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I'm trying to find two statistics $T_1$, $T_2$ such that $(T_1, T_2)$ is jointly sufficient for $(\lambda, \theta)$ for a random sample $X_1, \dots, X_n$ from a two parameter exponential distribution.

$f(x) = \begin{cases} \lambda e^{-\lambda (x-\theta)}, & \theta < x < \infty, \\ 0, & \text{elsewhere}. \end{cases}$

Thanks.

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2 Answers 2

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First note that $$ f(x_1 ,x_2 , \ldots ,x_n |\lambda ,\theta ) = \lambda ^n e^{\lambda \theta n} \exp \bigg( - \lambda \sum\limits_{i = 1 }^n {X_i } \bigg) \prod\limits_{i = 1}^n {{\mathbf 1}(x_i > \theta )}, $$ where $\mathbf 1$ denotes the indicator function. Then note that $$ \prod\limits_{i = 1}^n {{\mathbf 1}(x_i > \theta )} = {\mathbf 1}(\min \lbrace x_1 , \ldots ,x_n \rbrace > \theta ), $$ and apply Theorem 2 given here.

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  • $\begingroup$ Here we need to factorize $f(x_1 ,x_2 , \ldots ,x_n |\lambda ,\theta ) = u(x_1,x_2,\ldots,x_n) v(r_{1,\ldots,k}(x_1,x_2,\ldots,x_n | \lambda,\theta)$ right ? but I don't see a function $u(x_1,x_2,\ldots,x_n)$. Isn't $\prod\limits_{i = 1}^n {{\mathbf 1}(x_i > \theta )} = {\mathbf 1}(\min \lbrace x_1 , \ldots ,x_n \rbrace > \theta )$ dependent on $\theta$ ? $\endgroup$
    – 0x0
    Feb 28, 2011 at 1:37
  • $\begingroup$ You can take $u(x_1,\ldots,x_n)=1$ (it is nonnegative and does not depend on the parameters). As for the second question, consider the example given on p. 3 in that link (Uniform population). $\endgroup$
    – Shai Covo
    Feb 28, 2011 at 2:00
  • $\begingroup$ Also, note the last sentence in Theorem 2. $\endgroup$
    – Shai Covo
    Feb 28, 2011 at 2:07
  • $\begingroup$ Further, see the examples in en.wikipedia.org/wiki/Sufficient_statistic, in particular the example "Uniform distribution (with two parameters)". $\endgroup$
    – Shai Covo
    Feb 28, 2011 at 2:20
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    $\begingroup$ BTW your answer and comments helped me a lot to think about the problem rather than giving the answers straight. Thanks. $\endgroup$
    – 0x0
    Feb 28, 2011 at 2:57
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by neyman factorization theorem X_bar and X(1) are jointly sufficient stats

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  • $\begingroup$ You should include the picture in your post or, better yet, write what it says using MathJax. And please include some words of explanation as well. $\endgroup$ Mar 6, 2016 at 15:43

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