 Solution

The exponential distribution prior has rate $\lambda =1/3$ and support $\mu \ge 0$, so its prior density is $(1/3)e^{-\mu /3}{1}_{\{\mu \ge 0\}}$. By Bayes' theorem,where $C$ is a normalizing constant independent of $\mu$.

Completing the square gives a useful exact reference for the numerical sampling. PutThen this normal mean posterior with an exponential prior is a normal distribution with location $m$ and variance $s^2$, truncated below at zero. With $\varphi$ the standard normal distribution density and $\Phi$ its cumulative distribution function,The exponential rate is $1/3$, not three; the prior mean is three. The support restriction must be retained in both the formula and the sampler.