A sufficient statistic for a normal mean

Question

You have independent observations X1, \dots, Xn from a normal distribution with unknown mean \mu but known variance \sigma^2. Using the factorization idea, find a sufficient statistic for \mu.

Accepted Answer

The plan. Split the joint density into a piece that involves \mu (touching the data only through some summary) and a piece with no \mu in it. The summary in the first piece is sufficient. Expand the exponent. The joint density is proportional to \exp\!\Big(-\frac{1}{2\sigma^2}\sumi (xi - \mu)^2\Big). Expand the square: \sumi (xi - \mu)^2 = \sumi xi^2 - 2\mu \sumi xi + n\mu^2. Only the middle term mixes \mu with the data, and it touches the data solely through \sumi xi. So the density factors as f(x; \mu) = \underbrace{\exp\!\Big(\frac{\mu}{\sigma^2}\sumi xi - \frac{n\mu^2}{2\sigma^2}\Big)}{g(…

A sufficient statistic for a normal mean

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