2
$\begingroup$

5. Find the MLE for $\theta$ based on a random sample of size $n$ from a distribution wit pdf $$f(x; \, \theta) = \begin{cases} 2\theta^{2} x^{-3} & \theta \leqslant x \\ 0 & x < \theta;\, 0 < \theta \end{cases}$$

I'm practicing Maximum likelihood estimators for an upcoming test. The question that has stumped is the one above.

I attempted the following:

$$\ln(L(\theta)) = n\ln2\, +\,2n\ln(\theta)\,-\,3\sum_{i=1}^n x_i, $$ and it follows that $$\frac{d\,\ln(\theta)}{d\theta} = \frac{2n}{\theta}.$$ This wouldn't give any important information when set to zero. I checked the solution and it seems that the correct answer is $\hat\theta = X_{1:n}$. Could you please explain to me how this has come about?

$\endgroup$
  • $\begingroup$ check out why you should add the "self-study" tag: stats.stackexchange.com/tags/self-study/info $\endgroup$ – Taylor Sep 12 '18 at 21:52
  • 1
    $\begingroup$ My first advice is always *draw the log-likelihood" and "pay attention carefully to not just the functional form of the likelihood but the interval/region for the parameters over which that holds" (i.e. take care over where likelihood is zero) $\endgroup$ – Glen_b Sep 13 '18 at 8:25
3
$\begingroup$

First, you have the wrong log-likelihood. You might want to go back and check that part.

Second, this likelihood, when thought of as a function in $\theta$, is only nonzero whenever $\theta$ is less than all of the data points. This means $\theta \le x_i$ for all $i$, which is equivalent to $$ \theta \le \min_i\{x_i\}. $$ Also keep in mind $\theta$ is positive. These points are critical when considering the derivative, because you may not be able to find a root for the derivative on this interval. In other words, you might not be able to just set it equal to zero and solve for $\theta$. If that's the case, try drawing a picture of the likelihood with $\theta$ on the independent axis, paying special attention to the bounds.

$\endgroup$
  • $\begingroup$ Oh my god...I can't believe I didn't see that. I focused mostly on the variables not the constraints. thank you! $\endgroup$ – Destro Sep 12 '18 at 22:16

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.