Transmutation (Harder Problem)

The approach to solving this problem is the same as in the easier version of Transmutation, with the exception that this problem uses a 5x5 matrix instead of a 2x2 matrix. Simply find the eigenvalues and eigenvectors of the matrix, diagonalize the matrix, and find the limit as $t$ goes to infinity. The code at the very bottom does this process, substituting infinity for a really large value. This isn't a formal proof of the answer, but it gets the job done.

Running the code yields the following matrix:

$\begin{bmatrix} 0.01186944 & 0.01186944 & 0.01186944 & 0.01186944 & 0.01186944 \\ 0.02077151 & 0.02077151 & 0.02077151 & 0.02077151 & 0.02077151 \\ 0.0652819 & 0.0652819 & 0.0652819 & 0.0652819 & 0.0652819 \\ 0.0652819 & 0.0652819 & 0.0652819 & 0.0652819 & 0.0652819 \\ 0.83679527 & 0.83679527 & 0.83679527 & 0.83679527 & 0.83679527 \end{bmatrix}$

Which corresponds to the following set of equations:

$\Sigma_{\infty} = C_{\infty} + T_{\infty} + S_{\infty} + L_{\infty} + G_{\infty}$

$C_{\infty} = 0.01186944\Sigma_{\infty}$

$T_{\infty} = 0.02077151\Sigma_{\infty}$

$S_{\infty} = 0.0652819\Sigma_{\infty}$

$L_{\infty} = 0.0652819\Sigma_{\infty}$

$G_{\infty} = 0.83679527\Sigma_{\infty}$

Thus, the answer is $\left \lfloor{\frac{G_{\infty}}{C_{\infty} + T_{\infty} + S_{\infty} + L_{\infty}}} \right \rfloor = \boxed{5}$

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18

import numpy as np

A = np.asarray([[0.2,0.3,0.025,0.025,0],
                [0.5,0.4,0.025, 0.075,0],
                [0.2,0.2,0.8,0.1,0],
                [0.075,0.05,0.05,0.6,0.025],
                [0.025,0.05,0.1,0.2,0.975]])

eigenvalues, eigenvectors = np.linalg.eig(A)

def f(t):
    D = np.diag([x ** t for x in eigenvalues])
    P = eigenvectors
    P_inv = np.linalg.inv(P)
    return np.matmul(P, np.matmul(D, P_inv))

inf_matrix = f(10000000)
print(inf_matrix)

The problem immediately reduces to finding the stationary distribution of a Markov chain with the kernel given by $\bm{P}=\bm{Q}^T$ , where $\bm{Q}$ is the given coefficient matrix. It can be easily verified that the chain is ergodic (i.e., irreducible and aperiodic). Hence, the unique stationary distribution is obtained by solving the balance equation $\bm{\pi} \bm{P}=\bm{\pi}.$ By solving the above system of linear equations with the constraint that $\sum_i \pi_i=1$ , we obtain the following stationary distribution $\bm{\pi}= [0.0119, 0.0208, 0.0653, 0.0653, 0.8368].$ From which, the answer follows.

I got an answer of 5.12727273.