= Solution
Let $\phi$ denote the standard normal density. Under the shifted sampling density $q(y)=\phi(y-c)$, the <likelihood ratio> is $\phi(y)/q(y)=e^{-cy+c^2/2}$. Simulate independent $Y_k\sim N(c,1)$ and use
$$
\boxed{\widehat z_c=\frac1n\sum_{k=1}^n\mathbf1_{\{Y_k>c\}}e^{-cY_k+c^2/2}.}
$$
Multiplying the weight by $q$ restores the original density, so each summand $H$ has <expectation> $\int_c^\infty\phi(y)\,dy=z_c$. The estimator is unbiased. For the second moment,
$$
\mathbb E_qH^2=\int_c^\infty e^{-cy+c^2/2}\phi(y)\,dy
\leq e^{-c^2/2}z_c,
$$
because $y\geq c$. Since $nv_c=\mathbb E_qH^2-z_c^2$,
$$
\boxed{nv_c+z_c^2\leq e^{-c^2/2}z_c.}
$$
For $c>0$, $e^{-c^2/2}<1$ and $z_c>0$, so $nv_c<z_c-z_c^2=nv_0$, proving $v_c<v_0$. Nonnegativity of <variance> also implies $z_c^2\leq e^{-c^2/2}z_c$; division by the positive $z_c$ gives
$$
\boxed{z_c\leq e^{-c^2/2}.}
$$
Back to article page
