1
$\begingroup$

Suppose the data consist of a single number $X$, from the following probability density:

$$f(x|θ) = \begin{cases} \frac{1+xθ}{2} & & \text{for } -1 \leqslant x \leqslant 1, \\[6pt] 0 & & \text{otherwise}, \end{cases}$$

where $-1 \leqslant \theta \leqslant 1$. Find the maximum likelihood estimate (MLE) $\hat{\theta}$ of the parameter $\theta$ and find its (exact) probability distribution. Is the MLE unbiased? Find its bias and MSE. [Hint: First find the MLE for a few sample values of $X$; that should suggest to you the general solution. Drawing a graph helps! The distribution of the MLE will of course depend upon $\theta$.]

How do I begin?

$\endgroup$
2
$\begingroup$

Begin with the hint given. Take $x=0.3$. Can you find the value of $\theta\in[-1,1]$ that maximizes $\frac{1+0.3\theta}{2}$?

Take $x=0.8$. Can you find the value of $\theta\in[-1,1]$ that maximizes $\frac{1+0.8\theta}{2}$?

Take $x=-0.5$. Can you find the value of $\theta\in[-1,1]$ that maximizes $\frac{1-0.5\theta}{2}$?

Take $x=-0.9$. Can you find the value of $\theta\in[-1,1]$ that maximizes $\frac{1-0.9\theta}{2}$?

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

The first step of finding the MLE is to write out the likelihood function (or log-likelihood function). In the present problem, with only a single observed data point, the latter is given by:

$$\ell_x(\theta) = \ln(1 + x \theta) + \text{const.} \quad \quad \quad \text{for } -1 \leqslant \theta \leqslant 1.$$

This is the function you need to maximise to obtain the MLE. Maximising this function is a univariate calculus problem, which will require you to derive the first and second derivatives of the function. By maximising this function you will obtain the MLE as a function of the observed data value $x$, and you can then proceed to the next parts of the question. (In this case, you will see that the maximising value is a "boundary solution" to the problem.)

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.