I have carried out a simulation experiment where $n$ out of $N$ items $x_i$ were drawn with replacement with unequal probabilities $z_i$ from a finite population. (If it matters: $z_i = 1/(Nx_i)$ in my case, but perhaps there are more general results.)
Then, I estimated the population totals of the indicators $$ y_i := \begin{cases}1: y_i = m \\ 0: \text{otherwise}\end{cases} $$ for each $m$ in two ways:
- By applying the Hansen-Hurwitz estimator: $\hat{Y}_I = n^{-1} \sum_i (y_i/z_i)$.
- By first removing the duplicates from the sample and then applying the Horvitz-Thompson estimator for the selection probabilites $\pi_i = 1 - (1 - n/N)^{y_i}$ (which is the probability of selecting the item $i$ at least once, i.e. one minus the probability of not selecting it at all, approximable by $x_i$ Bernoulli trials with success probability $1 - n/N$ each if $x_i \ll n$): $\hat{Y}_{II} = \sum_i(y_i/\pi_i)$.
This was repeated 1000 times.
I compared the estimates with the true values and observed the ratio between estimate and true value. I can clearly see from my experiments that the second estimate is better in terms of sampling variance. How to support this with theoretical results?
From Cochran (1977), Chapter 9, I see the variance of the HH estimator as $$ V(\hat{Y}_I) = n^{-1}\sum_i^N z_i (y_i/z_i - Y)^2 $$ and that of the HT estimator as $$ V(\hat{Y}_{II}) = \sum_i^N \frac{1-\pi_i}{\pi_i}y_i^2 + 2\sum_i^N\sum_{j>i}^N \frac{\pi_{ij}-\pi_i\pi_j}{\pi_i\pi_j}y_iy_j $$
but I have trouble applying this to my case. First, does it mean that for HH the variance depends on $Y$ and for HT it doesn't? Second, I would assume $\pi_{ij} = \pi_i\pi_j$ since the inclusion of items is independent, and then the second term of the HT variance vanishes -- or am I wrong?
I appreciate any help.