The weather on any given day in a particular city can be sunny, cloudy, or rainy. It has been observed to be predictable largely on the basis of the weather on the previous day.
Observations have shown that:
if it is sunny on one day, it will be sunny the next day $\frac25$ of the time, and be cloudy the next day $\frac25$ of the time.
If it is cloudy on one day, it will be sunny the next day $\frac25$ of the time, and be cloudy the next day $\frac35$ of the time.
If it is rainy on one day, it will be sunny the next day $\frac35$ of the time, and be cloudy the next day $\frac15$ of the time.
Using 'sunny', 'cloudy', and 'rainy' (in that order) as the states in a system, set up the transition matrix for a Markov chain to describe this system.
Find the proportion of days that have each type of weather in the long run.
I have found the matrix, I can't seem to find the proportions for sunny, cloudy and rainy
$$\begin{bmatrix}\frac25&\frac25&\frac35\\
\frac25&\frac35&\frac15\\
\frac15&0&\frac15\end{bmatrix}.$$
Best Answer
Note that usually the probability $p_{i,j}$ in transition matrix denotes the probability to move from state $i$ to state $j$. And also the transition matrix should be stochastic: $\sum_j p_{i,j}=1$ for all $i$. Therefore your matrix should looks like: $$ \begin{bmatrix}\frac25&\frac25&\frac15\\ \frac25&\frac35&0\\ \frac35&\frac15&\frac15\end{bmatrix}. $$ And then the vector of stationary probabilities (long-term proportions of states) will be the solution of the system of equations: $$ \begin{bmatrix}P_s&P_c&P_r\end{bmatrix} \begin{bmatrix}\frac25&\frac25&\frac15\\ \frac25&\frac35&0\\ \frac35&\frac15&\frac15\end{bmatrix}=\begin{bmatrix}P_s&P_c&P_r\end{bmatrix} $$ In your matrix everything is the other way around: $p_{i,j}$ is the probability to move from $j$ to $i$. So, it is transposed. If you prefer to use your martix anyway, you should multiply matrix to a vector, not vice versa: $$ \begin{bmatrix}\frac25&\frac25&\frac35\\ \frac25&\frac35&\frac15\\ \frac15&0&\frac15\end{bmatrix}\begin{bmatrix}P_s\\P_c\\P_r\end{bmatrix}=\begin{bmatrix}P_s\\P_c\\P_r\end{bmatrix}. $$
Both cases the solution is $P_s=\frac8{19}$, $P_c=\frac9{19}$, $P_r=\frac2{19}$.