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Math Help - Eigenvectors and transition matrix of markov chains.

  1. #1
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    Eigenvectors and transition matrix of markov chains.

    Consider a Markov chain { X_n : n = 0,1,2,...} with states S=(1,2) and transition matrix

    P= \begin{pmatrix} \frac{1}{4} & \frac{3}{4} \\ \frac{1}{2} & \frac{1}{2}  \end{pmatrix}

    Let  \mathbf{x} be a vector such that  \mathbf{x}P= \mathbf{x}

    where x_1 > 0, x_2 > 0 and x_1 + x_2 = 1

    Let \lambda and  \mathbf{y} satisfy

     \mathbf{y}P=\lambda \mathbf{y}, where |\lambda|<1.

    Let  \mathbf{\rho} = (\rho_1,\rho_2) be such that \rho_1 > 0, \rho_2 > 0 , \rho_1+\rho_2=1 . Show that \rho can be written as

     \mathbf{\rho}= \mathbf{x}+K \mathbf{y} for some constant K

    I get:

     \mathbf{x}, \mathbf{y} \ are \ eigenvectors , therefore
     \mathbf{\rho}=\alpha \mathbf{x}+ \beta \mathbf{y}

    However I cannot see how to make \alpha = 1
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Eigenvectors and transition matrix of markov chains.

    The eigenvalues of P are \lambda_1=1,\lambda_2=-1/4 . Using the definition of eigenvector, you'll find that necessarily x=(2/5,3/5) and y=(-\gamma,\gamma)\;\;(\gamma \neq 0) . Now, using \rho_1+\rho_2=1 it is easy to prove that (\rho_1,\rho_2)=(2/5,3/5)+K(-\gamma,\gamma) for some K .
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    Re: Eigenvectors and transition matrix of markov chains.

    I am confused as to how you got

    y=(-\gamma, \gamma);\;(\gamma \neq 0)
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Eigenvectors and transition matrix of markov chains.

    Quote Originally Posted by FGT12 View Post
    I am confused as to how you got y=(-\gamma, \gamma);\;(\gamma \neq 0)
    yP=\lambda y implies y is an eigenvector of P associated to \lambda . If |\lambda|<1 necessarily \lambda=-1/4 . Solve the system y(P+(1/4)I)=0 .
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