Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 5 (d)

MATH152 April 2015

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Question B 5 (d)

The percentage of people with the disease, March Madness, is recorded every week. Note that it is possible to recover from March Madness one week and catch it again the following week. Records indicate that the disease can be modelled by a random walk and that if 50% of the population is infected with March Madness one week, then 60% of the population will be infected the next week. Records also indicate that if 100% of the population is infected one week, then 90% of the population will be infected the next week. It is known that 10% of the population has March Madness this week. (d) Approximately what will be the percentage of people with March Madness many weeks from now?

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Hint
Find the equilibrium probability of the probability transition matrix.

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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 part a), we know our transition matrix is: $$M={\begin{bmatrix}0.7&0.1\\0.3&0.9\end{bmatrix}}$$ To find the limit distribution, we need to find the eigenvector associated with eigenvalue ${\textstyle \lambda =1}$. This is the vector ${\textstyle {\mathbf {v}}}$ such that $$(M-I){\mathbf {v}}={\begin{bmatrix}-0.3&0.1\\0.3&-0.1\end{bmatrix}}{\begin{bmatrix}v_{1}\\v_{2}\end{bmatrix}}=0$$ Row-reducing by adding the first row to the second row, then multiplying the first row by ${\textstyle -1/0.3}$ (to get leading 1) we get $${\begin{bmatrix}1&-1/3\\0&0\end{bmatrix}}$$ Meaning that ${\textstyle v_{1}-(1/3)v_{2}=0}$ or ${\textstyle v_{2}=3v_{1}}$. Note that because we’re working with percentages we also require ${\textstyle 1=v_{1}+v_{2}=4v_{1}}$. Thus our limiting distribution is $$\color {blue}{\begin{bmatrix}1/4\\3/4\end{bmatrix}}$$