When the true mean is known, which divisor is right?

Question

A sensor is calibrated so that its true long-run mean reading is a known target $\mu$ (say the machine is set to exactly $100$ grams). You take $n$ readings $X_1, \dots, X_n$ and want to estimate the noise variance $\sigma^2$ .

Because $\mu$ is known rather than estimated, what should you divide $\sum_i (X_i - \mu)^2$ by to get an unbiased estimate of $\sigma^2$ ?

Accepted Answer

No degree of freedom is spent. When the mean is known, you center on the true \mu instead of the data-driven \bar X. Each squared deviation now has the full expected size: \mathbb{E}[(Xi - \mu)^2] = \operatorname{Var}(Xi) = \sigma^2. Summing over n independent readings, \mathbb{E}\!\left[\sumi (Xi - \mu)^2\right] = n\sigma^2. The estimator. Divide by n, not n-1: \hat\sigma^2 = \frac{1}{n}\sumi (Xi - \mu)^2, \qquad \mathbb{E}[\hat\sigma^2] = \sigma^2. Why the difference. Nothing was estimated from the data before computing deviations, so no degree of freedom is used up, all n deviations are fr…

When the true mean is known, which divisor is right?

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