0
$\begingroup$

Suppose $X = \theta{Y}$ for some $\theta>0$ with $Y∼Beta(8,1)$. What is the maximum likelihood estimator for $\hat{\theta}=\hat{\theta}(x)$?

The distribution function I got is $f_\theta(x) = \frac{8x^7}{\theta^8}$ on range$[0,\frac{1}{\theta}]$, but by taking the derivative of the log likelihood function for this distribution by $\theta$ does not work.

$\endgroup$
  • $\begingroup$ Taking a derivative is not going to work because the support of the random variable $X$ depends on the unknown parameter $\theta.$ // A somewhat similar task is to find the MLE for $\theta$ based on data from $\mathsf{Unif}(0,\theta)$. In that case the MLE is the largest observation out of $n$. That case is widely discussed in statistics books and on this site. $\endgroup$ – BruceET Feb 17 '18 at 21:41
  • $\begingroup$ thank you for your reply! i got the idea of finding MLE for $\theta$ but is it ok not to take Y into consideration at all? $\endgroup$ – Serena Feb 17 '18 at 23:44
  • $\begingroup$ Not sure what you mean. $X$ is defined in terms of $Y,$ so in that sense $Y$ can't be ignored. However both shape parameters of $Y$ are known, so there is nothing about its distribution to be estimated. $\endgroup$ – BruceET Feb 18 '18 at 8:10
  • $\begingroup$ Taking derivative is fine as long as you are aware of that the maximum may not be in the interior point - as in this case it is located in the boundary. $\endgroup$ – BGM Feb 18 '18 at 8:55
0
$\begingroup$

In problems like these you can still use ordinary calculus methods; you have a monotonic likelihood function with the MLE occurring at a boundary point. This gives rise to a biased estimator of the parameter value, which can then be adjusted to obtain a non-biased estimator. In your particular problem (with a single observation $x$) you have the log-likelihood function:

$$\ell_x (\theta) = -8 \ln \theta <0 \quad \quad \text{for all }\theta \geqslant x.$$

Since the log-likelihood is a decreasing function, the MLE occurs at the boundary point $\hat{\theta} = x$. It can easily be shown that this is a biased estimator with mean $\mathbb{E}(\hat{\theta}) = \tfrac{8}{9} \cdot \theta$, and so you can obtain an unbiased adjusted-MLE using the estimator $\tilde{\theta} = \tfrac{9}{8} \cdot x$.

$\endgroup$

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.