# Server Bug Fix: Asymptotic distribution for MLE of shifted exponential distribution

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).