+0

# 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 a

0
1838
2
+1793

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
+26971
+11

Starting at 1 there is probability 1/2 that it moves to 0 in 1 step.  There is probability 1/22 that it moves to 0 in 3 steps (1 → 2 → 1 → 0).  There is probability 1/ 23 that it moves to 0 in 5 steps (1 → 2 → 1 → 2 → 1 → 0), etc.

The expected number of steps is therefore t1 = 1/2 + 3/22 +  5/23 + ...  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 t2 = 1+3 or t2 = 4

So the ordered pair is (t1, t2) =  (3, 4)

.

Alan  Apr 18, 2015
#1
+93342
0

I do not understand what i am being asked to do :/

Melody  Apr 18, 2015
#2
+26971
+11

Starting at 1 there is probability 1/2 that it moves to 0 in 1 step.  There is probability 1/22 that it moves to 0 in 3 steps (1 → 2 → 1 → 0).  There is probability 1/ 23 that it moves to 0 in 5 steps (1 → 2 → 1 → 2 → 1 → 0), etc.

The expected number of steps is therefore t1 = 1/2 + 3/22 +  5/23 + ...  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 t2 = 1+3 or t2 = 4

So the ordered pair is (t1, t2) =  (3, 4)

.

Alan  Apr 18, 2015