Two-State Chain Return Time

Let $T$ be the number of days until the next bad day, counting from today (a bad day). Condition on tomorrow.

• With probability $\frac{7}{10}$ tomorrow is bad, and then $T=1$.
• With probability $\frac{3}{10}$ tomorrow is good. From a good day, each subsequent day is bad with probability $\frac{2}{5}$, independently of the past. So the further wait is $\mathrm{Geom}\!\left(\frac{2}{5}\right)$, with mean $\frac{5}{2}$. In this case $\mathbb{E}[T \mid \text{tomorrow good}] = 1 + \frac{5}{2}$.

Hence

Alternative. The mean return time to a state of an irreducible finite chain is $1/\pi$ of that state. Balancing the flow between the two states gives

So $\pi_G:\pi_B = 3:4$, and therefore $\pi_B=\frac{4}{7}$. The mean return time to $B$ is then $\frac{1}{\pi_B}=\frac{7}{4}$.

Generalisation. Write $p=\mathbb{P}(G\to B)$ and $q=\mathbb{P}(B\to G)$. Then $\pi_B=\frac{p}{p+q}$, and the mean return time to $B$ is $1+\frac{q}{p}$.