Past exam of the mathematics course of the University of Cambridge 2015 ii Paper 4 10J Solution Created 2026-09-24 Updated 2026-10-06
Write . Minimizing the residual sum of squares first over the coefficients of reduces the problem to minimizing . Full column rank ensures . Thus the partial regression formula and its variance areThis uses and the isotropic error covariance matrix .
Since the span of is contained in the column space of , orthogonal projection onto the latter removes at least as much squared norm. Consequentlyand inversion proves the requested lower bound. Full rank makes the denominator positive.
Adding to column leaves the column space unchanged. Every old fitted vector is reproduced by keeping all slopes unchanged and replacing the regression intercept by . Uniqueness of the ordinary least squares estimators then proves that slope is unchanged. In particular all columns can be centered by such changes. Applying the previous projection argument to the centered design proves the bound with and ; the normalized inner product is now the usual sample correlation.
Each reported coefficient test is the two-sided Student's t-test of against , conditional on the other predictors. Under the null, has a Student's t-distribution with degrees of freedom. At five percent the intercept is significant, but neither individual test slope is. The final F-test compares the full model against an intercept-only model: . It has degrees of freedom and rejects at five percent, since its -value is . Thus the predictors jointly explain variation, without strong evidence separating their individual conditional effects.
The two test predictors have sample correlation , although each has only moderate correlation with the final exam response. The variance inflation factor is . Their nearly shared direction produces substantial multicollinearity, inflating the individual slope standard errors. The joint F-test can detect this common predictive direction even though either coefficient, given the other, has a large -value.