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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)= ?
stud81
answered6 years ago
Anonymous
answered6 years ago
can anyone please answer the question?
stud81
answered6 years ago
Hi Anonymous,
Can you share answers to any of the other questions ....?
Anonymous
answered6 years ago
E[Ii]= p^2
E[H]-n.p^2
thats all i can calculate. stuck after this.
who cares
answered6 years ago
you are supposed to try and find your own solutions for the edx probability course
Explain Bot
answered11 months ago
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.