# Mathematical Statistics and Data Analysis - Solutions

### Chapter 7, Survey Sampling

#### (b)

In part (a), we know $\, r \,$ and as well as $\, p \,$(known from the structure of the randomized device). Thus we can get $\, q = \frac {p+r-1} {2p-1} \,$.

#### (c)

Let $\, X_i = 1 \,$ if $\, i^{th} \,$ member of the random sample answers yes. Then $\, R = \frac 1 n \sum_{i=1}^{n} X_i \,$. Thus $\, \Exp R = \frac 1 n \sum_{i=1}^{n} \Exp(X_i) = \frac n n \Exp(X_i) = r \,$.

Lets propose $\, Q = \frac {p+R-1} {2p-1} \,$ as an estimate of $\, q \,$. Since it is a linear expression in $\, R \,$, it follows that $\, \Exp Q = \frac {p+\Exp R-1} {2p-1} = \frac {p+r-1} {2p-1} \,$ which is same as $\, q \,$.

#### (d)

Since $\, X_1 \,$ is a bernoulli’s trial, $\, \Var(X_1) = r(1-r) \,$.

We have $\, \Cov(X_1, X_2) = \Exp(X_1 X_2) - \Exp X_1 \Exp X_2 \,$. Now, $\, \Exp(X_1 X_2) = 0 + 0 + 0 + 1 \times 1 \times P(X_1=1, X_2=1) = \frac {nr} N \times \frac {nr-1} {N-1} \,$. Thus $\, \Cov(X_1, X_2) = \frac {nr} N \times \frac {nr-1} {N-1} - r^2 \,$.

Putting the values in $\, \Var(R) \,$, we get $\, \frac {r(1-r)} n + \frac {n-1} n \Prn{ \frac {nr} N \times \frac {nr-1} {N-1} \,-\, r^2 } \,$. Approximating $\, \frac {nr-1} {N-1} \approx \frac {nr} N \,$, we get: $\, \Var(R) = \frac {r(1-r)} n + \frac {r^2(n-1)} n \Prn{\frac {n^2} {N^2} - 1} \,$.

TODO: Not sure how if the extra term can be ignored using finite population correction as expected from the answer.

#### (e)

$$\tag*{\blacksquare}$$