Jointly Sufficient Statistics for Pareto

Question

We have a sample drawn from a pareto distribution with pdf: $$f_X(x) = ab^ax^{-(a+1)} , x>b$$ We want to get jointly sufficient statistics for $a$ and $b$. So I find the likelihood function of a sample of size $n$: $$L(a,b)=a^nb^{an}(\prod_{i=1}^nX_i)^{-(a+1)}$$

So isn't just the $\prod_{i=1}^nX_i$ part a sufficient statistic by itself? Why do I need multiple statistics at this point?

Answer 1

You've got the likelihood function wrong. The PDF, written explicitly, is

$$ f_X(x)=ab^ax^{-(a+1)}[x>b] $$

(where $[I]$ is the indicator function for $I$), and thus the likelihood function is

\begin{eqnarray} L(a,b) &=& a^nb^{an}\left(\prod_{i=1}^nX_i^{-(a+1)}\left[X_i\gt b\right]\right) \\ &=& a^nb^{an}\left(\prod_{i=1}^nX_i\right)^{-(a+1)}\left[X_{(1)}\gt b\right]\;, \end{eqnarray}

where $X_{(1)}$ is the first order statistic of the $X_i$. Thus jointly sufficient statistics for $a$ and $b$ are $\prod_{i=1}^nX_i$ and $X_{(1)}$.

Answer 2

\begin{align} & f(x_1,\ldots,x_n) \\[8pt] = {} & \begin{cases} a^nb^{an} \left(\prod_{i=1}^n x_i\right)^{-(a+1)} & \text{if all of }x_1,\ldots,x_n \text{ are} \ge b, \\ 0 & \text{otherwise} \end{cases} \\[10pt] = {} & \begin{cases} a^nb^{an} \left(\prod_{i=1}^n x_i\right)^{-(a+1)} & \text{if } \min\{x_1,\ldots,x_n\} \ge b, \\ 0 & \text{otherwise.} \end{cases} \end{align}

This depends on $x_1,\ldots,x_n$ through the pair $\big( x_1\cdots x_n, \min\{x_1,\ldots,x_n\}\big).$