Suppose we have $X_1,...,X_n$ iid the shifted exponential distribution:

$$f(x)=\lambda e^{-\lambda(x-\theta)}, x\ge \theta$$

I have figured out both the MLE for $\lambda$ and $\theta$, which are $\hat \lambda = \frac{1}{\bar X - X_{min}}$ and $\hat \theta =X_{min}$.

I also found the asymptotic distribution of $\hat \theta$:

$$\sqrt{n}(\hat \theta-\theta) \rightarrow 0$$

Now I'm stuck at deriving the asymptotic distribution of $\hat \lambda$ and showing that it is a consistent estimator. How do you do this?


  • 2
    $\begingroup$ The asymptotic distribution of $\hat\theta$ is using the wrong scale: it should be $n$ not $\sqrt n$. See this answer. $\endgroup$
    – Xi'an
    Jun 4 '20 at 4:35
  • $\begingroup$ To arrive at a non-degenerate limiting distribution of $\hat\theta$, you ought to use $n(\hat\theta-\theta)\sim \mathsf{Exp}(1)$ as mentioned above. This is an exact distribution which is naturally also the asymptotic distribution. $\endgroup$ Jun 4 '20 at 7:27

For consistency, by the weak law of large numbers $\bar X_n \stackrel{\text p}\to \frac 1\lambda + \theta$ and $X_\min \stackrel{\text p}\to \theta$ so by Slutsky $$ \bar X_n - X_\min \stackrel{\text p}\to \frac 1\lambda. $$ By assumption $\lambda > 0$ so the map $x \mapsto x^{-1}$ is continuous, and the continuous mapping theorem finishes the job.

For the asymptotic distribution, by the standard CLT we know $\sqrt n (\bar X_n - \theta -\lambda^{-1}) \stackrel{\text d}\to \mathcal N(0, \lambda^{-2})$. Let $Y_n = \sqrt n (\bar X_n - \theta - \lambda^{-1})$ and consider $$ \sqrt n (\bar X_n - X_{\min,n} - \lambda^{-1}) = \sqrt n ([\bar X_n - \theta - \lambda^{-1}] - [X_{\min,n} - \theta])\\ = Y_n - Z_n $$ where $Z_n := \sqrt n (X_{\min,n} - \theta)$. You already worked out the asymptotic distribution of $Z_n$ so we can use that along with Slutsky again to conclude $$ Y_n - Z_n \stackrel{\text d}\to \mathcal N(0, \lambda^{-2}). $$

You can now finish this off with the delta method.


Although you are also asking about the estimator $\hat{\lambda}$, I am going to note some things about $\hat{\theta}$. In this particular case it is quite easy to obtain the exact distribution of this estimator. Since you have a series of shifted exponential random variables, you can define the values $Y_i = X_i - \theta$ and you then have the associated series $Y_1,Y_3,Y_3 ... \sim \text{IID Exp}(\lambda)$. This gives the exact distribution:

$$\hat{\theta} = X_{(1)} = \theta+ Y_{(1)} \sim \theta + \text{Exp}(n \lambda).$$

Note that this gives the pivotal quantity $n(\hat{\theta} - \theta) \sim \text{Exp}(\lambda)$. You can prove that $\hat{\theta}$ is a consistent estimator by computing the probability of a deviation larger than a specified level. For all $\varepsilon >0$ we have:

$$\begin{aligned} \mathbb{P}(|\hat{\theta} - \theta| < \varepsilon) = \mathbb{P}(\hat{\theta} - \theta< \varepsilon) = \exp(-n \lambda \varepsilon). \\[6pt] \end{aligned}$$

We therefore get the limiting result:

$$\begin{aligned} \lim_{n \rightarrow \infty} \mathbb{P}(|\hat{\theta} - \theta| < \varepsilon) = \lim_{n \rightarrow \infty} \exp(-n \lambda \varepsilon) = 0, \\[6pt] \end{aligned}$$

which is the required condition for weak consistency (i.e., convergence in probability of the estimator to the parameter it is estimating).

  • $\begingroup$ any idea why exactly does the asymptotic normality of MLE not hold in this case? Any regularity condition broke? $\endgroup$ Nov 21 '20 at 19:07
  • 1
    $\begingroup$ @MaverickMeerkat: The MLE occurs at a boundary point of the likelihood function, which breaks the ordinary regularity conditions. $\endgroup$
    – Ben
    Nov 21 '20 at 20:51

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