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

  • 4
    $\begingroup$ 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

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 $min({x}_{i}/i)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.