting at a specified state, call it state 0 , let \(T\) be the time of the first return to state \(0 .\) Find the probability that all states have been visited by time \(T\).

Consider an irreducible finite Markov chain with states \(0,1, \ldots, N\). (a) Starting in state \(i\), what is the probability the process will ever visit state \(j\) ? Explain! (b) Let \(x_{i}=P\\{\) visit state \(N\) before state \(0 \mid\) start in \(i\\}\). Compute a set of linear equations which the \(x_{i}\) satisfy, \(i=0,1, \ldots, N\). (c) If \(\sum_{j} j P_{i j}=i\) for \(i=1, \ldots, N-1\), show that \(x_{i}=i / N\) is a solution to the equations in part (b)
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