Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 28K / c

Solution

θ

W_{n} (θ) = n (\overline{X} - θ)^{T} i_{n} (\overline{X} - θ) .

Because

i_{n} \to I_{p}

and

n (\overline{X} - θ_{0}) \sim N_{p} (0, I_{p}),

W_{n} (θ_{0}) d χ_{p}^{2} .

q_{p, 1 - α}

is the

(1 - α)

p

(1 - α)

confidence region is

{θ : n (\overline{X} - θ)^{T} i_{n} (\overline{X} - θ) \leq q_{p, 1 - α}} .

In this known-identity-covariance model one may replace

i_{n}

I_{p}

; then the confidence ball is exact. For an individual coordinate, the corresponding exact confidence interval is

\overline{X}_{j} \pm z_{1 - α /2} / n .

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