Past exam of the probability, statistics and options I course of the University of Oxford / 2008 / S / 1 / d / Solution

For the shifted normal proposal, $w(x)=e^{-\mu x+{\mu }^2/2}$ and the estimator isLet $\bar{\Phi }=1-\Phi$. Completing the square gives its unaveraged second momentThe variance is $[I(\mu )-\bar{\Phi }(z{)}^2]/N$, so only $I$ needs minimization. It tends to infinity at both ends of the real $\mu$ axis and has a finite minimizer. Differentiation at a minimizer yieldsFor a standard normal variable, integration of $x\phi (x)=-{\phi }^{\prime }(x)$ shows $E_p[X\mid X>a]=\phi (a)/\bar{\Phi }(a)$. ConsequentlyThe shift is chosen to reduce the second moment of the weighted rare-event indicator, rather than merely to increase the proposal's tail probability.