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Assume there are two sticks with lengths \(l1\) and \(l2\). You have an instrument that can measure the length of a stick with an error \(e \in N(0,\sigma)\). Now assume that your budget constrains you to use the instrument only twice. What is the best way to use your budget in order to get the most accurate measurements?

The tricky part here is to realize that your goal is to minimize the variance of the measurement. The naive approach would be to measure \(l1\) and then \(l2\). In that case:

\[ \begin{align*} \hat{l1} &= l1 + e_1 \\ \hat{l2} &= l2 + e_2 \\ Var(\hat{l1}) &= Var(e_1)= \sigma^2 \\ Var(\hat{l2}) &= Var(e_2)= \sigma^2 \end{align*} \]

A better way to do this would be to measure the sum and the differences of the two:

\[ \begin{align*} \hat{m1} &= l1 + l2 + e_1 \\ \hat{m2} &= l2 - l1 + e_2 \\ \hat{l1} &= \frac{1}{2}(\hat{m1} - \hat{m2}) = l1 + \frac{1}{2}(e_1-e_2) \\ \hat{l2} &= \frac{1}{2}(\hat{m1} + \hat{m2}) = l2 + \frac{1}{2}(e_1+e_2) \\ Var(\hat{l1}) &= \frac{1}{4} [Var(e_1) + Var(e_2)] = \frac{1}{2}\sigma^2 \\ Var(\hat{l2}) &= \frac{1}{4} [Var(e_1) + Var(e_2)] = \frac{1}{2}\sigma^2 \end{align*} \]

In the above, the covariance of the two errors is zero since they are independent measurements.

Variance

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