# Mathematical Statistics and Data Analysis - Solutions

### Chapter 7, Survey Sampling

#### (a)

We need $\, \Exp X = \mu \,$. Since $\, X = \alpha \bar X_1 + \beta \bar X_2 \,$, it follows that $\, \Exp X = \alpha \Exp X_1 + \beta \Exp X_2 \,$. Since $\, \bar X_1 \,$ and $\, \bar X_2 \,$ are unbiased estimates of the population mean, $\, \mu \,$, it follows $\, \Exp X_1 = \Exp X_2 = \mu \,$. Thus $\, \Exp X = \alpha \mu + \beta \mu \,$. Thus to make $\, EX = \mu \,$, we need $\, \alpha + \beta = 1 \,$.

#### (b)

We have $\, \Var(X) = \Var(\alpha \bar X_1 + \beta X_2) \,$. Since $\, X_1 \,$ and $\, X_2 \,$ are independent it follows that $\, \Var(X) = \alpha^2 \sigma^2_{\bar X_1} + \beta^2 \sigma^2_{\bar X_2}\,$.

Thus we have to minimize the function, say f, $\, \Var(X) = \alpha^2 \sigma^2_{\bar X_1} + \beta^2 \sigma^2_{\bar X_2}\,$ under the constraint, say g, $\, \alpha + \beta = 1 \,$. We can use Lagranges multuplier here:

$\, f = \lambda \nabla g \,$, which gives $\, \sigma^2_{\bar X_1} \times 2 \alpha = \lambda \,$ and $\, \sigma^2_{\bar X_2} \times 2 \beta = \lambda \,$. We also have $\, \alpha + \beta = 1 \,$. Thus by solving these equations we get:

$$\, \alpha = \frac {\sigma^2_{\bar X_2} } {\sigma^2_{\bar X_1} + \sigma^2_{\bar X_2} } \,$$
$$\, \beta = \frac {\sigma^2_{\bar X_1} } {\sigma^2_{\bar X_1} + \sigma^2_{\bar X_2} } \,$$
$$\tag*{\blacksquare}$$