I am assuming that 2 is to the right of 1. Then N, 1, 2, 3 are seated from left to right next to each other. Let's write RLRL for the event that N turns right, 1 turns left, 2 turns right, 3 turns left (in this case 1 and 2 both have partners). Let's also write XRRX for the event that 1 turns right, 2 turns right, and N and 3 turn either way, and so on.Originally Posted by andrew2002
Find the joint distribution of I(1) and I(2) by considering all possible cases:
Prob(I1 = I2 = 1) = pq + p^2q^2 [XRLX and RLRL]
Prob(I1 = 1, I2 = 0) = q^3p + p^2q [RLLL and RLXR]
Prob(I1 = 0, I2 = 1) = p^3q + pq^2 [RRRL and LXRL]
Prob(I1 = I2 = 0) = q^3 + p^3 + p^2q^2 [LLLX and XRRR and LLRR].
Then find marginal distributions, means and covariances. Of course the expected values for 1 and 2 should be the same.
For the second part, consider persons N, 1, 2, 3, 4 sitting next to each other and use the same approach.