= Solution
{c}
Write $s=\sigma^2\geq1$. The two <multivariate normal distribution> <probability density functions> have ratio
$$
w(x)=\frac{p(x)}{q(x)}=s^{d/2}\exp\!\left[-\frac12(1-s^{-1})\|x\|^2\right].
$$
Here $\|x\|$ is the <Euclidean norm>. Consequently the smallest <rejection sampling> envelope constant, and its acceptance <probability>, are
$$
\boxed{M_*=\sup_xw(x)=s^{d/2}=\sigma^d,\qquad \mathbb P(\mathrm{accept})=s^{-d/2}.}
$$
The supremum is attained at $x=0$; hence no smaller constant can dominate the ratio. Draw $Y\sim N(0,sI_d)$ and an <independent> $U$ from the <uniform distribution> on $(0,1)$. Accept $Y$ when
$$
\boxed{U\leq\exp\!\left[-\frac12(1-s^{-1})\|Y\|^2\right];}
$$
otherwise repeat with fresh <independent random variables>. One can generate $Y=\sqrt s\,G$ using a <random vector> $G$ with <independent> <standard normal distribution> coordinates.
For any <measurable set> $B$, the joint acceptance <probability> is $\int_Bq(y)p(y)/(M_*q(y))\,dy=M_*^{-1}\int_Bp(y)\,dy$. Dividing by $1/M_*$ proves that the returned <random vector> has <probability density function> $p$. The number of proposals has a <geometric distribution> on the positive integers, with <expected value> $M_*$. When $s=1$, every proposal is accepted.
Back to article page
