Past exam of the options I course of the University of Oxford / 2010 / O / O2

Full column rank gives the least-squares estimatorWrite $A=(X^{T}X{)}^{-1}{X}^T$. Since $AX=I$ and $\mathrm{Cov}(ϵ)={\sigma }^{2}I$, we have $\hat{\theta }=\theta +Aϵ$ andThis calculation needs the stated uncorrelated equal-variance errors, but not normality.

Put $z_i=e^{-t_i}-c$ and take independent measurement errors ${ϵ}_i\sim N(0,{\sigma }^2)$, independent across the prescribed measurement times. ThenAssume the errors have zero mean so the stated activity curve is the observation mean. Also assume at least two distinct times, making the two design columns linearly independent. The times and the centering constant $c$ are treated as known.

By the definition of $c$, ${\sum }_i{z}_i=0$. Put $S={\sum }_i{z}_i^2>0$. ThenPart (a) therefore givesIn particular the two estimates are uncorrelated, and under the assumed normal-error model they are independent.

The expected activity is $E_t=\alpha +\beta (e^{-t}-c)$. Under normal errors, least squares is maximum likelihood, and invariance of maximum-likelihood estimation givesThe covariance term vanishes by part (ii), soThis is uncertainty in the estimated mean activity; the variance of a future individual observation would include its additional measurement-error variance.