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The likelihood function for the shifted exponential distribution isIt increases strictly up to the smallest observation and is zero beyond it. Thus shifted exponential maximum likelihood givesFor $t\ge 0$, independence and the survival function of the exponential distribution give the exact order statistic lawIn particular, ${\hat{\theta }}_n\ge \theta$ almost surely and ${\mathbb{P}}_{\theta }(|{\hat{\theta }}_n-\theta |>\epsilon )=e^{-n\epsilon }\to 0$. This proves statistical consistency. The exact endpoint limit for a shifted exponential distribution is stronger than an asymptotic approximation:The limit is one-sided and nonnormal; the moving support endpoint explains why a regular square-root-$n$ asymptotic normality of a maximum likelihood estimator theorem does not apply here.

Let $M={\min }_i{x}_i$ and $S={\sum }_i(x_i-M)$. On the likelihood support $0<\mu \le M$, the log likelihood is $n\log \lambda -\lambda {\sum }_i(x_i-\mu )$. For fixed $\lambda$ it increases with $\mu$, so $\hat{\mu }=M$. Maximizing in $\lambda$ then gives the shifted exponential maximum likelihood estimatesThese finite estimates exist almost surely for $n\ge 2$. For one observation, or an all-equal sample, the likelihood is unbounded as $\lambda \to \infty$.

The minimum has $M-\mu \sim \mathrm{Exp}(n\lambda )$, so $E\hat{\mu }=\mu +1/(n\lambda )$: its bias is positive. By exponential memorylessness, $S\sim \mathrm{Gamma}(n-1,\text{rate }\lambda )$. For $n>2$, integration of the gamma density gives $E(1/S)=\lambda /(n-2)$, and henceAt $n=2$ this inverse moment is infinite, so a finite bias does not exist. Finally $E\overline{X}=\mu +1/\lambda =\beta$, so $\hat{\beta }$ is unbiased. The positivity assertions are therefore subject to the sample-size qualifications above.