= Solution
On $x>0$, the exponential proposal is the proper density $g(x)=e^{-x}\mathbf1_{\{x>0\}}$; its integral is $1$. The word “improper” in the question is therefore not applicable on its stated positive support.
The target-to-proposal ratio is
$$
\frac{f(x)}{g(x)}=\sqrt{\frac2\pi}\exp(-x^2/2+x)
=\sqrt{\frac{2e}{\pi}}\exp\left[-\frac{(x-1)^2}{2}\right].
$$
Its maximum occurs at $x=1$, so the smallest valid envelope constant is
$$
\boxed{M=\sqrt{2e/\pi},\qquad P(\mathrm{accept})=1/M=\sqrt{\pi/(2e)}\approx0.76017>0.75.}
$$
One implementation draws $Z=-\log U_1$ and accepts it when $U_2\leq e^{-(Z-1)^2/2}$, for <independent random variables> with the <uniform distribution> on $(0,1)$. Accepted observations have the positive half-normal density by the usual <rejection sampling> calculation.
Back to article page
