+49 Fred the ant is on the real number line, and Fred is trying to get to the point $0.$ If Fred is at $1,$ then on the next step, Fred moves to either $0$ or $2$ with equal probability. If Fred is at $2,$ then on the next step, Fred always moves to $1.$ Let $e_1$ be expected number of steps Fred takes to get to $0,$ given that Fred starts at the point $1.$ Similarly, let $e_2$ be expected number of steps Fred takes to get to $0,$ given that Fred starts at the point $2.$ Determine the ordered pair $(e_1,e_2)$.
Tysm!! :D
+9673 If Fred starts on 1,
There's a 1/2 probability that it moves to 0 directly, and a 1/2 probability that it moves to 2.
Once it moves to 2, it must move back to 1.
And when it moves back to 1 again, there are again 1/2 probability that it moves to 0, and 1/2 probability that it moves to 2.
This repeats again and again.
So that means there are 1/2 probability it reaches 0 in 1 step, 1/4 probability it reaches 0 in 3 steps, 1/8 probability it reaches 0 in 5 steps, etc.
Because the e_1 = sum of all possible (probability * steps),
\(e_1 = \dfrac12 + \dfrac34 + \dfrac58 + \dfrac7{16} + \cdots = \displaystyle\sum_{n = 1}^\infty \dfrac{2n -1}{2^n}\) --- (1)
Now, we consider (e_1)/2. (The standard way of evaluating an A.G.S. (arithmetico-geometric series))
\(e_1 \cdot \dfrac12 = \dfrac14 + \dfrac38 + \dfrac5{16}+\cdots\) --- (2)
Subtracting (2) from (1),
\(\dfrac{e_1}2 = \dfrac12 + \left(\dfrac{2}4 + \dfrac28 + \dfrac2{16}+\cdots\right) = 1 + \left(\dfrac14 + \dfrac18 + \cdots\right) = \dfrac32\)
\(e_1 = 3\)
Therefore, if Fred starts at 1, then it is expected to reach 0 in 3 steps.
By a similar idea as above, we can get \(e_2 = 2\cdot \dfrac12 + 4\cdot \dfrac14 + 6\cdot \dfrac18 + \cdots = \displaystyle \sum_{n = 0}^\infty \dfrac{n + 1}{2^n} \)
By a similar approach, we can evaluate e_2 to be 4.
Therefore, (e_1, e_2) = (3, 4).
