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.

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    $\begingroup$ Well, you set up the problem correctly. You can use "Gaussian elimination" if you know it. Or just substitute one variable in and keep going. A hint is to just consider the first, second, and fourth equations. $\endgroup$ – Michael Oct 24 '17 at 17:01
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    $\begingroup$ @Michael 1 is $π_0$ = 18/55, $π_1$ = 16/55, $π_2$ = 21/55? $\endgroup$ – AmaC Oct 24 '17 at 18:09
  • $\begingroup$ Looks good. You can substitute these numbers into the equations yourself to verify. $\endgroup$ – Michael Oct 24 '17 at 18:12
  • $\begingroup$ Oh, of course. Thank you very much! $\endgroup$ – AmaC Oct 24 '17 at 18:15
  • $\begingroup$ A better way to verify your answer is to compute $\pi P$ to cover the possibility that you set the equations up incorrectly. $\endgroup$ – amd Oct 24 '17 at 18:23

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 $$ \pi P = \pmatrix{\frac{18}{55}&\frac{16}{55}&\frac{21}{55}} \pmatrix{ \frac{1}{2} & \frac{3}{10} & \frac{1}{5} \\ \frac{3}{10} & \frac{2}{5} & \frac{3}{10} \\ \frac{1}{5} & \frac{1}{5} & \frac{3}{5} \\ } = \pmatrix{\frac{18}{55}&\frac{16}{55}&\frac{21}{55}} = \pi. $$


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