The sample mean is sufficient for a normal mean

Question

Let X1, \dots, Xn \overset{iid}{\sim} \mathcal{N}(\mu, \sigma^2) with \sigma^2 known and \mu unknown. Show that T = \sumi Xi (equivalently, the sample mean) is sufficient for \mu using the factorization theorem.

Accepted Answer

Fisher-Neyman factorization. A statistic T(X) is sufficient for \mu if and only if the joint density factors as f(x; \mu) = g\big(T(x), \mu\big)\,h(x), where h does not involve \mu and g touches the data only through T. Write the joint density. f(x; \mu) = (2\pi\sigma^2)^{-n/2}\exp\!\left(-\frac{1}{2\sigma^2}\sum{i=1}^n (xi - \mu)^2ight). Expand the exponent to separate the \mu-dependence: \sumi (xi - \mu)^2 = \sumi xi^2 - 2\mu\sumi xi + n\mu^2. So f(x; \mu) = \underbrace{\exp\!\left(-\frac{1}{2\sigma^2}\big(n\mu^2 - 2\mu	extstyle\sumi xi\big)ight)}{g(T,\ \mu),\ \ T = \sumi xi}\ \cdot\ \…

The sample mean is sufficient for a normal mean

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