Science:Math Exam Resources/Courses/MATH307/April 2012/Question 08 (c)

MATH307 April 2012

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Question 08 (c)
Consider the stochastic matrix $P={\begin{bmatrix}0&1/4&1/3\\1&1/2&1/3\\0&1/4&1/3\end{bmatrix}}$ (c) What is $\lim _{n\to \infty }P^{n}[1,0,0]^{T}$? Give a reason.

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Hint

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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. Regardless, of the initial vector ($\mathbf {x_{0}} =[1,0,0]^{T}$ in this case), the probability vector found from $\lim _{n\rightarrow \infty }P^{n}\mathbf {x_{0}}$ will converge to the eigenvector with the eigenvalue pair of 1 scaled by a constant $a$ such that the sum of the elements in the vector sums to 1. It was found in part (a) the following is this particular eigenvalue/vector pair: $\mathbf {v} ={\begin{bmatrix}3\\8\\3\\\end{bmatrix}},\lambda =1$ Therefore, we want to scale the vector $[3,8,3]$ to have the three elements sum to 1. We can do that easily using basic algebra: ${\begin{aligned}3a+8a+3a&=1\\14a&=1\\a&=1/14\end{aligned}}$ Therefore, the $\lim _{n\rightarrow \infty }P^{n}[1,0,0]^{T}$ would converge to the matrix: $\mathbf {x_{\infty }} ={\frac {1}{14}}{\begin{bmatrix}3\\8\\3\end{bmatrix}}$ Note: Another way you can look at this is that we are using the power method to converge towards the eigenvector with the largest absolute eigenvalue. We already showed in (b) that there would be no eigenvalue greater than 1. Therefore, if we keep multiplying by P and normalize the vector every time, we would converge to the eigenvector $[3,8,3]^{T}$ (Notice that we don’t actually have to normalize it every time since a stochastic matrix will already cause the values of the matrix to sum to 1). Lastly, we know that any constant multiple of the eigenvector is still the same eigenvector, therefore we can scale $[3,8,3]^{T}$ to what the answer that we found above.