A space ship walks randomly between 3 planets: A. B and C. It does countless trips among then, always going from planet X to planet Y. The probabilities (are constants) and travel times are on the table below. Example: leaving from planet B, there is a 50% chance that it arrives on planet A, and 50% at planet B (yes, it can go back to the same place it left). ORIGIN (row) DESTINY (columns)
The problem: there is a space ship leving from A that will only stop the journey when it reaches C. What will be the expected mean time (in months) to reach C?
| PROBABILITY MATRIX | DESTINATION | ||
|---|---|---|---|
| ORIGIN | A | B | C |
| A | 0.05 | 0.90 | 0.05 |
| B | 0.50 | 0.50 | 0 |
| C | 0 | 0 | 1 |
| TRAVEL TIME MATRIX (in moths) | DESTINATION | ||
|---|---|---|---|
| ORIGIN | A | B | C |
| A | 5 | 2 | 10 |
| B | 3 | 3 | 0 |
| C | 0 | 0 | 0 |
[markov chain sketch with the probabilities and travel time][1] [1]: https://i.stack.imgur.com/XTKz7.png
What I have so far:
let "a" be the number of times it will bounce on a journey from A to C; "b", the number of times it will bounce on a journey from B to C; and "c", the number of times it will bounce on a journey from C to C.
a = 1 + 0.05a + 0.9b + 0.05c (c is the absorbing state, so c=0)
thus: a = 1 + 0.05a + 0.9b
b = 1+ 0.5a + 0.5b
I came up with a = 56 and b = 58.
BUT, these numbers represents the number of times it moves from one planet to another. It does not give me the times in months that I need. (I'm stuck here)
Can't you do essentially the same thing by taking the travel time into account? Let $m_A$ be the expected number of months of the journey if the ship is at $A$, and let $m_b$ be the expected number of months if the ship is at $B$.
Then we have $$m_A=.05(5+m_A)+.9(2+m_B)+.05\cdot10$$ We get a similar equation for $m_B$, and solve for $m_A$ and $m_B$. Then $m_A$ is the answer to the question.