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I'm trying to find two statistics T1, T2 such that (T1, T2) is jointly sufficient for (λ, θ) for a random sample $X_1,\ldots,X_n$ from a two parameter exponential distribution.

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


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1 Answer 1

up vote 3 down vote accepted

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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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 $\theta$ ? –  Sunil Feb 28 '11 at 1:37
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). –  Shai Covo Feb 28 '11 at 2:00
Also, note the last sentence in Theorem 2. –  Shai Covo Feb 28 '11 at 2:07
Further, see the examples in en.wikipedia.org/wiki/Sufficient_statistic, in particular the example "Uniform distribution (with two parameters)". –  Shai Covo Feb 28 '11 at 2:20
BTW your answer and comments helped me a lot to think about the problem rather than giving the answers straight. Thanks. –  Sunil Feb 28 '11 at 2:57

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