Science:Math Exam Resources/Courses/MATH152/April 2011/Question A 23

MATH152 April 2011

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Question A 23
Questions A22-A24 concern a random walk with three states that has transition matrix $\left[{\begin{array}{c c c}8/10&3/10&1/10\\2/10&3/10&7/10\\0&4/10&2/10\end{array}}\right]$ If the walker starts in state 2, which state will she most likely be in after 2 time steps?

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
We know that if we start from x then Px gives us the probability of 1 transition. How do we get to two transitions?

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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. We are starting in state 2 so there is 100% probability of starting in state 2 and zero elsewhere, therefore our input vector is $\mathbf {x} =\left[0,1,0\right].$ If we wanted to know the probabilities for moving one transition we would compute t1 = Px, $\mathbf {t} _{1}=P\mathbf {x} ={\begin{bmatrix}8/10&3/10&1/10\\2/10&3/10&7/10\\0&4/10&2/10\end{bmatrix}}{\begin{bmatrix}0\\1\\0\end{bmatrix}}={\begin{bmatrix}3/10\\3/10\\4/10\end{bmatrix}}.$ However we want to know the probabilities for a second transition. We can think of this as a first transition of our new vector t1. Therefore the second transition probabilities are t1 = P t1, $\mathbf {t} _{2}=P\mathbf {t} _{1}={\begin{bmatrix}8/10&3/10&1/10\\2/10&3/10&7/10\\0&4/10&2/10\end{bmatrix}}{\begin{bmatrix}3/10\\3/10\\4/10\end{bmatrix}}={\begin{bmatrix}37/100\\43/100\\20/100\end{bmatrix}}.$ Therefore, starting in state 2, it is most likely that after two transitions, the person will still be in state 2 since 43/100 is the highest of the three probabilities. Notice, we could also think of the problem as, $\mathbf {t} _{2}=P\mathbf {t} _{1}=P(P\mathbf {x} )=P^{2}\mathbf {x}$ so that if we wanted the nth transition probabilities we'd have tn = Pn x.