Past exam of the mathematical and Computational Biology course of the University of Cambridge / 2024 / Paper 1 / B3

Choose the diagonal matrix $W=\mathrm{diag}(w_1,\dots ,w_m)$. Multiplying a vector by $W$ scales its $i$th component by $w_i$. ThereforeThe Euclidean norm objective for ordinary linear least squares on $(WA,Wb)$ is exactly the weighted least squares objective, so the sets of minimizing vectors coincide. Zero weights simply discard the corresponding residuals.

Apply the least-squares normal equations to the transformed matrix and vector:or equivalentlyHere $W^{T}W=\mathrm{diag}(w_1^2,\dots ,w_m^2)$. A unique minimizing vector exists when the weighted design matrix has full column rank; the equations themselves remain valid without that assumption.

Using the paper's named weights $w_i$, takeThe least-squares normal equations for weighted least squares becomeMultiplying by four gives $14{\hat{\beta }}_0+33{\hat{\beta }}_1=59$ and $33{\hat{\beta }}_0+97{\hat{\beta }}_1=143$. Their determinant is $269$, givingThus, to two significant digits, $\boxed{{\hat{\beta }}_0=3.7,{\hat{\beta }}_1=0.20}$, with the slope measured in expression units per hour.

There is a wording convention to make explicit. Because the stated objective squares $w_i$, the choice above gives the final two squared residuals one-quarter of the influence of the others. If “half as much” instead means half the coefficient of each squared residual, use $W=\mathrm{diag}(1,1,1,1/\sqrt{2},1/\sqrt{2})$. Then the least-squares normal equations aregiving ${\hat{\beta }}_0=150/37\approx \boxed{4.1}$ and ${\hat{\beta }}_1=1/37\approx \boxed{0.027}$. Stating which quantity is halved resolves the ambiguity.