This is my first time looking at this material
On any given day, a student is either healthy or ill. Of the students who are healthy today, 95% will be healthy tomorrow. Of the students who are ill today, 60% will still be ill tomorrow.
- Find the stochastic matrix:
$$ \begin{bmatrix}
.95 & .40 \\
.05 & .60 \\
\end{bmatrix}$$
- Suppose 20% of the students are ill on Monday. What percentage of the students are likely to be ill on Tuesday? On Wednesday?
No clue what the arithmetic is. I think I take the stochastic matrix and multiply it by:
$$
\begin{bmatrix}
.20 \\
.80 \\
\end{bmatrix}
$$
which would give :
$$
\begin{bmatrix}
.51 \\
.49 \\
\end{bmatrix}
$$
Then rounded to the nearest tenth the percentage of students ill on monday would be .5? correct?
Is this correct? Am I doing this right?
Then to find out the likely to be ill on Wed I just take stochastic matrix from part one and multiply it by the resulting matrix of part two?
Best Answer
Your matrix is transposed from the usual form; see a description of how to build the stochastic matrix associated with a Markov chain here, for instance. In the typical setting, probability distributions are treated as row vectors. For your version of the transition matrix, probability distributions are treated as column vectors. I will stick to your convention.
Given a probability distribution $p = (p_1,p_2)^T$ (proportion of healthy/sick students), the distribution after one day (i.e. after one step of the Markov chain) is given by $Mp$, which is what you computed. In other words, the product you computed is correct.
The product for Wednesday will be $M(Mp) = M^2p$, which seems to be what you describe in your post.