Find the maximum likelihood estimator for $\theta$ when $f(x)=\frac{2x}{\theta^2}, 0\leq x \leq \theta$.
This should be a really easy question but I somehow cannot seem to get the right answer. My likelihood function is $\prod\limits_{i=1}^n \frac{2x_i}{\theta^2}$ which gives me after some simplification $n\ln(2\theta^{-2}) + \sum\limits_{i=1}^n ln(x_i)$ as my log-likelihood function. Then the derivative becomes $\frac{-2n}{\theta}$. If I equate this to $0$ to obtain the maximum I obtain $\hat{\theta}\rightarrow\infty$. However, this answer does not seem plausible. What did I do wrong?
Best Answer
There's an issue that arises here. Your distribution is supported (nonzero) for $0\leq x\leq\theta$. In other words, it's zero outside this interval. That means when you look at something like $f(x_i)=2\theta^{-2}x_i$, it should actually be $f(x_i)=2\theta^{-2}x_i\mathbb{1}_{0\leq x_i\leq \theta}$, where the indicator is zero when $x_i$ is outside the interval. When you are maximizing $f(x_1)\cdots f(x_n)$, this product is zero if $\theta<x_i$ for some $i$, so you can just assume that $\theta\geq x_i$ for each $i$.
So think about how you would maximize such a function. Don't deal with derivatives because the function won't have a zero derivative, but will have a maximum. As a hint, try looking at $\hat{\theta}=\max_i x_i$. Can you do better than that?