Normal mean posterior with an exponential prior ($\mu \mid x\sim N(\bar{x}-\lambda {\sigma }^2/n,{\sigma }^2/n)\text{ truncated to }[0,\infty )$)

For independent and identically distributed observations from a normal distribution with known variance ${\sigma }^2$, an exponential distribution prior of rate $\lambda$ on their nonnegative mean gives this truncated normal distribution posterior. Complete the square in $-n(\mu -\bar{x}{)}^2/(2{\sigma }^2)-\lambda \mu$; the truncation remains essential even when the untruncated normal mean is negative.