i have stumbled upon the following problem and have absolutely no idea how to approach it.the problem is as follows- Suppose we have a circle, centered at the origin, with a circumference of $12$ units. Both of the $x$-intercepts serve as the starting point for a particle, each of which, each second, moves (independently) a distance of one unit either clockwise or counterclockwise around the circle with equal probability. What is the expected number of seconds to elapse until the two particles simultaneously occupy the same point on the circle for the first time?
if it can be done using conditional expectation or otherwise,then please help me on this.thank you
Let $E(x)$ be the expected number of seconds to pass until a collision, when the particles are $x$ units apart. It can be seen that at all times we must have $x \leq 6$ for a circle with circumference $12$.
$$E(6) = 1 + \frac{1}{2}E(6) + \frac{1}{2}E(4)$$
$$E(4) = 1 + \frac{1}{2}E(4) + \frac{1}{4}E(6) + \frac{1}{4}E(2)$$
$$E(2) = 1 + \frac{1}{2}E(2) + \frac{1}{4}E(4) + \frac{1}{4}E(0)$$
$$E(0) = 0$$
So we get, using simple arithmetic, $E(2) = 10$, $E(4) = 16$ and $E(6) = 18$ so the answer is $18$.
We can generalise this: for a circle with circumference $4n$, it takes an average of $2n^2$ seconds for the first collision to occur.
We claim that:
$$E(2k) = 2\left(n^2 - (n-k)^2\right)$$ for all $k$ with a circle of circumference $4n$.
$E(0) = 0$, you can prove the rest!