Krdytkyu

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 $vec{x_0}=begin{bmatrix}1 & 0end{bmatrix}^T$. What happens when $x_{infty}$ ? What are probabilities for random day when $nrightarrow infty$

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.

$$ begin{cases}
lambda_1 + lambda_2 = text{Tr}(textbf{A}) \
lambda_1cdot lambda_2= det(textbf{A}) \
end{cases} $$

$$ implies begin{cases}
lambda_1 + lambda_2 = 0.9 + 0.5 \
lambda_1cdot lambda_2= 0.9cdot 0.5 - 0.5 cdot 0.5 \
end{cases} $$

$$ implies begin{cases}
lambda_1 + lambda_2 = 1.4 \
lambda_1cdot lambda_2= 0.4 \
end{cases} $$

$$ implies lambda_1 = 0.4,lambda_2 = 1 $$

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 ?

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 $$lim_{ntoinfty}w_2,lambda_2^n,vec{x}_{lambda_2} = w_2,(5,1)=(5/6, 1/6)$$