A point is on a number line, but it is limited to being at 0, 1, or 2. At each step, the point moves 1 unit left or right. If the point is at 1, it moves to either 0 or 2 with equal probability. However, if the point is at 2, it must move to 1 with its next step. For a point currently at 1, let $t_1$ be the expected number of steps the point will take before it reaches 0 for the first time. Similarly, for a point currently at 2, let $t_2$ be the expected number of steps for the point to first reach 0. Determine the ordered pair $(t_1,t_2)$.

Mellie
Apr 18, 2015

#2**+10 **

Starting at 1 there is probability 1/2 that it moves to 0 in 1 step. There is probability 1/2^{2} that it moves to 0 in 3 steps (1 → 2 → 1 → 0). There is probability 1/ 2^{3} that it moves to 0 in 5 steps (1 → 2 → 1 → 2 → 1 → 0), etc.

The expected number of steps is therefore t_{1} = 1/2 + 3/2^{2} + 5/2^{3} + ... or

$$t_1=\sum_{k=1}^\infty\frac{2k-1}{2^k}=3$$

Starting at 2 it takes 1 step to 1 from where the expected number of steps is 3 as obtained above, so t_{2} = 1+3 or t_{2} = 4

So the ordered pair is (t_{1}, t_{2}) = (3, 4)

.

Alan
Apr 18, 2015

#2**+10 **

Best Answer

Starting at 1 there is probability 1/2 that it moves to 0 in 1 step. There is probability 1/2^{2} that it moves to 0 in 3 steps (1 → 2 → 1 → 0). There is probability 1/ 2^{3} that it moves to 0 in 5 steps (1 → 2 → 1 → 2 → 1 → 0), etc.

The expected number of steps is therefore t_{1} = 1/2 + 3/2^{2} + 5/2^{3} + ... or

$$t_1=\sum_{k=1}^\infty\frac{2k-1}{2^k}=3$$

Starting at 2 it takes 1 step to 1 from where the expected number of steps is 3 as obtained above, so t_{2} = 1+3 or t_{2} = 4

So the ordered pair is (t_{1}, t_{2}) = (3, 4)

.

Alan
Apr 18, 2015