# Mathematical Statistics and Data Analysis - Solutions

### Chapter 7, Survey Sampling

#### (a)

We need to show that $\, \bar X = n^{-1} \sum_{i=1}^{N} U_i x_i \,$. Since we have exactly $\, n \,$ terms in the sum $\, \sum_{i=1}^{N} U_i x_i \,$, where $\, U_i x_i \ne 0 \,$, we can denote those terms by $\, X_j = U_i x_i \,$, where $\, j \,$ varies from $\, 1 \,$ to $\, n \,$. Thus we have $\, \sum_{i=1}^{N} U_i x_i = \sum_{j=1}^{n} X_j \,$. But $\, n^{-1} \sum_{j=1}^{n} X_j = \bar X \,$, it follows that $\, \bar X = n^{-1} \sum_{i=1}^{N} U_i x_i \,$.

#### (b)

$\, P(U_i = 1) \,$ means the probability that $\, i^{th} \,$ element of the population is present in a random sample of size n. This is equivalent to the probability of selecting an element in a random sample of size $\, n \,$. Thus $\, P(U_i = 1) = \frac n N \,$.

Since $\, U_i \,$ is either $\, 0 \,$ or $\, 1 \,$, by fundamental bridge, it follows $\, \Exp(U_i) = P(U_i=1) = \frac n N \,$.

#### (c)

$\, \Var(U_i) = P(U_i=1)(1-P(U_i=1)) = \frac n N(1-\frac n N) = \frac n N \frac (N-n) N \,$.

#### (f)

$$\tag*{\blacksquare}$$