Determining maximum likelihood estimator for scaled beta distribution

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.

• 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. – BruceET Feb 17 '18 at 21:41
• 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? – Serena Feb 17 '18 at 23:44
• 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. – BruceET Feb 18 '18 at 8:10
• 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. – BGM Feb 18 '18 at 8:55

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 $\tilde{\theta} = \tfrac{9}{8} \cdot x$.