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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?

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

1 Answer 1

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Do you mean to put minimum $n$ in your definition of $T^0$ and $T^1$?

Let $p(t)=P[T^1=t]$.
The total time to hit 0 starting from 2 is equal to the time to hit 1 starting from 2 plus the time to hit 0 starting from 1.
Also, notice that expected time to hit 0 starting at 1 is the same as the expected time to hit 1 starting at 2. Condition on the time to hitting 1.

$$E T^0=\sum_{t=1}^\infty P[T^1=t] E[t+\text{time to hit 0 starting from 1}|T^1=t]$$ $$=\sum_{t=1}^{\infty}(p(t) (t+E[\text{time to hit 0 starting from 1}]))$$

$$=\sum_{t=1}^{\infty}(p(t) t)+\sum_{t=1}^{\infty}(p(t) E[T^1])$$ $$=E[T^1]+E[T^1]\sum_{t=1}^{\infty}p(t) \\=2ET^1$$

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