Mathematical Statistics and Data Analysis - Solutions

Chapter 7, Survey Sampling

Solution 28

(a)

$$ \, \begin{align*} r = P(yes) \\ &= P(yes \,\vert\, \text{Q1 was asked})P(\text{Q1 was asked}) + P(yes \,\vert\, \text{Q2 was asked})P(\text{Q2 was asked}) \\ &= qp + (1-q)(1-p) \\ &= (2p-1)q + (1-p) \end{align*} \, $$

(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)

$$ \, \begin{align*} \Var(R) \\ &= \Var(\frac 1 n \sum_{i=1}^{n} X_i) \\ &= \frac 1 {n^2} \Var(\sum_{i=1}^{n} X_i) \\ &= \frac 1 {n^2} (\sum_{i=1}^{n} \Var(X_i) + \sum_{i=1}^{n} \sum_{j=1}^{n} \Cov(X_i, X_j)) \\ &= \frac {n\Var(X_i)} {n^2} + \frac {n(n-1)Cov(X_i,X_j)} {n^2} \\ &= \frac {\Var(X_1)} n + \frac {(n-1)Cov(X_1,X_2)} {n} && \text{ Since } \Cov(X_i, X_j) = \Cov(X_1, X_2) \\ \end{align*} \, $$

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)

$$ \, \begin{align*} \Var(Q) \\ &= \Var\Prn{\frac {p+R-1} {2p-1}} \\ &= \Var\Prn{\frac {p-1} {2p-1} + \frac R {2p-1}} \\ &= \frac 1 {(2p-1)^2} \Var(R) && \text{Since $p$ is constant, first term was zero} \\ &= \frac {r(1-r)} {n(2p-1)^2} && \text{ Using part-d result } \end{align*} \, $$
$$\tag*{$\blacksquare$} $$