# Markov's Chain stationary matrix determination

Mathematics Asked by Ignacio Toro on January 12, 2021

I have the following transition matrix $$P$$

$$begin{equation} P= begin{pmatrix} 1/2 & 1/3 & 1/6\ 1/4 & 3/4 & 0\ 1/5 & 2/5 & 2/5 end{pmatrix} end{equation}$$

I have tried to calculate the stationary matrix as
$$begin{equation} P^n= C D^n C^{-1} end{equation}$$

With $$D=C^{-1} P C$$ the diagonal matrix. Calculating $$C$$ with the eigenvalues and eigenvectors

$$begin{equation} C= begin{pmatrix} v_1 & w_1 & z_1\ v_2 & w_2 & z_2\ v_3 & w_3 & z_3 end{pmatrix} end{equation}$$

where the eigenvectors are

begin{align} vec{v} =langle (v_1,v_2,v_3)rangle &= langle(1,1,1)rangle \ vec{w} =langle(w_1,w_2,w_3)rangle &= langleleft(frac{2isqrt{39}+3}{6},frac{-3isqrt{39}-7}{16},1right)rangle \ vec{v} =langle(v_1,v_2,v_3)rangle &= langleleft(frac{-2isqrt{39}+3}{6},frac{3isqrt{39}-7}{16},1right)rangle \ end{align}

so the diagonal matrix is
$$begin{equation} D= begin{pmatrix} 1 & 0 & 0\ 0 & frac{-isqrt{39}+39}{120} & 0\ 0 & 0 & frac{isqrt{39}+39}{120} end{pmatrix} end{equation}$$

how can i calculate $$D^{n}$$? is the diagonalization correct?

I tried by other way using $$BP=B$$ where B is the final vector state, then

begin{align} [p_1 quad p_2 quad p_3] begin{pmatrix} 1/2 & 1/3 & 1/6\ 1/4 & 3/4 & 0\ 1/5 & 2/5 & 2/5 end{pmatrix} = [p_1 quad p_2 quad p_3] end{align}

using Gauss-Jordan elimination for solving this system it gives me infinite solutions.

I dont know if it is a problem with the transition matrix.

Since $$1$$ is the dominant eigenvalue of the Markov Chain (the derivation of that fact can be found elsewhere, thus I leave that up to the reader to investigate if desired), all we need to solve is

$$begin{bmatrix} 1/2 & 1/4 & 1/5 \1/3 & 3/4 & 2/5 \ 1/6 & 0 & 2/5 end{bmatrix} begin{bmatrix} x \y \ z end{bmatrix}= begin{bmatrix} x \y \ z end{bmatrix}$$ Solving gives infinite (why?) solutions as follows: $$x=3.6z,y=6.4z$$ with $$z$$ free. Using the constraint $$x+y+z=1$$ we find $$z=18/55,y=32/55,z=1/11$$ and thus we find for the Stationary matrix: $$begin{bmatrix} 18/55 & 18/55 & 15/55 \32/55 & 32/55 & 32/55 \ 1/11 & 1/11 & 1/11 end{bmatrix}$$ The rudiments of the calculations is nothing but row operations (Gauss Elimination)

Answered by imranfat on January 12, 2021

If you want to get a stationary distribution then it's needed to solve the system, consisting of these relations: $$(p_1, p_2, p_3) P = (p_1, p_2, p_3)$$ $$p_1 + p_2 + p_3 =1$$ $$p_i ge 0, 1 le i le 3.$$ If you will get more than one solution it's normal: e.g. if $$P$$ is identity matrix then every distribution is stationary.

Everything that you did with $$D^n$$ may be useful for limit distribution but not for stationary one.

Answered by Botnakov N. on January 12, 2021

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