= Solution
For $U\sim N(\mu,s^2)$ and $a=(T-\mu)/s$, substitution $U=\mu+sZ$ gives
$$
\mathbb E[U\mathbf1_{\{U\geq T\}}]
=\mu[1-\Phi(a)]+s\int_a^\infty z\phi(z)\,dz
=\mu[1-\Phi(a)]+s\phi(a),
$$
since $\phi'(z)=-z\phi(z)$. Divide by $1-\Phi(a)$ to obtain the upper-truncated normal mean.
For $U=\widehat\delta_1$, 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 yields
$$
\boxed{\mathbb E_\delta(\widehat\delta_1\mid C)
=\delta+\frac1{\sqrt{I_1}}
\frac{\phi(\delta\sqrt{I_1}-f_1)}{\Phi(\delta\sqrt{I_1}-f_1)}.}
$$
The ratio $r(x)=\phi(x)/\Phi(x)$ is an <Inverse Mills ratio>. It is positive, and the pooled <estimator>'s conditional bias is therefore
$$
b(\delta)=\frac{n_1}{n_2\sqrt{I_1}}r(\delta\sqrt{I_1}-f_1).
$$
This identity supplies both correction procedures below.
Back to article page
