Let ${X_1},...,{X_n}$ be a random sample from uniform distribution   $U\left[ {0,\theta } \right]$. Find the MLE of $\theta $.   I wrote the likelihood function of this distribution, that being a uniform one has $f\left( x \right) = {1 \over {b - a}} = {1 \over \theta }$ as $$l\left( \theta  \right) = \prod\limits_{i = 1}^n {{1 \over \theta }} $$  which has an equivalent loglikelihood function  $$\log \left( l \right) = \sum\limits_{i = 1}^n {\left[ {\log \left( {{\theta ^{ - 1}}} \right)} \right]}  =  - n\log \left( \theta  \right)$$  Now if I want to find the MLE, I'll set its first derivative to zero: $$\eqalign{   &amp; {\partial  \over {\partial \theta }}\left( { - n\log \theta } \right) = 0  \cr    &amp; {{ - n} \over \theta } = 0 \cr} $$  And that's where things stopped making sense for me, for I can't solve for $\theta $ like I normally can (for other distributions). Checking a bunch of different threads in this community, apparently writing $\theta  = {X_{(n)}}$, that is, the highest observation from the sample, is enough as an answer.   However, I can't stop thinking about the fact that no ${X_{(n)}}$ on the planet would make the first derivative be equal to zero, so how can that be a valid estimator?
There is a classic problem:  Suppose that $X_1,\ldots,X_n$ form an i.i.d. sample from a uniform distribution on the interval $(0,\theta)$, where $\theta&gt;0$ is unknown. I would like to find the MLE of $\theta$.  The pdf of each observation will have the form: $$ f(x\mid\theta) =  \begin{cases} 1/\theta\quad&amp;\text{for }\, 0\leq x\leq \theta\\ 0 &amp;\text{otherwise}. \end{cases} $$ The likelihood function therefore has the form: $$ L(\theta) = \begin{cases} 1/\theta^n \quad&amp;\text{for }\; 0\leq x_i \leq \theta\;\; \text{for all }i,\\ 0 &amp;\text{otherwise}. \end{cases} $$ The general solution is usually that the MLE of theta must be a value of $\theta$ for which $\theta \geq x_i$ and which maximizes $1/\theta^n$ among all such values.   The reasoning is that since $1/\theta^n$ is a decreasing function of $\theta$, the estimate will be the smallest possible value of $\theta$ such that $\theta\geq x_i$.  Therefore, the mle of $\theta$, $\hat{\theta}$, is $\max(X_1,\ldots,X_n)$.  Here, I do not understand why we cannot just differentiate the likelihood function with respect to theta and then set it equal to $0$?  Thanks!
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