4
$\begingroup$

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?

Thanks!

$\endgroup$
2
  • 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
6
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.