3
$\begingroup$

Suppose $X_1,X_2,\ldots,X_n$ are i.i.d variables having a two-parameter exponential distribution with common location and scale parameter $\theta$ :

$$f_{\theta}(x)=\frac{1}{\theta}e^{-(x-\theta)/\theta}\mathbf1_{x>\theta}\quad,\,\theta>0$$

I am wondering if it is possible to derive a maximum likelihood estimator (MLE) of $\theta$.

The likelihood function given the sample $x_1,\ldots,x_n$ is

$$L(\theta)=\frac{1}{\theta^n}e^{-n(\bar x-\theta)/\theta}\mathbf1_{x_{(1)}>\theta}\quad,\,\theta>0$$

, where $\bar x=\frac{1}{n}\sum\limits_{i=1}^n x_i$ and $x_{(1)}=\min\limits_{1\le i\le n} x_i$.

Since $L(\theta)$ is not differentiable at $\theta=x_{(1)}$, I cannot apply the second-derivative test here.

Even if I could say that $L(\theta)$ is increasing and/or decreasing in $\theta$ under the constraint $\theta<x_{(1)}$, I am not sure what choice of $\theta$ maximises $L(\theta)$. Differentiation is not valid as I understand. I think it is safe to assume $x_{(1)}<\bar x$, so the constraint is actually $\theta<x_{(1)}<\bar x$.

If MLE is unique, then it is likely to be a function of the sufficient statistic $(\overline X,X_{(1)})$. However I don't see how to derive it in this particular model. Any suggestion would be great.

$\endgroup$
  • $\begingroup$ @angryavian I just ran a small simulation using Python, and it gives a unique maximizer not equal to $x_{(1)}$. Not sure what gives, just thought I'd mention it. $\endgroup$ – David M. May 20 at 23:54
  • $\begingroup$ you can not generate observations less than $\theta$, hence the MLE of $\theta$ is most probably the minimum order statistic $\endgroup$ – Ahmad Bazzi May 21 at 5:51
3
$\begingroup$

The joint likelihood is $$\mathcal L (\theta \mid \boldsymbol x) = \theta^{-n} e^{-n(\bar x - \theta)/\theta} \mathbb 1(x_{(1)} \ge \theta).$$ Since the unique critical point of the unrestricted likelihood occurs for $$0 = \frac{d}{d\theta} \left[\log \mathcal L \right] = -\frac{n}{\theta} + \frac{n\bar x}{\theta^2} = \frac{n}{\theta^2}(\bar x - \theta)$$ or $\theta = \bar x \ge x_{(1)}$, and the likelihood is increasing on $\theta \in (0, x_{(1)}]$, it follows that the global maximum must be the minimum order statistic $\hat \theta = x_{(1)}$.

It is worth noting that while the MLE can exceed the true value of $\theta$, it cannot be larger than the minimum order statistic, since in such a case, such an estimate could not generate at least one observation in the sample. Also note that the MLE cannot be smaller than the true value of $\theta$, since such an estimate could only have been generated from a sample with an observation that is smaller than $\theta$, which is impossible. In short, we must always have $$0 < \theta \le \hat \theta \le x_{(1)} \le \bar x.$$

$\endgroup$
  • $\begingroup$ So is the unique mle $\hat\theta=x_{(1)}$ or is any $\hat\theta$ satisfying $0<\hat\theta< x_{(1)}$ a possible answer? (you have included zero as a value of $\theta$ in the last line) If you have any confirmation using simulation, please share that as well. Also, is differentiating the likelihood necessary? $\endgroup$ – StubbornAtom May 21 at 6:11
  • $\begingroup$ @StubbornAtom The inclusion of $0$ was an oversight, apologies. The increasing nature of $\mathcal L$ w.r.t. $\theta$ would preclude choosing any $\hat \theta \in (0, x_{(1)})$. Indeed, since you do not know $\theta$, but you know $\theta \le \hat \theta$, in a sense, you cannot choose any other value except $\hat \theta = x_{(1)}$. $\endgroup$ – heropup May 21 at 6:47

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.