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Let be $X_1,X_2,..X_n$ independent random variables such that $f_k(x;\theta)=e^{k\theta-x}$ for $x>k\theta$. The problem is to find a sufficient statistic for $\theta$. I don't achieve to replace $I_{(n\theta,\infty)}(x_i)$ (that is 1 when $x_i>i\theta$ $\forall{i}\in\{1,2,...,n\}$ and $0$ in any other case) from $f(x_1,x_2,...;x_n;\theta)=exp\{{\theta\frac{n(n+1)}{2}-\displaystyle\sum_{i=1}^{n}}x_i\}I_{(i\theta,\infty)}(x_i)$, for use the factorization theorem

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    $\begingroup$ You can write the product $I_{(i\theta, \infty)}(x_i)$ as $I_{(\theta, \infty)}(\min x_i/i).$ Pull the part with $e^{\theta n(n+1)/2}$ away from the exponential, and this gives you the required factorisation. $\endgroup$ – stochasticboy321 Mar 11 '18 at 1:57
  • $\begingroup$ OP: You were on the site since @stochasticboy321 gave you this precise suggestion. What are your plans about it? $\endgroup$ – Did Mar 11 '18 at 16:52
  • $\begingroup$ About the suggestion, is correct that $I_{(\theta,\infty)}(x_i)$ is equivalent to $I_{(\theta,\infty)}(min(\frac{x_i}{i}))$, I just need to replace in the function $\endgroup$ – The Student Mar 11 '18 at 20:06
  • $\begingroup$ ^Grand. Once you do that, please add your solution as an answer, to serve as reference to future visitors with a similar question. $\endgroup$ – stochasticboy321 Mar 11 '18 at 22:38
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Note that the condition $x_i>i\theta$ is equivalent to ${x}_{i}/i>\theta$ then you can replace the indicator function with this last and take as suficient statistic the $min({x}_{i}/i)$

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