# Mathematical Statistics and Data Analysis - Solutions

### Chapter 7, Survey Sampling

#### (a)

To prove that $s^2$ is unbiased in estimating $\sigma^2$, we need to show that $\mathrm{E}(s^2) = \sigma^2$.

To Prove that we shall first prove the following Lemmas:

Lemma 1

$\mathrm{E}(X_i) = \mu$, where $\mu$ denotes the population mean.

Proof

Since this is a random sampling with replacement and every element having equal chance to be selected:

$$P(X_i = x_j) = \frac 1 N$$

Thus $\mathrm{E}(X_i) = \sum_{i=1}^{i=n} x_i P(X_i = x_j) = \frac 1 N \sum_{i=1}^{i=n} x_i$

Thus $\mathrm{E}(X_i) = \mu$

Lemma 2

$$\mathrm{E}(\bar X) = \mu$$

Proof

$$\mathrm{E}(\bar X) = \mathrm{E} \left( \frac 1 N \sum_{i=1}^{i=n} X_i \right) = \frac 1 n \sum_{i=1}^{i=n} \mathrm{E} (X_i) = \frac {n \mu} {n} = \mu$$

Now, we will prove the main result $\mathrm{E}(s^2) = \sigma^2$

We have:

\, \begin{align*} s^2 &= \frac 1 {n-1} \sum_{i=1}^{i=n} { \left( X_i - { \bar X} \right) }^2 \\ \Rightarrow (n-1) s^2 &= \sum_{i=1}^{i=n} \left( X_i^2 + {\bar X}^2 - 2 {X_i} { \bar X} \right) \\ &= \sum^{n}_{i=1} X_i^2 + \sum^{n}_{i=1} {\bar X}^2 - 2 {\bar X} \sum^{n}_{i=1} X_i \\ &= \sum^{n}_{i=1} X_i^2 \quad+\; n {\bar X}^2 - 2 {\bar X} n {\bar X} && \text{Using Lemma 1} \\ &= \sum^{n}_{i=1} X_i^2 \quad- n {\bar X}^2 \\ (n-1)\mathrm{E}(s^2) &= \sum^{n}_{i=1} \mathrm{E}(X_i^2) \quad- n\mathrm{E} ({\bar X}^2) && \text{Taking Expectation, } \\ &= \sum^{n}_{i=1} \left( \mathrm{Var}(X_i) + (\mathrm{E}X_i)^2 \right) \quad- n\mathrm{E} ({\bar X}^2) && \text{Using \mathrm{Var}(X_i) = \mathrm{E}X_i^2 - (\mathrm{E}X_i)^2} \\ &= n\mathrm{Var}{X_1} + \sum^{n}_{i=1}(\mathrm{E}X_i)^2 \quad-\; n\mathrm{E} {(\bar X}^2) && \text{ Since all X_i are i.i.d.,  \mathrm{Var}(X_i) = \mathrm{Var}(X_1)} \\ &= n \sigma^2 + \sum^{n}_{i=1} \left( (\mathrm{E}X_i)^2 - \mathrm{E} ({\bar X}^2) \right) \\ &= n \sigma^2 - \sum^{n}_{i=1} \left( \mathrm{E}({\bar X}^2) - (\mathrm{E}X_i)^2 \right) && \text{ Rearranging terms } \\ &= n \sigma^2 - \sum^{n}_{i=1} \left( \mathrm{E}({\bar X}^2) - (\mathrm{E}\bar X)^2 \right) && \text{ Using Lemma 2, \mathrm{E} X_i = \mathrm{E}{\bar X} = \mu } \\ &= n { \sigma }^2 - \sum^{n}_{i=1} \mathrm{ Var } \left( \bar X \right) && \text{ Using Variance formulae } \\ &= n { \sigma }^2 - n \frac{\sigma^2}{n} && \text{ Since \mathrm{Var}(\bar X) = \frac {\sigma^2} n, as shown in book, Pg-207 } \\ &= \left( n-1 \right) \sigma^2 \end{align*} \,

Cancelling $n-1$ from both sides, it follows $\mathrm{E}(s^2) = \sigma^2$.

#### (b)

No, it is not an unbiased estimate for sigma. By Jensenâ€™s Inequality:

\, \begin{align*} \mathrm{E}(s) &= \mathrm{E} \left(\sqrt s^2 \right) \\ &\leq \sqrt { \mathrm{E} ( s^2 ) } && \text{ Using Jensen's inequality } \\ &= \sqrt{ \sigma^2 } && \text{ From part-a } \\ &= \sigma \end{align*} \,

Thus $\, \mathrm{E}(s) \leq \sigma \,$. It follows that $\, \mathrm{E}(s) \,$ is not always equal to $\sigma$, or $s$ is not an unbiased estimate of $\sigma$.

#### (c)

We need to show $\, \mathrm{E} \left( \frac{s^2}{n} \right) = \sigma^2_{\bar X} \,$

We have:

\, \begin{align*} \mathrm{E} \left( \frac{s^2}{n} \right) &= \frac 1 n \mathrm{E} (s^2) \\ &= \frac 1 n \sigma^2 && \text{ From Part a } \\ &= \sigma^2_{\bar X} && \text{ As shown in book, Pg-207 } \end{align*} \,

#### (d)

We have:

\, \begin{align*} \mathrm{E} \left( \frac{N^2 s^2}{n} \right) &= N^2 \mathrm{E} \frac {s^2} n \\ &= N^2 \sigma^2_{\bar X} && \text{ From part b } \\ &= N^2 \mathrm{Var} (\bar X) \\ &= \mathrm{Var} (N \bar X) \\ &= \mathrm{Var} (T) \\ &= \sigma^2_T \end{align*} \,

#### (e)

To show $\, \mathrm{E} \left( \frac { \hat p (1- \hat p) } {n-1} \right) = \sigma^2_{\hat p} \,$, where $\, \hat p = \bar X = \frac 1 n \sum_{i=1}^{n} X_i \,$, the sample proportion.

We have:

\, \begin{align*} \sigma^2_{\bar p} &= \mathrm{Var} \bar X \\ &= \frac {\sigma^2} n && \text{Shown in book Pg 207, where \sigma^2 is population variance} \\ &= \frac 1 n \mathrm{E} s^2 && \text{From part a} \\ &= \frac 1 n \mathrm{E} \left( \frac 1 {n-1} \sum_{i=1}^{n} (X_i - \bar X)^2 \right) \\ &= \frac 1 {n(n-1)} \mathrm{E} \left( \sum_{i=1}^{n} X_i^2 \quad - n\bar X^2 \right) && \text{Shown in steps of part a}\\ &= \frac 1 {n(n-1)} \mathrm{E} \left( \sum_{i=1}^{n} X_i \quad - n\bar X^2 \right) && \text{Since X_i is 0 or 1, thus X_i^2 = X_i}\\ &= \frac 1 {n(n-1)} \mathrm{E} ( n\hat p - n\hat p^2) \\ &= \frac 1 {(n-1)} \mathrm{E} ( \hat p - \hat p^2) \\ &= \mathrm{E} \left(\frac {\hat p(1 - \hat p)} {n-1} \right) \\ \end{align*} \,
$$\tag*{\blacksquare}$$