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?
Best Answer
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.