 Solution

For normal rejection sampling with an exponential envelope, complete the square in the probability density function ratio:The maximum occurs at $x=\lambda$, which lies in the support because $\lambda >0$. ThusThe PDF exponent is $e^{{\lambda }^2}$ inside the square root; this distinction is lost in the converted TeX.

The acceptance rule simplifies toAn exponential proposal can be generated as $Y=-{\lambda }^{-1}\log V$ from an independent uniform distribution value $V$. Since the target is normalized, the overall acceptance probability is $1/M(\lambda )$. To maximize it,Therefore the optimal rate is $\lambda =1$, with acceptance probability $\sqrt{\pi /(2e)}$.

Finally, give each accepted half-normal distribution value $Y$ an independent fair sign $R\in \{-1,1\}$ and output $Z=RY$. On each half-line its probability density function is half the reflected half-normal distribution probability density function:Consequently $Z\sim N(0,1)$. The sign must be drawn independently; rejection itself does not supply it.