Problem
State of the weather in city can be modeled with simple probability. After rainy day it will rain the next day with probability of $0.5$ and after sunny day it will be sunny next day with probability of $0.9$. Let vector $\vec{x_n}$ be
$$ \vec{x_n} = \begin{bmatrix} \text{probability of sunny weather at day $n$} \\ \text{probability of rainy day at day $n$} \end{bmatrix} $$
Probability of rainy and sunny weather for $n+1$:th day can be solved with matrix equation
$$ x_{n+1} = \begin{bmatrix} 0.9 & 0.5 \\ 0.1 & 0.5 \end{bmatrix} x_n $$
We can assume that $\vec{x_0}=\begin{bmatrix}1 & 0\end{bmatrix}^T$. What happens when $x_{\infty}$ ? What are probabilities for random day when $n\rightarrow \infty$
Attempt to solve
First we compute eigenvalues and eigenvectors for our matrix. We can compute eigenvalues easily by utilizing two facts about eigenvalues. $1)$ sum of eigenvalues is same as trace. $2)$ product of eigenvalues is same as determinant. Let matrix $A$ be
$$ \textbf{A} = \begin{bmatrix} 0.9 & 0.5 \\ 0.1 & 0.5 \end{bmatrix}$$
Then we can note the equation as
$$ x_{n+1}=\textbf{A}x_n $$
Then we can solve eigenvalues.
$$ \begin{cases}
\lambda_1 + \lambda_2 = \text{Tr}(\textbf{A}) \\
\lambda_1\cdot \lambda_2= \det(\textbf{A}) \\
\end{cases} $$
$$ \implies \begin{cases}
\lambda_1 + \lambda_2 = 0.9 + 0.5 \\
\lambda_1\cdot \lambda_2= 0.9\cdot 0.5 – 0.5 \cdot 0.5 \\
\end{cases} $$
$$ \implies \begin{cases}
\lambda_1 + \lambda_2 = 1.4 \\
\lambda_1\cdot \lambda_2= 0.4 \\
\end{cases} $$
$$ \implies \lambda_1 = 0.4,\lambda_2 = 1 $$
Eigenvectors can be acquired by solving $\vec{x}$ from equation $(\textbf{A}-\lambda I)\vec{x}=\vec{0}$
$$
\begin{bmatrix} 0.9 – \lambda & 0.5 \\ 0.1 & 0.5 – \lambda \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}
$$
Now we have eigenvectors
$$ \lambda_1 \text{ gives } \vec{x_{\lambda1}} = \begin{bmatrix} 5 \\ 1 \end{bmatrix}, \lambda_2 \text{ gives } \vec{x_{\lambda2}} = \begin{bmatrix} -1 \\ 1 \end{bmatrix} $$
Eigenvectors are linearly independent so we can solve
$$ w_1 \vec{x_{\lambda1}} + w_2 \vec{x_{\lambda2}} = \vec{x_0} $$
$$ \implies w_1 \begin{bmatrix} 5 \\ 1 \end{bmatrix} + w_2 \begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} $$
$$ \implies \begin{bmatrix} 5 & -1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \end{bmatrix} = \begin{bmatrix} 1 \\0 \end{bmatrix} $$
$$ \implies w_1=1/6, w_2=-1/6 $$
I can write the original equation as:
$$ x_{n+1}=\textbf{A}^n \vec{x_{0}} = w_1 \cdot \lambda_1^n \cdot \vec{x_{\lambda1}}+w_2 \cdot \lambda_2^n \cdot \vec{x_{\lambda2}}$$
$$ x_{n+1} = w_1 \cdot 0.4^n \cdot \begin{bmatrix} 5 \\ 1 \end{bmatrix} + w_2 \cdot 1^n \cdot \begin{bmatrix} -1 \\ 1 \end{bmatrix} \implies w_2 \begin{bmatrix} -1 \\ 1 \end{bmatrix}, \text{ when } k \rightarrow \infty $$
Then we have $P("\text{sunny}")= -1w_1$ and $p(\text{"it is raining"})=w_1$
$$ P("\text{sunny}") = \frac{1}{6} \approx 16.67\% $$
$$ P("\text{raining}") = -\frac{1}{6} \approx -16.67\% $$
Which is quite confusing that i have negative probability ? I think something went wrong with these calculations but i cannot see what ?
Best Answer
It's all good except for the fact that the eigenvalue $\lambda_1=0.4$ yields as eigenvector $(-1,1)$ and eigenvalue $\lambda_2=1$ the eigenvector $(5,1)$ (that is, the other way around).
Being so, at the end you would have that $$\lim_{n\to\infty}w_2\,\lambda_2^n\,\vec{x}_{\lambda_2} = w_2\,(5,1)=(5/6, 1/6)$$