Science:Math Exam Resources/Courses/MATH152/April 2012/Question 07 (d)

MATH152 April 2012

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[hide] Question 07 (d)
Consider a random walk that has transition matrix: $P={\begin{bmatrix}1/4&1/4&1/4\\3/4&3/4&1/4\\0&0&1/2\end{bmatrix}}$ Based on the results in (b), explain why the probability vector always approaches the equilibrium found in (c), regardless of the initial probability vector $\mathbf {x} (0)$.

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[show] Hint
Remember that for any matrix $\displaystyle P$, all eigenvector-eigenvalue combinations $(\mathbf {v} ,\lambda )$ satisfy $P\mathbf {v} =\lambda \mathbf {v} \quad \rightarrow \quad P^{n}\mathbf {v} =\lambda ^{n}\mathbf {v} .$ What happens when $\displaystyle \lambda =1$ versus when $\displaystyle |\lambda |<1$ as $n\rightarrow \infty$?

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[show] Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The stationary distribution is the eigenvector corresponding to eigenvalue 1. The other two eigenvectors correspond to eigenvalues with absolute value strictly less than 1. This means that only the component of the initial distribution that is parallel to the stationary distribution will maintain its length over repeated applications of P. Since the 3x3 matrix P has three distinct eigenvalues, every vector of initial conditions can be composed to a sum of eigenvectors. Hence, all components of the initial distribution that are not parallel to the stationary distribution will decay with repeated applications of P. Hence, all initial distributions eventually converge to the stationary distribution computed in part (c).