On any given day Eric is either cheerful (C), so-so (S), or glum (G). If he is
cheerful today, then he will be C, S, or G tomorrow with respective probabilities 0.5, 0.3,
0.2. If he is feeling so-so today, then he will be C, S, or G tomorrow with probabilities 0.3,
0.4, 0.3. If he is glum today, then he will be C, S, or G tomorrow with probabilities 0.2, 0.2,
0.6. Write a one step transition matrix for the Eric’s mood. Label the states 0, 1 and 2, and find the long-run proportion for this process.
So far, I have:
State 0: cheerful,
State 1: so-so,
State 2: glum.
$$P = \left[
\begin{array}{ccc}
0.5&0.3&0.2\\
0.3&0.4&0.3\\ 0.2&0.2&0.6
\end{array}
\right] $$
When solving for the long-run proportion (πP = π):
$π_0 = 0.5π_0 + 0.3π_1 + 0.2π_2$
$π_1 = 0.3π_0 + 0.4π_1 + 0.2π_2$
$π_2 = 0.2π_0 + 0.3π_1 + 0.6π_2$
$π_0 + π_1 + π_2 = 1$
Now, I am a little confused as how to solve for this system of equations.
Best Answer
Note that this system of linear equations is redundant since exactly three equations are needed to solve for three variables. However, the three equations arising from $\pi P=\pi$ form a homogenous system (the RHS is equal to zero), so we cannot use these; instead, we two of the equations as well as the equation $\sum_{i=0}^2 \pi_i=2$. This gives us
\begin{align} -\frac12\pi_0 +\frac3{10}\pi_1+\frac15\pi_2 &= 0\\ \frac35\pi_0 - \frac35\pi_1 +\frac15\pi_2 &= 0\\ \pi_0+\pi_1+\pi_2&=0 \end{align} which indeed has unique solution $$ \left(\pi_0,\pi_1,\pi_2\right) = \left(\frac{18}{55},\frac{16}{55},\frac{21}{55}\right). $$ Indeed, we have $$ \pi P = \pmatrix{\frac{18}{55}&\frac{16}{55}&\frac{21}{55}} \pmatrix{ \frac{1}{2} & \frac{3}{10} & \frac{1}{5} \\ \frac{3}{10} & \frac{2}{5} & \frac{3}{10} \\ \frac{1}{5} & \frac{1}{5} & \frac{3}{5} \\ } = \pmatrix{\frac{18}{55}&\frac{16}{55}&\frac{21}{55}} = \pi. $$