Question
A conservator wants to measure how much the true weights of objects vary across a large museum collection. Four objects sampled independently from the collection give scale readings of 99, 100, 100 and 101 grams. Each reading equals that object's true weight plus a fresh, independent scale error. The errors have mean zero and a known variance of 1 g² and are independent of the true weights. Treat true object weights as independent draws with a common, unknown variance \(\sigma^2\).
Using the ordinary sample variance (dividing by \(n-1\)), what is the unbiased algebraic estimate of \(\sigma^2\)? What nonnegative estimate could the conservator report instead? Explain why that truncation cannot generally remain unbiased, even though a negative true variance is impossible.
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Hint 1
Write reading \(Y=\text{true weight }X+\text{independent error }\varepsilon\). How does \(\operatorname{Var}(Y)\) relate to \(\operatorname{Var}(X)\) and \(\operatorname{Var}(\varepsilon)\)?
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Solution
For independent weights \(X_i\) and scale errors \(\varepsilon_i\), readings \(Y_i=X_i+\varepsilon_i\) have \(\operatorname{Var}(Y_i)=\sigma^2+1\) g². The sample variance \(S_y^2=(1/(n-1))\sum(Y_i-\bar Y)^2\) is unbiased for \(\operatorname{Var}(Y_i)\). Consequently \(U=S_y^2-1\) g² has \(\mathbb{E}[U]=\sigma^2\), even though individual realized values of \(U\) may be negative.
Here \(\bar Y=100\) g, \(n=4\), and \(\sum(Y_i-\bar Y)^2=1+0+0+1=2\) g². Thus \(S_y^2=2/3\) g² and \(U=-1/3\) g². If the estimate is required to lie in the physical parameter space \(\sigma^2\ge0\), a simple constrained alternative is \(\tilde U=\max(0,U)=0\) g². This truncation is not the same estimator and is not generally unbiased, since \(\tilde U-U=\max(0,-U)\ge0\), with strictly positive expectation whenever \(P(U<0)>0\).
For an exact illustration, suppose all true weights equal 100 g \((\sigma^2=0)\), take just two readings, and independently let each noise error be \(+1\) or \(-1\) g with equal probability. With probability 1/2 the signs match, so \(S_y^2=0\) and \(U=-1\); with probability 1/2 they differ, so \(S_y^2=2\) and \(U=+1\). Hence \(\mathbb{E}[U]=0\), while \(\mathbb{E}[\max(0,U)]=(0+1)/2=1/2\) g²: a positive bias despite zero true variation. A negative raw correction signals sampling uncertainty, not a physically negative variance.
Assumptions
The four measured objects are independent draws from one population and the measurement errors are independent across objects.
The noise variance of 1 g² is known externally rather than estimated from these same four observations.
The corrected estimator is for population variance, not the average squared error of an individual object's estimate.
Final Answer
The unbiased variance-corrected estimate is \(-1/3\) g²; the nonnegative truncated estimate is \(0\) g². Truncation avoids an inadmissible reported variance but introduces upward bias in general.