# 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)

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

#### (d)

\, \begin{align*} \Exp(U_i U_j) \\ &= \sum_{m=0}^{1} \sum_{n=0}^{1} m n P(U_i = m, U_j = n) \\ &= 0 + 0 + 0 + 1 \cdot 1 \cdot P(U_i=1, U_j=1) \\ &= \frac n N \frac (n-1) (N-1) \end{align*} \,

#### (e)

\, \begin{align*} \Cov(U_i, U_j) \\ &= \Exp(U_i U_j) - \Exp(U_i) \Exp(U_j) \\ &= \frac n N \frac (n-1) (N-1) - \Prn{\frac n N}^2 \\ &= \frac {-n} N \frac {N-n} {N(N-1)} \end{align*} \,

#### (f)

\, \begin{align*} \Var(\bar X) \\ &= \Var \Prn{\frac 1 n \sum_{i=1}^{N} U_i x_i} \\ &= \frac 1 {n^2} \sum_{i=1}^{N} U_i x_i \\ &= \frac 1 {n^2} \Prn{ \sum_{i=1}^{N} x_i^2 \Var(U_i) \; + \; \sum_{i=1}^{N} \sum_{j=1,\;j \ne i}^{N} x_i x_j \Cov(U_i, U_j) } \\ &= \frac 1 {n^2} \Prn{ \sum_{i=1}^{N} x_i^2 \Prn{\frac n N \frac {N-n} N} \; + \; \sum_{i=1}^{N} \sum_{j=1,\;j \ne i}^{N} x_i x_j \Prn{\frac {-n} N \frac {N-n} {N(N-1)} } } \\ &= \frac 1 {n^2} \frac n N \frac {N-n} N \Prn{ \sum_{i=1}^{N} x_i^2 \; - \; \frac 1 {N-1} \sum_{i=1}^{N} \sum_{j=1,\;j \ne i}^{N} x_i x_j } \\ &= \frac 1 {n^2} \frac n N \frac {N-n} N \frac 1 {N-1} \Prn{ (N-1)\sum_{i=1}^{N} x_i^2 \; - \; \sum_{i=1}^{N} \sum_{j=1,\;j \ne i}^{N} x_i x_j } \\ &= \frac 1 {n^2} \frac n N \frac {N-n} N \frac 1 {N-1} \Prn{ (N-1)\sum_{i=1}^{N} x_i^2 \; - \; \Prn{\sum_{i=1}^{N} \sum_{j=1}^{N} x_i x_j \; - \; \sum_{i=1}^{N} x_i^2 } } \\ &= \frac 1 {n^2} \frac n N \frac {N-n} N \frac 1 {N-1} \Prn{ N\sum_{i=1}^{N} x_i^2 \; - \; \sum_{i=1}^{N} \sum_{j=1}^{N} x_i x_j } \\ &= \frac 1 {n^2} \frac n N \frac {N-n} N \frac 1 {N-1} N^2 {\sigma^2} && \text{Using variance formulae} \\ &= \frac 1 n \frac {N-n} {N-1} {\sigma^2} \end{align*} \,
$$\tag*{\blacksquare}$$