Suppose I take two independent random samples from a population of size $N$: the first is a simple random sample of size $n$ and the second is a simple random sample of size $m$. Let set $S$ contain all of the records selected from the two samples of size $n$ and $m$ less any duplicates. Thus, $S$ should consist of $n_s$ records where $n_s$ $\le ($$m$+$n$) since $S$ does not contain duplicates.

I am trying to create an unbiased estimator for the population total that only depends on the records in $S$ in terms of only all $y_i$, $n$, $m$, and $N$.

I tried to mimic the unbiased Horvitz Thompson estimator $\hat{t} = \sum_{iϵS}y_i/\pi_i$, where $\pi_i$ is the probability that $y_i$ is in $S$. Here, I believe $\pi_i$ is the probability of the union of the event that $y_i$ is selected in the first sample of $n$ people and the event that $y_i$ is selected in the second sample of $m$ people. However, I am unsure of how to express the upper limit of the sum without using $n_s$.

Is there a way to mimic the Horvitz Thompson estimator in terms only of all $y_i$, $n$, $m$, and $N$? Or should I be taking a different approach to creating an unbiased estimator?

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    $\begingroup$ If you have all the $y_i$, surely you also know $n_s$? ie just by counting the $y$s. $\endgroup$ – Peter Ellis Jan 8 '14 at 7:14
  • $\begingroup$ To expand on my first comment - it's not clear whether the $y_i$ you have is just the set S and you have just $n_s$ of them, with the duplicates removed; or whether you have all $m+n$ values of $y_i$ including the duplicates. $\endgroup$ – Peter Ellis Jan 8 '14 at 20:45

An option is to derive two independent HT estimates from each sample: $s_1$ and $s_2$:

$\hat{t}_1 = \sum_{i \in s_1} \frac{y_i}{\pi_i} = \frac{N}{n} \sum_{i \in s_1} y_i$,

$\hat{t}_2 = \sum_{i \in s_2} \frac{y_i}{\pi_i} = \frac{N}{m} \sum_{i \in s_2} y_i$.

And then you can use average of both estimates to derive another unbiased estimate of $t$:

$\hat{t} = \frac{1}{2} \left( \hat{t}_1 + \hat{t}_2 \right) = \frac{N}{2} \left( \frac{1}{n} \sum_{i \in s_1} y_i + \frac{1}{m} \sum_{i \in s_2} y_i \right)$.

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    $\begingroup$ This is unbiased but probably isn't efficient if n and m are different because $y_i$ in the smaller subsample are given more weight than those in the bigger subsample. $\endgroup$ – Peter Ellis Jan 8 '14 at 7:05

There is other solution.

You can compute the sampling probability for each individual to be sampled in combined sample $S$ by

$$\pi_i = \frac{n}{N} + \frac{m}{N} - \frac{mn}{N^2} = \frac{n + m - \frac{mn}{N}}{N}$$

You can apply the standard HT-estimator for the population total now:

$$\hat{t} = \sum_{i \in S} \frac{y_i}{\pi_i} = \frac{N}{n + m - \frac{mn}{N}} \sum_{i \in S} y_i$$


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