I have the following problem:

Let $X_1,..., X_n$ be a sample of independent, identically distributed random variables, with density $$f_{\theta}(x)=\begin{cases} e^{\theta-x}, & \text{if } x\geq \theta\\ 0, & \text{elsewhere}\\ \end{cases}$$

Let $\hat\theta_n$ be the maximum-likelihood estimator (MLE) of $\theta$. I am asked to find the asymptotic distribution of $\sqrt n (\hat\theta_n-\theta)$ and $\hat\theta_n$ efficiency.

So far, I computed the MLE, which is $\hat\theta_n=min\{x_i\}$, for $i=0,...,n.$. I also know that if I find that $\sqrt n (\hat\theta_n-\theta)$ converges in distribution to $N(0,I(\theta))$, we would have both the asymptotic distribution and efficiency of the estimator.

Central Limit Theorem could be a way to do it, but I'm really struggling on it.

Any hints or solutions would be appreciated.

  • $\begingroup$ This MLE is not asymptotically normal. Also pdf should have $x>\theta$ instead. $\endgroup$ Oct 22, 2019 at 15:38
  • $\begingroup$ There was a mistake in the pdf, the first case is if $x\geq\theta$. If it's not normal, which is the asymptotic distribution then? $\endgroup$
    – eseou
    Oct 22, 2019 at 15:43
  • $\begingroup$ $n(\hat\theta_n-\theta)$ is exactly distributed as $\mathsf{Exp}(1)$, which is what I would also call the asymptotic distribution of the MLE $\hat\theta_n$. $\endgroup$ Oct 22, 2019 at 15:47
  • $\begingroup$ That, in fact, is something I also want to prove later. But how can I get there is my question. $\endgroup$
    – eseou
    Oct 22, 2019 at 15:50
  • $\begingroup$ Pdf of $\hat\theta_n$ is $ne^{n(\theta-x)}$, right? $\endgroup$
    – eseou
    Oct 22, 2019 at 16:11

1 Answer 1


To answer the question you are asked, find the distribution function of $T=\sqrt n(\hat\theta_n-\theta)$.

Now $\hat\theta_n$ has CDF $$P(\hat\theta_n\le t)=\begin{cases} 1-e^{-n(t-\theta)} &,\text{ if }t\ge\theta \\ 0&,\text{ if }t<\theta \end{cases}$$

So CDF of $T$ should be

\begin{align} F_n(t)&=P\left(\sqrt n(\hat\theta_n-\theta)\le t\right) \\&=P\left(\hat\theta_n\le \frac{t}{\sqrt n}+\theta\right) \\&=\begin{cases}1-e^{-\sqrt nt}&,\text{ if }t\ge 0 \\ 0&,\text{ if }t<0\end{cases} \end{align}

Since $n(\hat\theta_n-\theta)\sim \mathsf{Exp}(1)$, it is clear that $T$ is exponential with mean $1/\sqrt n$.

Taking limit as $n\to\infty$, show that $F_n(t)$ converges to another distribution function. This would give you a degenerate asymptotic distribution of the MLE $\hat\theta_n$.

To arrive at a non-degenerate limiting distribution of the MLE, an appropriate scaling of $\hat\theta_n$ is $T'=n(\hat\theta_n-\theta)$. We know $T'$ is exactly distributed as $\mathsf{Exp}(1)$, which is naturally also the asymptotic distribution.

Maximum likelihood estimators are expected to be (asymptotically) efficient in most cases.

Formally, if $T_0$ is the uniformly minimum variance unbiased estimator (UMVUE) of $\theta$, then (large sample) efficiency of $\hat\theta_n$ is defined by $\mathrm e=\frac{\operatorname{Var}(T_0)}{\operatorname{Var}(\hat\theta_n)}$. Find $T_0$ and hence conclude $\hat\theta_n$ is indeed efficient.

  • $\begingroup$ in this case is Var(T0) the fisher information? and var (MLE) the denominator, is that the variance of T' distributed as EXP(1) given previously? this is superefficiency principle right? $\endgroup$ Apr 26, 2021 at 15:58

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