91Ó°ÊÓ

Since we have four different values for a DNA nucleotide, let's consider them as states 1, 2, 3, and 4. The probability of not changing the nucleotide equals to stay within the same state (1-3α). If it changes, it will transition to one of the other three states with equal probability α. We can illustrate the transition probability matrix as follows: \[ P = \left[ {\begin{array}{cccc} 1-3\alpha & \alpha & \alpha & \alpha \\ \alpha & 1-3\alpha & \alpha & \alpha \\ \alpha & \alpha & 1-3\alpha & \alpha \\ \alpha & \alpha & \alpha & 1-3\alpha \\ \end{array}} \right] \] Now, we will find the probabilities as required. ##Step 2: Compute \(P_{1,1}^{n}\)##

To compute \(P_{1,1}^{n}\), we have to raise the transition probability matrix to the power of \(n\). Since this is a regular matrix, we can use the following formula to find the probability: \[ P^n = \lim_{k \to \infty} \sum_{j=0}^{k} P^j \] Now, we only need the first element of the matrix \(P^n\) which represents \(P_{1,1}^{n}\). Thus we need to find the first element of matrix \(P^n\) after the sum of all matrix powers. Since all rows of matrix \(P\) are the same: \[ P^j = \left[ {\begin{array}{cccc} P_{1,1}^{j} & P_{1,2}^{j} & P_{1,3}^{j} & P_{1,4}^{j} \\ P_{1,1}^{j} & P_{1,2}^{j} & P_{1,3}^{j} & P_{1,4}^{j} \\ P_{1,1}^{j} & P_{1,2}^{j} & P_{1,3}^{j} & P_{1,4}^{j} \\ P_{1,1}^{j} & P_{1,2}^{j} & P_{1,3}^{j} & P_{1,4}^{j} \\ \end{array}} \right] \] Now, we can apply the formula above to obtain: \[ P_{1,1}^{n} = \frac{1}{4} + \frac{3}{4}(1-4\alpha)^n \] ##Step 3: Calculate the long-run proportion##

The long-run proportion of time the chain is in each state is given by the stationary distribution, denoted by \(\pi\). We have to solve for the stationary distribution of our Markov chain model. A stationary distribution satisfies the following conditions: 1. \(\pi P = \pi\) 2. \(\sum_{i=1}^{4} \pi_i = 1\) Since this is a regular Markov chain, we need to find a probability vector \(\pi\) for which the limit as \(n\) approaches infinity is equal to the stationary distribution. In the limit, the proportions of time spent in each state will be the same, so the limit is a matrix with equal rows: \[ \lim_{n\to\infty} P^n = \left[ {\begin{array}{cccc} \pi_1 & \pi_2 & \pi_3 & \pi_4 \\ \pi_1 & \pi_2 & \pi_3 & \pi_4 \\ \pi_1 & \pi_2 & \pi_3 & \pi_4 \\ \pi_1 & \pi_2 & \pi_3 & \pi_4 \\ \end{array}} \right] \] So, we have: \[ \left[ {\begin{array}{c} \pi_1 \\ \pi_2 \\ \pi_3 \\ \pi_4 \\ \end{array}} \right] = \left[ {\begin{array}{cccc} 1-3\alpha & \alpha & \alpha & \alpha \\ \alpha & 1-3\alpha & \alpha & \alpha \\ \alpha & \alpha & 1-3\alpha & \alpha \\ \alpha & \alpha & \alpha & 1-3\alpha \\ \end{array}} \right] \left[ {\begin{array}{c} \pi_1 \\ \pi_2 \\ \pi_3 \\ \pi_4 \\ \end{array}} \right] \] Since the proportions of time spent in each state will be the same in the long run, we can assume that \(\pi_1 = \pi_2 = \pi_3 = \pi_4 = \pi\). Therefore, the long-run proportion of time the chain is in each state \(\pi = \frac{1}{4}\).