Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 32 / 3 / b / ii / Solution

For $U\sim N(\mu ,s^2)$ and $a=(T-\mu )/s$, substitution $U=\mu +sZ$ givessince ${\varphi }^{\prime }(z)=-z\varphi (z)$. Divide by $1-\Phi (a)$ to obtain the upper-truncated normal mean.

For $U={\hat{\delta }}_1$, set $\mu =\delta$, $s=1/\sqrt{I_1}$ and $T=f_1/\sqrt{I_1}$. Then $a=f_1-\delta \sqrt{I_1}$. Symmetry of the normal density and distribution yieldsThe ratio $r(x)=\varphi (x)/\Phi (x)$ is an Inverse Mills ratio. It is positive, and the pooled estimator's conditional bias is thereforeThis identity supplies both correction procedures below.