Question
A charity audits a random selection of expense receipts. Its computer is more likely to include certain transactions than others; each transaction is selected independently, and the probability of selection is recorded even when the receipt itself is not opened. The treasurer needs an estimate of total spending on four transactions, not merely the average among receipts examined.
| Transaction | Inclusion probability | This audit's result |
|---|---|---|
| 1 | \(1/2\) | selected; receipt shows $100 |
| 2 | \(1/4\) | selected; receipt shows $80 |
| 3 | \(3/4\) | not selected; amount unknown |
| 4 | \(1/2\) | not selected; amount unknown |
Treat each transaction's true dollar amount \(y_i\) as fixed, not random. Construct an unbiased estimator of total spending based only on selected receipts and calculate its value for this audit. Derive its design variance as a formula in the four amounts, then give an unbiased estimator of that variance which can be calculated from the same selected receipts. Why does multiplying the audited receipts' ordinary average by four not generally work?
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Hint 1
Let \(I_i\) be 1 if transaction i is audited and 0 otherwise. Divide each observed amount by the probability that it would be observed.
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Solution
Write \(T=\sum_{i=1}^{4}y_i\) for the fixed, unknown total, \(I_i\in\{0,1\}\) for independent selection indicators, and \(\pi_i=P(I_i=1)\). Because the treasurer observes \(y_i\) whenever \(I_i=1\), the inverse-probability-weighted (Horvitz–Thompson) estimator is
\[
\widehat T=\sum_{i=1}^4\frac{I_i y_i}{\pi_i}.
\]
It is design-unbiased: \(E[\widehat T]=\sum_i E[I_i]y_i/\pi_i=\sum_i y_i=T\), where expectation is over the audit draw, not a random model for the spending itself.
Only transactions 1 and 2 were observed, so the realised estimate is
\[
\widehat T=\frac{100}{1/2}+\frac{80}{1/4}
=200+320=\boxed{\$520}.
\]
Independence of the four inclusions gives zero cross-covariance, hence the true design variance for this fixed ledger is
\[
\operatorname{Var}(\widehat T)
=\sum_i\frac{y_i^2}{\pi_i^2}\operatorname{Var}(I_i)
=\boxed{\sum_i\frac{1-\pi_i}{\pi_i}y_i^2}.
\]
The treasurer cannot evaluate that numerical variance exactly without the missing \(y_3,y_4\). But \(E[I_i y_i^2(1-\pi_i)/\pi_i^2]=(1-\pi_i)y_i^2/\pi_i\), so an unbiased estimated variance is
\[
\widehat V=\sum_{i:I_i=1}\frac{1-\pi_i}{\pi_i^2}y_i^2.
\]
For this audit,
\[
\widehat V=\frac{1/2}{(1/2)^2}(100)^2
+\frac{3/4}{(1/4)^2}(80)^2
=20{,}000+76{,}800=\boxed{96{,}800\text{ dollars}^2}.
\]
It is a random estimate of variance, not a claim that the unknown realised ledger's variance must equal 96,800.
The two observed receipts average \(\frac{\$(100+80)}{2}=\$90\), and multiplying by four gives $360. That treats audited receipts as an equally representative sample of all four transactions, which they are not: their inclusion probabilities differ. If larger payments tend to be selected more often, an unweighted audited average especially overrepresents them; even without that trend, unequal inclusion probabilities invalidate its general unbiasedness.
Assumptions
The full ledger consists of four fixed nonnegative transaction amounts, but only the selected amounts are disclosed to the estimator.
The inclusion probabilities (1/2,1/4,3/4,1/2) are strictly positive, known in advance, and each inclusion indicator is independent.
There are no missing or misreported selected receipts and no requirement that at least one transaction be selected.
Variance refers to repeated draws of the audit sample over the same fixed ledger; its unbiased estimator need not be nonrandom.
Final Answer
Horvitz–Thompson total \(\widehat T=\sum_i I_i y_i/\pi_i\), here $520. Design variance \(\sum_i (1-\pi_i)y_i^2/\pi_i\). Unbiased variance estimator \(\widehat V=\sum_{i:I_i=1}(1-\pi_i)y_i^2/\pi_i^2\), here 96,800 dollars².