Mathematical Statistics and Data Analysis - Solutions

Chapter 7, Survey Sampling

Solution 24

Since this is a simple random sampling, we can use few results from the book:

$\, \Exp{X_i} = \mu \,$, where $\, \mu \,$ is the population mean.

$\, \Var{X_i} = \sigma^2 \,$, where $\, \sigma^2 \,$ is population variance.

$\, \Cov(X_i,X_j) = \frac {-\sigma^2} {N-1} \,$.

(a)

For $\, \bar X_c = \sum_{i=1}^{n} c_i X_i \,$, to be unbiased estimate of population mean, $\, \Exp {\bar X_c} = \mu \,$ must be true.

$$ \, \begin{align*} \Exp{\bar X_c} &= \Exp\Prn{\sum_{i=1}^{n} c_i X_i} \\ &= \sum_{i=1}^{n} \left( c_i \Exp{X_i} \right) \\ &= \sum_{i=1}^{n} (c_i \mu) \\ &= \mu \sum_{i=1}^{n} c_i \end{align*} \, $$

Thus for $\, \Exp {X_c} = \mu \,$ to be true, we must have $\, \sum_{i=1}^{n} c_i = 1 \,$.

(b)

Lets first find an expression for the variance of the estimate, i.e. $\, \Var(\bar X_c) \,$:

$$ \, \Var(\bar X_c) \\ \begin{align*} &= \Var\Prn{\sum_{i=1}^{n} c_i X_i} \\ &= \sum_{i=1}^{n} \sum_{j=1}^{n} \Prn{c_i c_j \Cov(X_i, X_j)} \\ &= \sum_{i=1}^{n} c_i^2 \Cov(X_i,X_i) + \sum_{i=1}^{n} \sum^{n}_{j=1,j \ne i} \Prn{c_i c_j \frac {-\sigma^2} {N-1} } \\ &= \sum_{i=1}^{n} c_i^2 \Var(X_i) - \frac {\sigma^2} {N-1}\sum_{i=1}^{n} \sum^{n}_{j=1,j \ne i} (c_i c_j) \\ &= \sigma^2 \sum_{i=1}^{n} c_i^2 - \frac {\sigma^2} {N-1}\sum_{i=1}^{n} \Prn{c_i \sum^{n}_{j=1,j \ne i} c_j} \\ &= \sigma^2 \sum_{i=1}^{n} c_i^2 - \frac {\sigma^2} {N-1}\sum_{i=1}^{n} (c_i(1-c_i)) && \text{Since }\sum_{i=1}^{n} c_i = 1 \\ &= \sigma^2 \sum_{i=1}^{n} c_i^2 \Prn{1+\frac 1 {N-1}} - \frac {\sigma^2} {N-1} \sum_{i=1}^{n} c_i \\ &= \sigma^2 \sum_{i=1}^{n} c_i^2 \Prn{1+\frac 1 {N-1}} - \frac {\sigma^2} {N-1} && \text{Since }\sum_{i=1}^{n} c_i = 1 \\ \end{align*} \, $$

Thus, to minimize $\, \Var(\bar X_c) \,$, we have to minimize $\, \sum_{i=1}^{n} c_i^2 \,$ under the given condition constraint $\, \sum_{i=1}^{n} c_i = 1 \,$.

This can be solved using Lagranges multiplier where $\, f = \sum_{i=1}^{n} c_i^2 \,$ is the function we have to minimize under the constraint $\, g = \sum_{i=1}^{n} c_i \,$.

We have:

$$ \, \nabla_c f = \lambda \nabla_c g \, $$

Let $\, f_{c_i} \,$ is the partial differentiation of $\, f \,$ w.r.t $\, c_i \,$ and similarly $\, g_{c_i} \,$ is partial differentiation of $\, g \,$ w.r.t. $\, c_i \,$.

Then by lagranges multiplier we should have $\, f_{c_i} = \lambda g_{c_i} \,$ for $\, i \in \{x \in \mathbb N \; \vert \; 1 \le x \le n \} \,$. It follows that $\, 2c_i = \lambda \,$, or $\, c_i = \frac {\lambda} 2 \,$. Now putting the value of $\, c_i \,$ in the constraint $\, g \,$, gives $\, \sum_{i=1}^{n} \frac {\lambda} 2 = 1 \,$. Thus $\, \lambda = \frac 2 n \,$. Now since $\, c_i = \frac {\lambda} 2 \,$, we get $\, c_i = \frac 1 n \,$.

Now to check that this value of $\, c_i \,$ minimizes the $\, f \,$, we can do the second derivative test, which I am skipping :)

Thus $\, c_i = \frac 1 n \,$ is the value that minimizes $\, \Var(\bar X_c) \,$.

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