Applied matrix problem with probabilities

Question

For city we have simplified its weather forecasting as such. If it rains then the probability for rain the next day is $0.2$. If its sunny then the probability for sunny day the next day is $0.7$.

Vector $$x_{k}=\begin{bmatrix}\text{probability for sunny weather at day } k \\ \text{probability for rainy weather at day } k\end{bmatrix}$$ is the probability for sunny and rainy weather. At day $k+1$ we get from $k$ days probabilities that $$x_{k+1}=\begin{bmatrix} 0.7 & 0.8 \\ 0.3 & 0.2 \end{bmatrix} x_k$$ Whats the probability that it rains on random day? I have a hint that I can assume that $x_0 = [1\;0]^T$. So

\begin{align} x_{0+1}&=\begin{bmatrix} 0.7 & 0.8 \\ 0.3 & 0.2 \end{bmatrix} \begin{bmatrix} 1 \\ 0\end{bmatrix} = \begin{bmatrix} 0.7 \\ 0.3 \end{bmatrix} \\ x_{1+1}&=\begin{bmatrix} 0.7 & 0.8 \\ 0.3 & 0.2 \end{bmatrix} \begin{bmatrix} 0.7 \\ 0.3\end{bmatrix} = \begin{bmatrix} 0.73 \\ 0.27 \end{bmatrix} \end{align}

Should I continue this for $k\to \infty$ or use some other method?

Answer 1

There is a simple way to compute the steady state probabilities $p$ (sunny) and $q$ (rainy), since by definition, one more transition from the steady state won't change the probabilities, so

$0.7p + 0.8q = p,$

$0.3p + 0.2q = q,$

$p+q = 1,$ which yields

$p = \dfrac8{11}, q = \dfrac3{11}$

Answer 2

\begin{align} x_{0+1}&=\begin{bmatrix} 0.7 & 0.8 \\ 0.3 & 0.2 \end{bmatrix} \begin{bmatrix} 1 \\ 0\end{bmatrix} = \begin{bmatrix} 0.7 \\ 0.3 \end{bmatrix} \\[10pt] x_{1+1}&=\begin{bmatrix} 0.7 & 0.8 \\ 0.3 & 0.2 \end{bmatrix} \begin{bmatrix} 0.7 \\ 0.3\end{bmatrix} = \begin{bmatrix} 0.73 \\ 0.27 \end{bmatrix} \end{align}

If there is a limit $\begin{bmatrix} p \\ q \end{bmatrix}$ as the index approaches $\infty$, then it satisfies $$ \begin{bmatrix} 0.7 & 0.8 \\ 0.3 & 0.2 \end{bmatrix} \begin{bmatrix} p \\ q \end{bmatrix} = \begin{bmatrix} p \\ q \end{bmatrix}. $$