= Solution
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> estimates
$$
\boxed{\hat\mu=M,\qquad\hat\lambda=n/S,\qquad
\hat\beta=M+S/n=\overline x.}
$$
These finite estimates exist almost surely for $n\ge2$. For one observation, or an all-equal sample, the likelihood is unbounded as $\lambda\to\infty$.
The minimum has $M-\mu\sim\operatorname{Exp}(n\lambda)$, so $E\hat\mu=\mu+1/(n\lambda)$: its bias is positive. By exponential memorylessness, $S\sim\operatorname{Gamma}(n-1,\text{rate }\lambda)$. For $n>2$, integration of the gamma density gives $E(1/S)=\lambda/(n-2)$, and hence
$$
\boxed{E\hat\lambda=\frac{n\lambda}{n-2},\quad
\operatorname{bias}(\hat\lambda)=\frac{2\lambda}{n-2}>0.}
$$
At $n=2$ this inverse moment is infinite, so a finite bias does not exist. Finally $E\overline X=\mu+1/\lambda=\beta$, so \b[$\hat\beta$ is unbiased]. The positivity assertions are therefore subject to the sample-size qualifications above.
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