Maximum likelihood estimator for $\theta$

Question

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?

Answer 1

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.