Consider a Markov chain Xnn 0 with three states 123 and the
     Consider a Markov chain (X_n)_n  0 with three states 1,2,3 and the transition matrix  A = [0.5 0.3 0.2 0.1 0.4 0.5 0 0.2 0.8]  (Python) For the Markov chain in Problem (1), choose the initial state using the uniform initial distribution [1/3,1/3, 1/3], and simulate ten steps of this chain. Repeat this procedure 1000 times and find approximately the distribution after 10 steps. Is it close to the stationary distribution from (1) (iii)?![Consider a Markov chain (X_n)_n 0 with three states 1,2,3 and the transition matrix A = [0.5 0.3 0.2 0.1 0.4 0.5 0 0.2 0.8] (Python) For the Markov chain in Pr  Consider a Markov chain (X_n)_n 0 with three states 1,2,3 and the transition matrix A = [0.5 0.3 0.2 0.1 0.4 0.5 0 0.2 0.8] (Python) For the Markov chain in Pr](/WebImages/8/consider-a-markov-chain-xnn-0-with-three-states-123-and-the-997056-1761513240-0.webp) 
  
  Solution
import numpy as np P = np.matrix([[0.5,0.3,0.2], [.1,.4,.5], [0,0.2,0.8], ) n = 10 Pn = P**n import numpy as np from matplotlib import pyplot P = np.matrix([[1., 0., 0., 0., 0., 0.], [1./4, 1./2, 0., 1./4, 0., 0.], [0., 0., 0., 1., 0., 0.], [1./16, 1./4, 1./8, 1./4, 1./4, 1./16], [0., 0., 0., 1./4, 1./2, 1./4], [0., 0., 0., 0., 0., 1.]]) v = np.matrix([[1/3,1/3,1/3]]) # Get the data plot_data = [] for step in range(20): result = v * P**step plot_data.append(np.array(result).flatten()) # Convert the data format plot_data = np.array(plot_data) # Create the plot pyplot.figure(1) pyplot.xlabel(\'Steps\') pyplot.ylabel(\'Probability\') lines = [] for i, shape in zip(range(6), [\'x\', \'h\', \'H\', \'s\', \'8\', \'r+\']): line, = pyplot.plot(plot_data[:, i], shape, label=\"S%i\" % (i+1)) lines.append(line) pyplot.legend(handles=lines, loc=1) pyplot.show()![Consider a Markov chain (X_n)_n 0 with three states 1,2,3 and the transition matrix A = [0.5 0.3 0.2 0.1 0.4 0.5 0 0.2 0.8] (Python) For the Markov chain in Pr  Consider a Markov chain (X_n)_n 0 with three states 1,2,3 and the transition matrix A = [0.5 0.3 0.2 0.1 0.4 0.5 0 0.2 0.8] (Python) For the Markov chain in Pr](/WebImages/8/consider-a-markov-chain-xnn-0-with-three-states-123-and-the-997056-1761513240-0.webp)
