MLE for case where parameter vanishes

Question

I'm to find MLE of

$$f_\beta(x)=3e^{-3(x-\beta)}, \quad x>\beta$$

Proceeding in typical fashion.

$$L(\beta)=\prod_{i=1}^n 3e^{-3(x-\beta)}=3^ne^{\sum-3(x-\beta)}$$

Taking log

$$\ell(\beta)=n\ln3+\sum_{i=1}^n-3(x-\beta)$$

Normally I would take derivative here but in doing so the parameter vanishes. I know the answer is supposed to be $X_{(1)}$ (the minimum from the data) but I can't find anywhere the rationale behind this decision.

Answer 1

Your approach is good for the so called "regular cases", i.e., when the support of $X$ and the parametric space $\Theta$ are independent. However, $x \ge \beta$, so your likelihood function should incorporate the indicator, namely $$ \mathcal{L}(\beta;X)=3^n\exp\{-3\sum(x_i - \beta\}\prod I\{X_i \ge X_{(1)} \ge \beta \}, $$ that is monotonic decreasing function that attains its maximum on the boundary, i.e., $$ \arg \max_{\beta \in \Theta}\mathcal{L}(\beta;X)=X_{(1)}. $$

Answer 2

Your likelihood is incomplete, because you have not accounted for the fact that the density is zero for $x \le \beta$. Consequently, the log-likelihood as you have written it does not suggest any restriction on $\beta$, when in fact $\beta$ cannot be larger than the minimum value observed in the sample.

If I asked you to find the maximum value of the function $f(z) = e^{-z}$ for $z \in \mathbb R$, what would you say?

Now, if I asked you to find the maximum value of the function $f(z) = e^{-z}$ for $z \ge 0$, what would you say?

Note that in both cases, $f$ is the same. Note also that $f$ has no critical points since $f'(z) < 0$ for all $z \in \mathbb R$. Yet, why does a maximum exist for the second case? This illustrates the simple truth from elementary calculus that a global maximum may exist at a boundary point of the domain, rather than at a critical point.