A squirrel walks down a path. Every minute it either moves one yard forward, one yard backward, or stands still. At its beginning point there is a wall so that the squirrel either moves forward or stands still from this point. The probability that the squirrel moves forward is 1/3, backwards is 1/2, and stays in the same position is 1/6. If the squirrel is in the starting position it moves forward with probability 1/3 and remains still with probability 2/3. What is the expected number of minutes it takes for the squirrel to return to this starting position?
I'm familiar with birth and death stationary distributions given it is only possible to move forward or backward; $$\pi[x] = (1 - \frac{p}{q})(\frac{p}{q})^x$$ Thus, the expected time to return is simply $\frac{1}{\pi[0]}$. However, this is a bit more tricky as the squirrel can remain in the same position with positive probability. I assume that here we still want to find a stationary distribution - I'm unsure of how to do that in this case.
I'm not looking for a straight answer, but a nudge in the right direction would be greatly appreciated!
Given that the squirrel has moved during a given minute, there is a $2/5$ chance it moved forward and a $3/5$ chance it moved back. I claim that the stationary distribution for this squirrel is the same as for a "restless" squirrel who always moves forward and back with probabilities $3/5$ and $2/5$. This is because the only difference between your squirrel and the restless one is that your squirrel will spend a little extra time at each point before moving, but this does not affect the ratios between the times it spends at each point. This gives $$\pi[x] = \left(1-\frac{2/5}{3/5}\right)\left(\frac{2/5}{3/5}\right)^x.$$