Rationale behind MLE of $f_{\theta}(x) = \frac{1}{\theta} I_{\{1, \dots,\theta\}}(x)$

Question

Our probability density for $\theta \in \{1,\dots,\theta_0\}$ is $$f_{\theta}(x) = \frac{1}{\theta} I_{\{1, \dots,\theta\}}(x)$$

Let $X_{(n)}$ be the largest order statistic. Acooding to the solution, the likelihood function is $$\frac{1}{\theta^n} I_{\{X_{(n)}, \dots,\theta_0\}}(x)$$

I don't get why. Also, $X_{(n)}$ is supposed to be the MLE of $\theta$. Why?

Answer 1

The likelihood based on the sample $(x_k)$ is by definition $L(\theta)=f_\theta(x_1)f_\theta(x_2)\cdots f_\theta(x_n)$. In the present case, $L(\theta)=\theta^{-n}\mathbf 1_{x_1\leqslant\theta}\mathbf 1_{x_2\leqslant\theta}\cdots\mathbf 1_{x_n\leqslant\theta}=\theta^{-n}\mathbf 1_{\max(x_k)\leqslant\theta}$ hence $L(\theta)=0$ if $\theta\lt\max(x_k)$ and $\theta\mapsto L(\theta)$ is decreasing on $\theta\geqslant\max(x_k)$. Thus, $L(\theta)$ is maximal at $\theta=\max(x_k)$.