= Solution
For an exponential proposal of rate $\lambda>0$,
$$
h_\lambda(x)=\lambda e^{-\lambda x}\mathbf1_{x\geq0},
$$
and the smallest valid rejection constant is
$$
\begin{aligned}
M(\lambda)
&=\sup_{x\geq0}\frac{f(x)}{h_\lambda(x)}\\
&=\frac1\lambda\sqrt{\frac2\pi}
\sup_{x\geq0}\exp\left(-\frac{x^2}{2}+\lambda x\right)\\
&=\sqrt{\frac2\pi}\,\frac{e^{\lambda^2/2}}{\lambda},
\end{aligned}
$$
because the exponent is maximized at $x=\lambda$. Differentiating its logarithm gives
$$
\frac d{d\lambda}\log M(\lambda)
=\lambda-\frac1\lambda.
$$
The unique minimum occurs at $\lambda=1$. Since rejection efficiency is $1/M(\lambda)$, choosing any $\lambda\ne1$ makes the algorithm less efficient:
$$
\boxed{\text{the rate-one proposal is optimal within the exponential family}.}
$$
Back to article page
