Maximum likelihood estimator for $\theta$ and $E[X]$ Let $X_1,..., X_n $ be a random sample of a variable with PDF:
$$f(x|\theta)=\frac{\theta}{x^2} I_{(\theta, \infty)}(x), \theta >0$$
Find the maximum likelihood estimator for $\theta$ and $ E[X]$
My attempt:
The likelihood function is:
$$L(\theta;x) = \theta^n I_{(\theta, \infty)}(x_{(1)}) \prod \frac{1}{x_i^2} = 
\theta^n I_{(0, x_{(1)}}(\theta) \prod \frac{1}{x_i^2}$$
Since the indicator function and the product are positive, the likelihood function is increasing. Also, since $\theta$ is on the interval given in the indicator, then $\theta$ is maximum when $\theta = X_{(1)}$. (Is this correct?)
The second doubt is about how to find an estimator of $E[X]$. Calculating it, we have that
$$E[X] = \infty$$
What should I answer, in this case? Or did I do something wrong?
 A: Let $X_1,\dots,X_n$ be a random sample from density $f(x_i)=(\theta/x_i^2)\,I_{(\theta,\infty)}(x_i)$, for $\theta>0$. Since $I_{(\theta,\infty)}(x_i)=I_{(0, x_i)}(\theta)$, writing $x=(x_1,\dots,x_n)$, the likelihood function is
$$
  L_x(\theta) = \frac{\theta^n}{\prod_{i=1}^n x_i^2} I_{(0,x_{(1)})}(\theta) \, , \qquad (*)
$$
in which $x_{(1)}=\min\{x_1,\dots,x_n\}$. The way the density is defined implies that there is no MLE in the usual sense, because the candidate $x_{(1)}\neq\arg\max_\theta L_x(\theta)$. In fact, $L_x(x_{(1)})=0$. If, for each $\theta>0$, we change the version of the density in just one point, and this doesn't change the family of sampling distributions, doing $f(x_i)=(\theta/x_i^2)\,I_{[\theta,\infty)}(x_i)$, then it's true that $\hat{\theta}_{\mathrm{MLE}}=X_{(1)}$. 
This is not a serious difficulty, but it's a curious case in which the particular versions of the sampling densities chosen for the problem change the answer. In the second edition of DeGroot's "Probability and Statistics" there is a similar example starting on page 343.
 
Also, since $\mathrm{E}_\theta[X_i]=\infty$, for every $\theta>0$, asking for an MLE of this quantity doesn't make sense.
A: One cannot provide an estimator for a distribution moment that is not finite. Here we have a case of a distribution that does not have a finite expected value.  The question that usually comes next is "then what does the sample mean from an i.i.d. sample estimates in such a case?" The answer is that the sample mean is a linear function of the random variables of the sample, and it too, has no finite expected value. So it is not an estimator of the expected value of the random variable.  
In such cases, we look for other centrality measures, like for example the median. Here we have
$$F_X(x) = \int_\theta^{x}\frac {\theta}{t^2}dt = 1-\frac {\theta}{x}$$
and denoting the median by $m$ we get
$$F_X(m) = \frac 12 \Rightarrow 1-\frac {\theta}{m} = \frac 12 \Rightarrow m=2\theta$$
Therefore an MLE for a centrality measure of this distribution is
$$\hat m_{MLE} = 2\hat \theta_{MLE} = 2X_{(1)}$$
