 Solution

On $x>0$, the exponential proposal is the proper density $g(x)=e^{-x}{1}_{\{x>0\}}$; its integral is $1$. The word “improper” in the question is therefore not applicable on its stated positive support.

The target-to-proposal ratio isIts maximum occurs at $x=1$, so the smallest valid envelope constant isOne implementation draws $Z=-\log U_1$ and accepts it when $U_2\le e^{-(Z-1{)}^2/2}$, 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.