Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 1 / 17H (Statistics)

With $P_0=X_0(X_0^T{X}_0{)}^{-1}{X}_0^T$, ${\hat{\beta }}_0=(X_0^T{X}_0{)}^{-1}{X}_0^{T}Y$ and ${\hat{\sigma }}^2=\Vert (I-P_0)Y{\Vert }^2/n$.

The statistic is $F=[(RS{S}_0-RS{S}_1)/p_1]/[RS{S}_1/(n-p)]$, with an $F_{p_1,n-p}$ null distribution.

Under the null, $R=(I-P_0)\epsilon$, so $R/\sigma \stackrel{d}{=}(I-P_0)Z$ for $Z\sim N(0,I)$.

The displayed linear form is centred normal with variance $n^{-1}{X}_{1j}^T(I-P_0)X_{1j}\le n^{-1}\Vert X_{1j}{\Vert }^2=1$. The normal tail bound proves the claim.

By (iv) and the union bound, $P(T\ge t)\le p_1{e}^{-t^2/2}$. Setting $t=\sqrt{2\log (p_1/\alpha )}$ bounds the size by $\alpha$.

Scale cancels inBecause $R=\sigma (I-P_0)Z$ under the null, it has exactly the requested distribution; its quantiles can be simulated without knowing $\sigma$.