Science:Math Exam Resources/Courses/MATH152/April 2011/Question A 24

MATH152 April 2011

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Question A 24
Questions A22-A24 concern a random walk with three states that has transition matrix $\left[{\begin{array}{c c c}8/10&3/10&1/10\\2/10&3/10&7/10\\0&4/10&2/10\end{array}}\right]$ What is the equilibrium probability vector for the random walk?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
An equilibrium probability vector is one such that if you were try to perform a transition, it wouldn't move. Physically it represents the final probabilities of ending up in a given state from any initial state.

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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. From the hint we see that a probability equilibrium vector gives us the probabilities of a final state from any initial state. This means that once we transition to a certain point, further transitions do not impact the probability any more. Let's assume we have reached this special equilibrium vector v then $P\mathbf {v} =\mathbf {v}$ which says exactly what we stated; attempting to transition v does nothing to it. Therefore the special vector v, will satisfy ${\begin{aligned}(P-I)\mathbf {v} &=\mathbf {0} \\{\begin{bmatrix}-2/10&3/10&1/10\\2/10&-7/10&7/10\\0&4/10&-8/10\end{bmatrix}}{\begin{bmatrix}v_{1}\\v_{2}\\v_{3}\end{bmatrix}}&={\begin{bmatrix}0\\0\\0\end{bmatrix}}\end{aligned}}$ which we can solve (using Gaussian elimination) to get, for a free parameter, t, $\mathbf {v} =\left[{\frac {7}{2}}t,2t,t\right].$ Now recall that the input vector v is a set of probabilities and so the entries of v must sum to 1. Therefore we have that, $v_{1}+v_{2}+v_{3}={\frac {7}{2}}t+2t+t={\frac {13}{2}}t=1$ and therefore, t=2/13 so that our equilibrium vector v is $\mathbf {v} =\left[{\frac {7}{13}},{\frac {4}{13}},{\frac {2}{13}}\right].$ Note (This is not necessary to answer the question): Our result means that starting from any initial state, we will end up in state 1 with a probability of 7/13, state 2 with a probability of 4/13 and state 3 with a probability of 2/13. If we try this with our initial state from A 23, [0,1,0] we notice that after 8 transitions, $\displaystyle {}P^{8}[0,1,0]=[0.5309904100,0.3114556300,0.1575539600]$ where our equilibrium vector is $\mathbf {v} =[0.5384615385,0.3076923077,0.1538461538]$ which is already a fairly good agreement! Note further: It may have crossed some minds that the equilibrium vector is always the eigenvector for the transition matrix corresponding to an eigenvalue of 1, whose entries sum to 1. For probability matrices where the columns sum to 1 there will always be such a vector.