# Mathematical Statistics and Data Analysis - Solutions

### Chapter 7, Survey Sampling

#### Solution 51

To prove that bias of $\, \hat \theta_J \,$ is of the order of $\, n^{-2} \,$, we need to show $\, \Exp(\hat \theta_J) \,$ contains one term equals to $\, \theta \,$ and remaining terms contains $\, n^{2} \,$ or higher powers in denominator.

Given:

$$\, \Exp(\hat \theta) = \theta + \frac {b_1} n + \frac {b_2} {n^2} + \frac {b_3} {n^3} + ... \,$$
$$\, \Exp(\hat \theta_j) = \theta + \frac {b_1} {m(p-1)} + \frac {b_2} {(m(p-1))^2} + ... \,$$
$$\, V_j = p{\hat \theta} - (p-1){\hat \theta_j} \,$$
$$\, \hat \theta_J = \frac 1 p \sum_{j=1}^{p} V_j \,$$

Proof:

\, \begin{align*} \Exp(\hat \theta_J) &= \Exp \Prn{ \frac 1 p \sum_{j=1}^{p} V_j } \\ &= \Exp \Prn{ \frac 1 p \sum_{j=1}^{p} (p{\hat \theta} - (p-1){\hat \theta_j}) } \\ &= \frac 1 p\sum_{j=1}^{p} \Exp (p{\hat \theta} - (p-1){\hat \theta_j}) \\ &= \frac 1 p\sum_{j=1}^{p} \Prn{p\Exp{\hat \theta} - (p-1)\Exp{\hat \theta_j} } \\ &= \frac 1 p\sum_{j=1}^{p} \Prn{p\Prn{\theta + \frac {b_1} n + \frac {b_2} {n^2} + ...} - (p-1)\Prn{\theta + \frac {b_1} {m(p-1)} + \frac {b_2} {(m(p-1))^2} + ...} } \\ &= \frac 1 p\sum_{j=1}^{p} \Prn{p\Prn{\theta + \frac {b_1} n + \frac {b_2} {n^2} + ...} - \Prn{(p-1)\theta + \frac {b_1} {m} + \frac {b_2} {m^2(p-1)^{2-1}} + ...} } \\ &= \frac 1 p\sum_{j=1}^{p} \Prn{ (p+1-p)\theta + \Prn{\frac p n - \frac 1 m}b_1 + \Prn{\frac p {n^2} - \frac 1 {m^2(p-1)} }b_2 + \Prn{\frac p {n^3} - \frac 1 {m^3(p-1)^2} }b_3 + ...} \\ &= \frac 1 p\sum_{j=1}^{p} \Prn{\theta + \Prn{\frac p {mp} - \frac 1 m}b_1 + \Prn{\frac {p m^2 (p-1) - n^2} {m^2 n^2 (p-1)} }b_2 + ...} \\ &= \frac 1 p\sum_{j=1}^{p} \Prn{\theta + 0 + \Prn{\frac {p m^2 (p-1) - m^2 p^2} {m^2 n^2 (p-1)} }b_2 + ...} \\ &= \frac 1 p\sum_{j=1}^{p} \Prn{\theta + \Prn{\frac {m^2 p} {m^2 n^2 (p-1)} }b_2 + ...} \\ &= \frac 1 p\sum_{j=1}^{p} \Prn{\theta + \Prn{\frac p {n^2 (p-1)} }b_2 + ...} \\ &= \theta + \Prn{\frac p {n^2 (p-1)} }b_2 + ... \\ \end{align*} \,

Last step is reduced by removing summation and division by $\, p \,$ because every terms occurs $\, p \,$ times in the sum.

Thus all the terms contains $\, n^2 \,$ or higher powers in the denominator.

$$\tag*{\blacksquare}$$