 Solution

Let $\varphi$ denote the standard normal density. Under the shifted sampling density $q(y)=\varphi (y-c)$, the likelihood ratio is $\varphi (y)/q(y)=e^{-cy+c^2/2}$. Simulate independent $Y_k\sim N(c,1)$ and useMultiplying the weight by $q$ restores the original density, so each summand $H$ has expectation ${\int }_c^{\infty }\varphi (y)dy=z_c$. The estimator is unbiased. For the second moment,because $y\ge c$. Since $n{v}_c={\mathbb{E}}_q{H}^2-z_c^2$,For $c>0$, $e^{-c^2/2}<1$ and $z_c>0$, so $n{v}_c<z_c-z_c^2=n{v}_0$, proving $v_c<v_0$. Nonnegativity of variance also implies $z_c^2\le e^{-c^2/2}{z}_c$; division by the positive $z_c$ gives