Finding the MLE of a function when $L'(\theta)$ doesn't depend on $\theta$

Question

Here's the problem:

Find the MLE of of $\theta$ when $f(x\mid\theta)=(1+x\theta)/2$ for $-1<x<1$, $=0$ otherwise.

$0<\theta<1$

Find the maximum likelihood of $\theta$ and find its exact probability distribution. Is the MLE unbiased? Find its bias and MSE.

So I'm having problems starting this. If I try to do it the usual way -- find $\theta$ to max the function -- $L'(\theta)=x/2$ which doesn't have a max. This is super obvious just from looking at the function (it's linear), but I'm not sure what to do when this way doesn't work. Thoughts?

Edit: What I come up with (just by looking at the graph and thinking about it) is that $MLE=0$ if $-1<x<0$ and $MLE=1$ if $0<x<1$, but that seems odd and a strange place to move forward from.

Answer 1

Work with:

$$L(\theta\mid-) = \prod_i \frac{(1+x_i\theta)}{2}$$

or equivalently with the log-likelihood:

$$LL(\theta\mid-) = \sum_i \log(1+x_i\theta) - n \log2$$

Your attempt assumes that you have a single observation in which case instead of using calculus think about when the likelihood is maximized by thinking about the cases when $x$ is greater or less than $0$. But, in general, you are likely to have more than one observation.

Answer 2

You're working with an endpoint maximum.

If $x>0$, then $\dfrac{1+x\theta}{2}$ increases as $\theta$ increases, so the value of $\theta$ that maximizes this is at the right endpoint: $\theta=1$. If $x<0$, the similarly it's at the left endpoint: $\theta=0$.