Results 1 to 4 of 4

Thread: Eigenvectors and transition matrix of markov chains.

  1. #1
    Member
    Joined
    Aug 2010
    Posts
    77

    Eigenvectors and transition matrix of markov chains.

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

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

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

    where $\displaystyle x_1 > 0, x_2 > 0$ and $\displaystyle x_1 + x_2 = 1$

    Let$\displaystyle \lambda$ and $\displaystyle \mathbf{y}$ satisfy

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

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

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

    I get:

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

    However I cannot see how to make$\displaystyle \alpha = 1$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,163
    Thanks
    46

    Re: Eigenvectors and transition matrix of markov chains.

    The eigenvalues of $\displaystyle P$ are $\displaystyle \lambda_1=1,\lambda_2=-1/4$ . Using the definition of eigenvector, you'll find that necessarily $\displaystyle x=(2/5,3/5)$ and $\displaystyle y=(-\gamma,\gamma)\;\;(\gamma \neq 0)$ . Now, using $\displaystyle \rho_1+\rho_2=1$ it is easy to prove that $\displaystyle (\rho_1,\rho_2)=(2/5,3/5)+K(-\gamma,\gamma)$ for some $\displaystyle K$ .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2010
    Posts
    77

    Re: Eigenvectors and transition matrix of markov chains.

    I am confused as to how you got

    $\displaystyle y=(-\gamma, \gamma);\;(\gamma \neq 0)$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,163
    Thanks
    46

    Re: Eigenvectors and transition matrix of markov chains.

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

Similar Math Help Forum Discussions

  1. Markov Chains - Working out the transition matrix
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: Mar 14th 2011, 04:38 PM
  2. Probability Transition Matrix and Markov Chains
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: Jan 23rd 2011, 07:45 AM
  3. Markov Transition Matrix
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: Oct 20th 2009, 04:22 AM
  4. Markov Chains & Transition Probabilities
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: Apr 26th 2009, 02:41 AM
  5. Markov Chain - Transition Matrix
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: Apr 9th 2009, 02:51 PM

Search Tags


/mathhelpforum @mathhelpforum