# Determine asymptotic distribution and efficiency of an estimator

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.

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

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.

• 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? Apr 26, 2021 at 15:58