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.
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• 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

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
• 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