Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 29K / c

Solution

Let

R = (I_{k} 0) \in R^{k \times p}, r = R θ^{*} .

R θ_{0} = r

W_{n, R} = n (R θ_{n} - r)^{T} [R I (θ_{n})^{- 1} R^{T}]^{- 1} (R θ_{n} - r) .

Under

H_{0}

n (R θ_{n} - r) = n R (θ_{n} - θ_{0}) d N_{k} (0, V_{0}),

where

V_{0} = R I (θ_{0})^{- 1} R^{T} .

V_{0}

is positive definite because

I (θ_{0})

is positive definite and

R

has full row rank. Consistency gives

R I (θ_{n})^{- 1} R^{T} p V_{0} .

Therefore Slutsky theorem yields

V_{0}^{- 1/2} n (R θ_{n} - r) d N_{k} (0, I_{k}),

and applying the continuous squared-norm map proves rigorously that

W_{n, R} d χ_{k}^{2} .

Let

q_{k, 1 - α}

be the

(1 - α)

quantile of

χ_{k}^{2}

. Rejecting

H_{0}

when

W_{n, R} > q_{k, 1 - α}

has rejection probability tending to

α

under every fixed parameter satisfying

H_{0}

, and is therefore an asymptotically valid level-

α

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