Question

Suppose that whether it rains tomorrow depends on past weather conditions only through the past 2 days. Specifically, suppose that if it has rained yesterday and today, then it will rain tomorrow with probability .8; if it rained yesterday but not today, then it will rain tomorrow with probability .3; if it rained today but not yesterday, then it will rain tomorrow with probability .4; and if it has not rained either yesterday or today, then it will rain tomorrow with probability .2. What proportion of days does it rain?

Short answer

The proportion of days that it rains is $\frac{2}{5}$.

Step-by-step solution

Given Information

We have to find the probability for the portion of days that it will rain.

Simplify

This Markov chain has 4 states depending on the weather yesterday and today and these states are

$\left(R,R\right),\left(R,NR\right),\left(NR,R\right),\left(NR,NR\right)$

where for example says that it has not rained the day before, but it rains today. So, the transition matrix is given in the text and it is equal to

$P=\left[\begin{array}{cccc}0.8& 0.2& 0& 0\\ 0& 0& 0.3& 0.7\\ 0.4& 0.6& 0& 0\\ 0& 0& 0.2& 0.8\end{array}\right]$

Calculating for stationary distribution. Solving $\pi =\pi P$which implies these set of equations

localid="1651821523899" $\begin{array}{l}{\pi }_{R,R}=0.8{\pi }_{R,R}+0.4{\pi }_{NR,R}\\ {\pi }_{R,NR}=0.2{\pi }_{R,R}+0.6{\pi }_{NR,R}\\ {\pi }_{NR,R}=0.3{\pi }_{R,NR}+0.2{\pi }_{NR,NR}\\ {\pi }_{NR,NR}=0.7{\pi }_{R,NR}+0.8{\pi }_{NR,NR}\end{array}$

Solving this system of equations using ${\sum }_i{\pi }_i=1$,

localid="1651821988029" $\pi =\left[\begin{array}{l}4/15\\ 2/15\\ 2/15\\ 7/15\end{array}\right]$

So, the proportion that it rains today is the sum of the first and the third row in $\mathrm{\pi }$as these states indicate that it rains today. Hence, the proportion is localid="1651822039229" $\frac{2}{5}$.