Suppose that $X_1,...X_n$ are independent observations from a uniform distribution on $[0,\theta] $ where $\theta \in N$. Then the maximum liklehood estimator of $\theta$
(A) may not exist in some cases.
(B) is $\lfloor X_n\rfloor$
(C) is $\lceil X_n\rceil$
(D) is $\lfloor X_n\rfloor$ if $(X_n-\lfloor X_n\rfloor)\leq\dfrac{1}{2}$, otherwise it is is $\lceil X_n\rceil$
My input: I took an example. Since domain of $X$ is natural number therefor considering ${(1,7,3,5,6,8)} $ as my sample Mle is Max of sample that is 8 and i tried to think of options B C and D. For B and C i am getting same answers. Is there any catch in this problem?Also i am a beginner in statistical inference portion please give me some extra knowledge if you can. Thank you!
MLE is defined as $\max$ (or sometimes $\sup$) over the parametric space, i.e., $$ \hat{\theta}_n = \arg \max_{\theta \in \Theta} \mathcal{L}(\theta; x_1,...,x_n), $$ in our case $\Theta$ is on the right to the support of $X$, i.e., $\theta \ge X_i$,$\forall i$. Next recall that $\mathbb{P}(X_i \in \mathbb{N})=0$, as such if you choose the flooring option, you'll have that your estimator will be almost surely smaller than the maximal sample value. Thus, the estimator won't be in the parametric space at all, hence cannot be an MLE. Assume that you have $\{ 1.3, 4.4,, 3.2, 4.5\}$, thus $x_{(n)} = 4.5$. By choosing the floor function you have that your estimator is $4$, which is ridiculous, as half of the sample is larger then your point estimator. So, you have to choose the ceiling function.