Derive exact normal-model inference

Try

You observe iid normal measurements with unknown mean μ and positive variance σ². The standardized mean Z=√n(X̄−μ)/σ is exactly standard normal, but σ is unknown. Replacing σ by the sample standard deviation S changes the distribution. The result is a t pivot, not still a standard normal. To understand why, separate the data vector into its average direction and the perpendicular residual directions. Their independence is the mechanism behind the exact ratio distribution.

Predict

Predict how many residual directions remain after estimating one mean from n observations. The residuals sum to zero, so they lie in an (n−1)-dimensional subspace. Their squared length, after division by σ², will have a chi-square law with n−1 degrees of freedom under normality. The count is geometric. It is not a general rule that every estimated quantity always costs one degree of freedom in every model; this result follows from this particular linear projection.

Experiment

For three standardized observations, use an orthonormal coordinate system. The first vector captures the mean; the other two span the residual plane.

Coordinate direction	Unit vector	Role
Average	(1,1,1)/√3	Standardized mean
Contrast one	(1,−1,0)/√2	Residual coordinate
Contrast two	(1,1,−2)/√6	Residual coordinate

Check every dot product and squared norm. An orthogonal transformation of a standard normal vector is still standard normal, with independent coordinates. Thus the square of the first coordinate and the sum of squares of the other two use independent pieces of the random vector. Repeat the dimension argument for n observations without needing to construct every basis vector explicitly.

Understand

Define S²=Σ(Xᵢ−X̄)²/(n−1), with n≥2. Orthogonal normal decomposition proves V=(n−1)S²/σ² has a chi-square distribution with n−1 degrees of freedom and is independent of Z. By the definition of the t law, Z divided by √(V/(n−1)) has t distribution with n−1 degrees of freedom. Simplifying the ratio gives √n(X̄−μ)/S. A pivot is a parameter-dependent quantity whose sampling distribution does not depend on the unknown parameter; it need not itself be a statistic.

The mean interval follows by placing this pivot between symmetric t quantiles and solving for μ. The variance interval follows from chi-square quantiles. If qₐ denotes the a quantile of χ²ₙ₋₁, then q_(α/2)≤(n−1)S²/σ²≤q_(1−α/2) inverts to lower endpoint (n−1)S²/q_(1−α/2) and upper endpoint (n−1)S²/q_(α/2). The reversed denominator order is essential.

Apply

For a synthetic sample of five measurements with X̄=12 and S=3, write the two-sided mean interval as 12±t_(4,1−α/2)·3/√5. Keep the quantile symbolic unless its numerical source is supplied. The expression is exact under iid normal sampling, not a numerical approximation created by rounding a quantile. For variance, the numerator is 4·9=36. Write both chi-square endpoints and explain why the interval is usually asymmetric.

Now replace normal observations by an arbitrary skewed distribution of finite variance. The residual coordinates can be uncorrelated without being independent, and the residual sum of squares need not be chi-square. A large-sample mean approximation may still be useful under suitable conditions, but it is a different theorem. The projection identity alone does not establish an exact t law.

Continue

Distinguish the data-computable standard error from the theoretical pivot that contains μ or σ². Inference works by inverting a known distributional statement, not by pretending unknown parameters were observed. Continue to Neyman–Pearson testing to derive tests from an optimization criterion. Later multivariate projection lessons reuse this same geometry for regression and analysis of variance, preserving the exact normal-model assumptions.