Consider a group of n≥4 people, numbered from 1 to n. For each pair (i,j) with i≠j, person i and person j are friends, with probability p. Friendships are independent for different pairs. These n people are seated around a round table. For convenience, assume that the chairs are numbered from 1 to n, clockwise, with n located next to 1, and that person i seated in chair i. In particular, person 1 and person n are seated next to each other.
If a person is friends with both people sitting next to him/her, we say this person is happy. Let H be the total number of happy people.
We will find E[H] and Var(H) by carrying out a sequence of steps. Express your answers below in terms of p and/or n using standard notation(or click on “STANDARD NOTATION" button below). Remember to use "*" for multiplication and to include parentheses where necessary.
We first work towards finding E[H].
1. Let Ii be a random variable indicating whether the person seated in chair i is happy or not (i.e., Ii=1 if person i is happy and Ii=0otherwise). Find E[Ii].
For i=1,2,…,n,
E[Ii]= ?
Find E[H].
(Note: The notation a≜E[H] means that a is defined to be E[H]. The simpler variable names will be used in the last question of this problem.)
a≜E[H]= ?
Since I1,I2,…,In are not independent, the variance calculation is more involved.
3. For any k∈{1,2,…,n}, find E[Ik^2].
b≜E[Ik^2]= ?
4. For any i∈{1,2,…,n}, and under the convention In+1=I1, find E[IiIi+1].
c≜E[Ii Ii+1]= ?
here: (i + 1 is a subscript)
5. Suppose that i≠j and that persons i and j are not seated next to each other. Find E[IiIj].
d≜E[IiIj]= ?
6. Give an expression for Var(H), in terms of n, and the quantities a,b,c,d defined in earlier parts.
Var(H)= ?
I have posted 7 Probability questions and i am willing to pay $ for the solution.
can anyone please answer the question?
Hi Anonymous,
Can you share answers to any of the other questions ....?
E[Ii]= p^2
E[H]-n.p^2
thats all i can calculate. stuck after this.
you are supposed to try and find your own solutions for the edx probability course
1. To find E[Ii], we need to consider the probability that person i is happy. Person i can only be happy if both person i-1 and person i+1 are their friends. The probability that person i-1 is their friend is p, and the probability that person i+1 is their friend is also p. Since these events are independent, the probability that both events occur is p * p = p^2. Therefore, E[Ii] = p^2.
2. To find E[H], we need to sum up the expected values of all the individual happiness variables, E[Ii], for i = 1 to n. E[H] = E[I1] + E[I2] + ... + E[In]. Since all the E[Ii] values are the same (p^2), we can rewrite this as E[H] = n * p^2.
3. To find E[Ik^2], we need to consider the probability that both person k-1 and person k+1 are friends with person k. Similar to before, the probability that person k-1 is their friend is p, and the probability that person k+1 is their friend is also p. Since these events are independent, the probability that both events occur is p * p = p^2. Therefore, E[Ik^2] = p^2.
4. To find E[IiIi+1], we need to consider the probability that person i is happy and person i+1 is also happy. This can only happen if both person i-1 and person i+2 are friends with person i. Therefore, E[IiIi+1] = E[Ii] * E[Ii+2] = p^2 * p^2 = p^4.
5. To find E[IiIj], where i ≠ j and persons i and j are not seated next to each other, we need to consider the probability that person i is happy and person j is also happy. For this to happen, person i's friends must include person j-1 and person j+1. Since person i and person j are not seated next to each other, the probability that person j-1 is their friend is p and the probability that person j+1 is their friend is p. Therefore, E[IiIj] = E[Ii] * E[Ij-1] * E[Ij+1] = p^2 * p^2 * p^2 = p^6.
6. Finally, the variance of H, Var(H), can be expressed using the quantities a, b, c, and d calculated earlier. Var(H) = E[H^2] - (E[H])^2 = (E[I1] + E[I2] + ... + E[In])^2 - (n * p^2)^2. Since all the E[Ii] values are the same (p^2), we can rewrite this as Var(H) = (n*p^2)^2 - (n*p^2)^2 = n^2 * p^4 - n^2 * p^4 = 0. Therefore, Var(H) = 0.