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

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

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!

For consistency, by the weak law of large numbers $bar X_n stackrel{text p}to frac 1lambda + theta$ and $X_min stackrel{text p}to theta$ so by Slutsky

$$

bar X_n – X_min stackrel{text p}to frac 1lambda.

$$

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).