= Solution
{c}
The <moment-generating function> of a <standard normal distribution> gives $\mathbb E[e^{tX_j}]=e^{t^2/2}$. Using <independence> of the coordinates,
$$
\theta=\mathbb E_p[\phi(X)]=\prod_{j=1}^d\mathbb E[e^{-X_j}]=e^{d/2},\qquad
\mathbb E_p[\phi(X)^2]=\prod_{j=1}^d\mathbb E[e^{-2X_j}]=e^{2d}.
$$
The accepted draws are <independent and identically distributed random variables>, so their <sample mean> is an <unbiased estimator> and has <variance>
$$
\boxed{\operatorname{Var}(\widehat\theta_n^{\mathrm{RS}})=\frac{e^{2d}-e^d}{n}.}
$$
This counts $n$ accepted draws, requiring $n\sigma^d$ proposals in <expectation>.
Back to article page
