Explain formally how expected time to hit 0 from two is the sum of the expected time to hit 1 from 2 and 0 from 1

I have a symmetric random walk on the integers with probability $$p$$ and $$q$$ of going up and down respectively started at $$X_0 = 2$$.

Let

$$T^0 = \min\{ n > 0: X_n = 0\}, T^1 = \min\{ n > 0: X_n = 1\}$$

I want to show that the expected time from 2 to hit zero is the expected time of hitting one plus the expected time of hitting 0 from 1. $$\mathbb{E}_2[T^0] = \mathbb{E}_2[T^1] + \mathbb{E}_1[T^0]$$

This seems intuitively obvious to me, but I would like to be able to show it formally. I think this is a job for the law of iterated expectations, but the conditioning on the starting position confuses me. Can anyone show how to prove this formally step by step?

• Isn't this just linearity of expectation? The total time is the sum of the two times.
– whuber
Mar 5, 2021 at 15:49

Do you mean to put minimum $$n$$ in your definition of $$T^0$$ and $$T^1$$?
Let $$p(t)=P[T^1=t]$$.
$$E T^0=\sum_{t=1}^\infty P[T^1=t] E[t+\text{time to hit 0 starting from 1}|T^1=t]$$ $$=\sum_{t=1}^{\infty}(p(t) (t+E[\text{time to hit 0 starting from 1}]))$$
$$=\sum_{t=1}^{\infty}(p(t) t)+\sum_{t=1}^{\infty}(p(t) E[T^1])$$ $$=E[T^1]+E[T^1]\sum_{t=1}^{\infty}p(t) \\=2ET^1$$