### Chapter 7, Survey Sampling

#### Solution 25

The joint distribution of any subcollection $\, Y_{i_1}, Y_{i_2}, Y_{i_3}, ... , Y_{i_n} \,$ of $\, Y_i \,$ is same as of $\, X_1, X_2, X_3, ..., X_n \,$:

Since each permutation has an equal probability, the joint distribution of the subcollections $\, Y_{i_1}, Y_{i_2}, Y_{i_3}, ... , Y_{i_n} \,$ must be identically distributed i.e. the joint distribution of every subcollection is same. Thus the joint probability $\, P(Y_{i_1} = a_1, Y_{i_2} = a_2, ... , Y_{i_n} = a_n) \; = \; P(Y_1 = a_1, Y_2 = a_2, ... , Y_n = a_n) \,$ for any subcollection $\, Y_{i_1}, Y_{i_2}, Y_{i_3}, ... , Y_{i_n} \,$ is same.

Now we can check easily $\, P(Y_1 = a_1, Y_2 = a_2, ... , Y_n = a_n) \,$ is equal to $\, P(X_1 = a_1, Y_2 = a_2, ... , X_n = a_n) \,$. For simplicity, consider for the case $\, n=2 \,$: $\, P(Y_1 = a_1, Y_2 = a_2) \,$ is the probability that first and second item of a random permutaion of $\, Y_1,Y_2, ..., Y_N \,$ are $\, a_1 \,$ and $\, a_2 \,$ respectively. This is clearly the same as drawing two items from $\, N \,$ items such that the first item drawn is $\, a_1 \,$ and second item drawn is $\, a_2 \,$. This in turn is same as $\, P(X_1 = a_1, X_2 = a_2) \,$. Similarly the result holds true for any size $\, n < N \,$.

Now, it is easy to see $\, \Var(Y_i) = \Var(X_i) = \sigma^2 \,$. Since $\, \Var(Y_i) = \Exp(Y_i^2) - (\Exp(Y_i))^2 \,$ and $\, \Exp(Y_i) = \Exp(X_i) \,$ because $\, Y_i \,$ and $\, X_i \,$ are identically distributed, the required result $\, \Var(Y_i) = \Var(X_i) \,$ follows.

Similarly, since the joint distribution of any subcollection of the permutations is identically distributed as that of a simple random sample, it follows that $\, \Cov(Y_i, Y_j) = \Cov(X_i, X_j) \,$.

Now, since $\, Y_1 + Y_2 + ... + Y_N = \tau \,$, or a constant, it follows that variance of $\, \Var\Prn{\sum_{i=1}^{n} Y_i} = 0 \,$

Using the variance formulae,

Thus we get $\, \gamma = \frac {-\sigma^2} {N-1} \,$

$$\tag*{$\blacksquare$} $$