# Sufficient statistic function for $f(x) = \theta x^{-2}, \; \; 0 < \theta \leq x < \infty$. [closed]

Exercise :

Let $X_1, \dots, X_n$ be a random sample from a distribution with probability density function $f(x) = \theta x^{-2}, \; \; 0 < \theta \leq x < \infty$, where $\theta$ unknown parameter.

(i) Check if the given distribution belongs to the Exponential Family of Distributions.

(ii) Find a sufficient statistic function $T$ of $\theta$.

Attempt :

(i) It is :

It does not belong to the Exponential Family of Distributions, since the support of $X$ depends on $\theta$.

(ii) To try and find a sufficient statistic function, we must write $p(x|\theta)$ in the form of $G(t,\theta)H(x)$.

Thus :

$$p(x|\theta) = \prod_{i=1}^n \theta x_i^{-2} \mathbb{I}_{[\theta, + \infty]}(x_i)$$

How should I continue to find a sufficient statistic function from here on?

## closed as off-topic by Jyrki LahtonenMay 27 '18 at 9:45

• This question does not appear to be about math within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

• Cross posting on two different SE-sites is against the rules. You exacerbated the problem by not crosslinking the two versions. The CV version has a higher upvoted answer, so I am migrating this there. The CV-moderators can then merge (and later remove this comment). – Jyrki Lahtonen May 27 '18 at 9:43

## 1 Answer

Using the factorization criteria you have $$\prod_{i=1}^n \theta x_i^{-2}I[x_i\ge \theta]=\theta^nI[x_{(1)}\ge\theta]\times \prod_{i=1}^n\frac{1}{x_i^2}I[x_i\ge x_{(1)}]=g(\theta ; T(X))H(X),$$ i.e., the MSS is $T(X)=X_{(1)}=\min\{X_1,...,X_n\}$. It is minimal as it one dimensional like $\theta$. The last step follows from the fact that $\prod_{i=1}^nI[x_i \ge \theta] = I[x_{(1)} \ge \theta]\prod_{i=2}^nI[x_i \ge x_{(1)}]$. And note that as the support of $X$ depends on $\theta$, thus it cannot belong to the exponential family.

Another approach is to verify that $X_{(1)}$ is the MLE, hence it must be a function of the sufficient statistics, and as $X_{(1)}$ is the identity function and it is one dimensional then it is the minimal sufficient statistic.

• What exactly is $X_{(1)}$ and how did you yield that result ? – Rebellos May 26 '18 at 23:15
• @Rebellos I edited the answer. Hope it helps – V. Vancak May 26 '18 at 23:20
• The expression that you yield has the factor $x_{(1)}$ in both parts, thus it cannot be a sufficient statistic function (I think), since you need to have the $x_i$ in a different place such as $H(x)$. – Rebellos May 26 '18 at 23:23
• Presumably you are saying $I[x_i\ge x_{(1)}]=1$ for all $i$. And I think you have shown $\min\{X_1,...,X_n\}$ is a sufficient statistic (as requested) rather than a minimal sufficient statistic – Henry May 27 '18 at 0:57
• @Henry It is minimal as $x_{(1)} \in \mathbb{R}$. – V. Vancak May 27 '18 at 8:01