# Calculating Maximum Likelihood

Suppose that $$Y_1,\dots,Y_n$$ is a random sample from the density function given by

$$f(y|\theta)=\begin{cases}\frac1\theta, &y\in(0,\theta), \\ 0, &\mathrm{otherwise.}\end{cases}$$

for some $$\theta>0$$. Let $$\hat\theta$$ be the Maximum Likekihood Estimator for $$\theta$$. Which of the following statements hold true?

• (A) $$\hat\theta=(y_1+\dots+y_n)/n$$.
• (B) $$\hat\theta=\min_{i=1,\dots,n} y_i$$.
• (C) $$\hat\theta=\max_{i=1,\dots,n} y_i$$.
• (D) $$\hat\theta=1/\bar\theta$$.
• (E) None of the above.

The solution says C is correct and gives a brief explanation (plot the likelihood). I tried to work it out by the following method:

L(y|$$\theta$$) = $$\frac{1}{\theta^n} = \theta^{-n}$$. Taking logarithms gives -nlog$$\theta$$ and differentiating this wrt theta gives $$\frac{-n}{\theta}=0$$.

Can my method be used, if not, why? How should I go about getting the correct answer?

You have to maximize $$L$$, or equivalently minimize $$\theta$$. You have no other condition from the expression of $$L$$, so any value of $$\theta$$ is valid, granted it's possible given the data.
The least possible value is obviously $$\max y_i$$, since any smaller value would be impossible: you can't get random samples outside of $$[0,\theta]$$.
Note that maximizing $$L$$ does not necessarily mean you have to find a zero of its derivative. This remark applies to all optimization problems. It's not uncommon that the optimum is found on some kind of boundary instead of a critical point.