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

MATH152 April 2015

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Question B 5 (a)

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. (a) What is the $2\times 2$ probability transition matrix for this system?

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
To find the matrix, it is enough to recall that matrix multiplication is a linear transformation and that the columns of a matrix are the images of the standard basis vectors under that 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. Represent the proportion of people who are infected as column vectors like so: ${\begin{pmatrix}{\text{ proportion of people who are not infected }}\\{\text{ proportion of people who are infected }}\end{pmatrix}}$ And let $M$ be the desired transition matrix. The data given in this question becomes, $M{\begin{pmatrix}0.5\\0.5\end{pmatrix}}={\begin{pmatrix}0.4\\0.6\end{pmatrix}}$ $M{\begin{pmatrix}0\\1\end{pmatrix}}={\begin{pmatrix}0.1\\0.9\end{pmatrix}}$ The first column of $M$ is $M{\begin{pmatrix}1\\0\end{pmatrix}}$ and the second column of $M$ is $M{\begin{pmatrix}0\\1\end{pmatrix}}$. So, it is enough to find $M{\begin{pmatrix}1\\0\end{pmatrix}}$. We have that $2{\begin{pmatrix}0.5\\0.5\end{pmatrix}}-{\begin{pmatrix}0\\1\end{pmatrix}}={\begin{pmatrix}1\\1\end{pmatrix}}-{\begin{pmatrix}0\\1\end{pmatrix}}={\begin{pmatrix}1-0\\1-1\end{pmatrix}}={\begin{pmatrix}1\\0\end{pmatrix}}$ So as $M$ is a linear transformation, $M{\begin{pmatrix}1\\0\end{pmatrix}}=M\left(2{\begin{pmatrix}0.5\\0.5\end{pmatrix}}-{\begin{pmatrix}0\\1\end{pmatrix}}\right)=2M{\begin{pmatrix}0.5\\0.5\end{pmatrix}}-M{\begin{pmatrix}0\\1\end{pmatrix}}=2{\begin{pmatrix}0.4\\0.6\end{pmatrix}}-{\begin{pmatrix}0.1\\0.9\end{pmatrix}}={\begin{pmatrix}0.7\\0.3\end{pmatrix}}$ Hence, $M={\begin{pmatrix}M{\begin{pmatrix}1\\0\end{pmatrix}}&M{\begin{pmatrix}0\\1\end{pmatrix}}\end{pmatrix}}=\color {blue}{\begin{pmatrix}0.7&0.1\\0.3&0.9\end{pmatrix}}$