Given the sample mean, the scatter forgets the center

Question

You have independent observations X1, \dots, Xn from a normal with unknown mean \mu and known variance \sigma^2, and you condition on the sample mean \bar X. Explain why the residuals Xi - \bar X have a distribution that does not depend on \mu, and connect this to the sample mean being a sufficient statistic for \mu.

Accepted Answer

Split each point into center plus scatter. Write every observation as the sample mean plus a residual: Xi = \bar X + (Xi - \bar X). The sample mean \bar X is where the "location" of the data lives; the residuals Xi - \bar X describe only the shape of the scatter around that location. Why the residuals forget \mu. Shifting the true mean \mu shifts every observation by the same amount, and therefore shifts \bar X by that same amount too. But the residuals are differences, so the common shift cancels: Xi - \bar X is unchanged when \mu moves. Their joint distribution is centered at zero with a sp…

Given the sample mean, the scatter forgets the center

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