I have to find optimum sample size in sampling on two occasions.
Suppose that the samples are of the same size n on both occasions. In selecting the second sample, $m$ of the units in the first sample are retained. The remaining $u$ units are discarded and replaced by a new selection from the units not previously selected.
Notation:
$\bar y_{hu}=$mean of unmatched portion on occasion $h$
$\bar y_{hm}=$mean of matched portion on occasion $h$
$\bar y_{h}=$mean of whole sample on occasion $h$
$$\mathbb V(\bar {y_2}\prime)=\frac{S_2^2(n-u\rho^2)}{n^2-u^2\rho^2}\ldots(1)$$
$\rho$ denotes correlation coefficient
The optimum value of $u$ is found by minimizing equation (1) with respect to variation in $u$.This gives $$\frac{u}{n}=\frac{1}{1+\sqrt{1-\rho^2}}\ldots (2)$$ How they derived equation (2) from (1)?
I supposed to differentiate (1) with respect to $u$ and set it to zero. But i couldn't remove $S_2^2$ of equation (1) to derive (2).
If I differentiate $\mathbb V(\overline{y}_2')$ it gives: $\frac{S_2^2\cdot \rho^2 \cdot (n^2-2nu + \rho^2u^2)}{(n^2-u^2\rho ^2)^2}$
To get the fraction equal to zero, $n^2-2nu + \rho^2u^2$ has to be zero, if $S_2^2\cdot \rho^2 \neq 0$ . To solve this equation, you can use the the general formula to solve an quadratic equation:
$x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$
Here is $a=\rho ^2, b=-2n$ and $c=n^2$