Clear Days, Rainy Days Markov Chain Problem This is a markov chain problem I came across that I had some difficulty with. There are two possible states: either the weather is clear, or it is rainy.
If today is a clear day, then tomorrow has a 60% chance of being a clear day (and 40% of being rainy). If today is a rainy day, then tomorrow has a 70% chance of being a rainy day (and 30% of being clear).
Without being told the initial state, what is the expected number of clear days in a time period of 365 days?
My initial thoughts were to calculate the conditional expected values for an initial state of clear or cloudy, weigh them each 50/50, and add them, but the calculations get messy quickly and require a sophisticated calculator. I think this problem can be approximated somehow (as somehow this is meant to be solvable on the spot), but I'm not sure how. Any thoughts?
 A: As discussed in the comments, the question asks you to find the stationary distribution of the Markov chain. The stationary distribution describes the "long-run" distribution of states of the Markov chain, which seems like a reasonable approximation for the length of time given (i.e. a year).
In this example, the stationary distribution $(\pi_C,\pi_R)$, where $\pi_C$ denotes the stationary probability that the weather is clear, satisfies
\begin{align}
0.6 \pi_C + 0.3 \pi_R &= \pi_C \\
\pi_C + \pi_R &= 1.
\end{align}
An easy calculation gives us that $(\pi_C,\pi_R)=(3/7,4/7)$. Hence, one would expect about $3/7$ of the days of the year to exhibit clear weather.
More generally, the stationary distribution $\pi$, whenever it exists must satisfy $$ \pi = \pi P, \label{1}\tag{1}$$ where $P$ is the matrix of transition probabilities between states. (In this formulation, the rows of $P$ must sum up to one.) I'll leave you to check that the stationary probabilities we computed above satisfy \eqref{1}.
A: This question is asking about the stationary distribution of a Markov Chain, in short, no matter what the initial state is, a Markov Chain will always converge to its stationary distribution. In this case the two states are $\{\text{clear}, \text{rainy}\}$, and the transition matrix is:
\begin{bmatrix}
0.6& 0.4 \\
0.3& 0.7
\end{bmatrix}
Calculating the 'left eigenvector' gives the stationary distribution is
\begin{bmatrix}\frac 37 & \frac47\end{bmatrix}.
Therefore, in expectation, there would be around $365 \times 3 \div 7 \approx 152.6$ clear days. 
