Question
A city must estimate attendance at an upcoming outdoor festival. Two surveys of partly overlapping visitors separately estimate the same unknown total \(T\), and each estimate is unbiased. Their reported estimates, in thousands of people, are \(X=40\) and \(Y=38\).
From the survey design, the city knows the error variances in squared thousands are \(\mathrm{Var}(X)=9\) and \(\mathrm{Var}(Y)=4\), while their covariance is \(\mathrm{Cov}(X,Y)=5\). The positive covariance reflects some shared respondent-related errors. The manager wants a combined estimator \(\widehat T=wX+(1-w)Y\) that remains unbiased, and initially allows any real weight \(w\), including negative values.
Find the minimum-variance weight, the resulting estimate from these two reports, and its variance. Would the answer change if the manager required both weights to lie between zero and one? Why might equal weighting be inefficient here?
Get hint
Hint 1
Because both surveys are unbiased for the same target, weights summing to one preserve unbiasedness even if a weight is negative.
Check your answer
Solution
For any real \(w\), \(E[wX+(1-w)Y]=wT+(1-w)T=T\), so the estimator stays unbiased. The error variance, including the surveys' covariance, is
\[
V(w)=9w^2+4(1-w)^2+2(5)w(1-w)
=3w^2+2w+4.
\]
Its second derivative is 6, which is positive, so the single stationary point is a global minimum. Setting \(V'(w)=6w+2=0\) gives
\[
\boxed{w^*=-\tfrac13},\qquad 1-w^*=\tfrac43.
\]
The combined attendance estimate is \(-40/3+4(38)/3=\boxed{112/3}\) thousand, about 37,333 people. Its minimum variance is
\[
V(-1/3)=3/9-2/3+4=\boxed{11/3}\ (\text{thousand people})^2.
\]
A negative weight is admissible because the task imposes unbiasedness, not convex averaging. It can subtract some of the correlated error shared with the other survey. The covariance matrix \(\begin{pmatrix}9&5\\5&4\end{pmatrix}\) is valid (determinant \(36-25=11>0\)), so the gain is not an artefact of impossible error parameters.
If weights must satisfy \(0\le w\le1\), the variance function is increasing throughout that interval (since \(V'(w)=6w+2>0\)). The constrained minimum is \(\boxed{w=0}\), using Y alone with variance 4. Equal weighting would give \(V(1/2)=3/4+1+4=\boxed{23/4}\), worse than either optimum. The general unrestricted formula, \(w^*=(\sigma_Y^2-\sigma_{XY})/(\sigma_X^2+\sigma_Y^2-2\sigma_{XY})\), becomes the familiar inverse-variance weight when covariance is zero; if two estimates have equal variance and less-than-perfect correlation, symmetry gives weights 1/2. Shared respondents can invalidate that simple equal-weighting intuition.
Assumptions
Both estimators target the same fixed total T and are unbiased.
Error variances and covariance apply to the same scale of thousands of people.
Unrestricted weights are any real numbers summing to one; the optional constrained policy restricts each to \([0,1]\).
Final Answer
Unrestricted optimum: \(w=-\frac13\) on X and \(\frac43\) on Y, estimate \(\frac{112}3\) thousand people, variance \(\frac{11}3\,(\text{thousand people})^{2}\). With nonnegative weights, use Y only (\(w=0\)), variance 4.