= Solution
The <likelihood function> for the shifted <exponential distribution> is
$$
L_n(\theta)=\exp\left(n\theta-\sum_{i=1}^nY_i\right)\mathbf1\{\theta\le Y_{(1)}\},\qquad Y_{(1)}=\min_iY_i.
$$
It increases strictly up to the smallest observation and is zero beyond it. Thus <shifted exponential maximum likelihood> gives
$$
\boxed{\widehat\theta_n=Y_{(1)}.}
$$
For $t\ge0$, <independence> and the survival function of the <exponential distribution> give the exact <order statistic> law
$$
\mathbb P_\theta(\widehat\theta_n-\theta>t)=\prod_{i=1}^n\mathbb P_\theta(Y_i-\theta>t)=e^{-nt}.
$$
In particular, $\widehat\theta_n\ge\theta$ almost surely and $\mathbb P_\theta(|\widehat\theta_n-\theta|>\varepsilon)=e^{-n\varepsilon}\to0$. This proves <statistical consistency>. The <exact endpoint limit for a shifted exponential distribution> is stronger than an asymptotic approximation:
$$
\boxed{n(\widehat\theta_n-\theta)\sim\operatorname{Exp}(1)\ \text{for every }n,\qquad n(\widehat\theta_n-\theta)\xrightarrow{d}\operatorname{Exp}(1).}
$$
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.
Back to article page
