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A two-dimensional Hotelling statistic and its exact F cutoff

Example

Assume AC and use the iid multivariate-normal sample model with p=2, n=4, and fixed positive-definite covariance Σ. For the observation x1=(1,0)T, x2=(−1,0)T, x3=(0,1)T, and x4=(0,−1)T, test H0:μ=μ0 with μ0=(1,1)T. The sample mean is zero, the sample covariance is (2/3)I2, the Hotelling statistic is T2=12, and its scaled F2,2 statistic is 4. For a chosen 0<α<1, the exact test rejects precisely when 4>F2,2−1(1−α).

Facts & Assumptions

Given: AC, the iid N2(μ,Σ) sample model with fixed Σ≻0, the four displayed observed vectors, n=4, p=2, μ0=(1,1)T, and 0<α<1.

[F1]

The sample covariance is S=∑i(xi−xˉ)(xi−xˉ)T/(n−1). The iid multivariate-normal sample and covariance conventions

[F2]

Under AC, in the iid multivariate-normal model with Σ≻0 and n>p, T2=n(Xˉ−μ0)TS−1(Xˉ−μ0) when S≻0 and is zero otherwise. Under μ=μ0, (n−p)T2/(p(n−1)) has law Fp,n−p. The one-sample Hotelling statistic has an exact F law

[F3]

Under the same model assumptions and 0<α<1, the test rejects when the scaled statistic exceeds the 1−α quantile and has exact null size α over all Σ≻0. The one-sample Hotelling F test has exact size alpha

[F4]

Fr,s−1(u) is the unique u-quantile of the F distribution for positive degrees of freedom. Student t and Fisher F laws Student t and Fisher F CDFs are continuous and strictly increasing on their supports

Verification

1.1givenF1algebra

The four vectors sum to zero, so xˉ=0. Their outer products sum to 2I2, hence R=2I2 and S=R/3=(2/3)I2 by [F1]. This matrix is positive definite and has inverse (3/2)I2.

2.1F2step 1.1algebra

The difference xˉ−μ0=(−1,−1)T has squared Euclidean norm 2. Therefore [F2] gives T2=4(3/2)⋅2=12, and the scale (n−p)/(p(n−1))=2/(2⋅3)=1/3 makes the F statistic equal 4.

3.1F3F4step 2.1∎

By [F3]–[F4], applying the exact test rule to this observation is exactly the stated comparison 4>F2,2−1(1−α). The example reports the rule as a function of α and does not infer a rejection decision without choosing α.

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