MLE of a parameter $\theta$ which is linear with the observations

Question

Let $Z$ be a single observation having density function $f(z;\theta) = (2\theta z -1 + \theta)$, $0\leq z\leq 1$. Assuming that $\theta$ is a nonrandom parameter and takes values $-1\leq \theta\leq 1$, I try to find MLE of $\theta$ as a function of $Z$. Finding log-likelihood and taking the derivative yields: \begin{equation} \hat{\theta} = \frac{\partial log f(z;\theta)}{\partial \theta} = \frac{2z - 1}{2\theta z + 1 - \theta} = 0 \end{equation} which gives $z = 0.5$. I'm stuck at this point since $\hat{\theta}$ does not depend on the observation. Any ideas how to proceed?

Answer 1

The likelihood is $\theta(2z-1)+1$. We wish to find the $\theta$ that maximizes the likelihood for a single observation $z$, which will be our maximum likelihood estimator.

We break into cases: If $2z-1>0$ or $z>\frac 12$, then the density is increasing in $\theta$ and we pick $\hat\theta=1$.

If $2z-1<0$ or $z<\frac 12$, the density is decreasing in $\theta$ and we pick $\hat\theta=-1$.

If $2z-1=0$ or $z=\frac 12$, the density does not depend on $\theta$. Thus it achieves its maximum at every point in the parameter space and $\hat\theta\in[-1,1]$ are all maximum likelihood estimates.

Thus, the MLE of $\theta$ for one observation is given by:

$$\hat\theta=\begin{cases}1&\text{ if } 1\ge z>\frac 12\\ -1&\text{ if } 0\le z<\frac 12\\ [-1, 1]&\text{ if }z=\frac 12\end{cases}$$