Problem 4 (Adapted from Naiman, 2015: Sampling distributions; relative efficienc

Question

Problem 4 (Adapted from Naiman, 2015: Sampling distributions; relative efficiency). A population consists of N individuals. Each individual has a certain number of friends. Suppose the count of individuals in the population with a given number of friends is as in the following table: number in population with k friends 0 N -4 2M-1 2M where M 2 3 is an unknown positive integer and N > 7. A sample of size 3 is drawn without replacement and the numbers of friends for thei sampled individual is denoted by Xi, for i 1,2,3. (a) Let S denote the 3-tuple of elements X1, X2, X3 that drawn but written in increasing order. For example, if (Xi, X2,Xs) (M, 0, 1) then S-(0,1, M). There are 15 possible values that S can take on. Write down a list of these 15 possibilities. (b) Make a table giving the probability mass function of S, that is, for each value of s in the list in b) compute P(S = s). Make sure that your probabilities add to 1.

Explanation / Answer

a)

The trick here is to realize that in the group of N individuals, there is ONLY ONE individual each that 0, 1, 2M-1 or 2M friends. But there are N-4 individuals that have M friends and as N is atleast 7, that means there are atleast 3 individuals that have M friends. Thus, out of the sample of 3 individuals, it is possible that all 3 of them have M friends i.e (X1, X2, X3) = (M,M,M).

Using the above, here is the list of 15 possibilities that S can take on:  

1. (0, 1, M)

2. (0, 1, 2M-1)

3. (0, 1, 2M)

4. (0, M, 2M-1)

5. (0, M, 2M)

6. (0, 2M-1, 2M)

7. (1, M, 2M-1)

8. (1, M, 2M)

9. (1, 2M-1, 2M)

10. (M, 2M-1, 2M)

11. (0, M, M)

12. (1, M, M)

13. (M, M, 2M-1)

14. (M, M, 2M)

15. (M, M, M )

b)

Probability mass function of S :

Let's see it on the basis of the number of possible outcomes for each configuartion.

Again, remember that except for those having M friends, only 1 individual each has 0, 1, 2M-1 and 2M friends.

So, there will always be only 1 outcome whenever an outcome involves : 0, 1, 2M-1 or 2M.

It gets complicated when you have to pick when there is M in the outcome. Know that there are N-4 individuals that that have M friends. Thus, the no. of ways of choosing an individual that has M friends = N-4 ways.

These are the total number of ways for all possible values of S. To find probability we have to divide each value by the total number of ways of choosing 3 out of N individuals = nC3 = n! / (n-3)!3!

Probability mass function for S :

Outcome(X1,X2,X3) : Probability(X1,X2,X3)

1. (0, 1, M) : N-4 / nC3

2. (0, 1, 2M-1) : 1 / nC3

3. (0, 1, 2M) : 1 / nC3

4. (0, M, 2M-1) : N-4 / nC3

5. (0, M, 2M) : N-4 / nC3

6. (0, 2M-1, 2M) : 1 / nC3

7. (1, M, 2M-1) : N-4 / nC3

8. (1, M, 2M) : N-4 / nC3

9. (1, 2M-1, 2M) : 1 / nC3

10. (M, 2M-1, 2M) : N-4 / nC3

11. (0, M, M) : N-4C2 / nC3

12. (1, M, M) : N-4C2 / nC3

13. (M, M, 2M-1) : N-4C2 / nC3

14. (M, M, 2M) : N-4C2 / nC3

15. (M, M, M ) : N-4C3 / nC3

To verify that the probabilities add to 1 :

We have :

4(1 / nC3 ) + 6( N-4 / nC3 ) + 4( N-4C2 / nC3)  + 1(N-4C3 / nC3) = 4 + (6N - 24) + (2(N-4)(N-5)) + ((N-4)(N-5)(N-6)/6) /  nC3

= 4 + 6N - 24 + 2N2 - 18N + 40 + 1/6(N3 - 11N2 + 30N - 4N2 + 44N - 120) /  nC3  =

= 1/6( 12N2 - 72N + 120 + (N3 - 11N2 + 30N - 4N2 + 44N - 120)) / (n(n-1)(n-2)/6)

= N3 - 3N2 + 2N / (N3 - 3N2 + 2N)

= 1

Hence, the probabilites from the probability mass function for S all add upto 1. The calculations might seem confusing so you can re-check by yourself but rest assured that the calculations are correct and the answer is right.

Cheers!

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