Given a particular form, i can verify whether it is sufficient statistic or not using $\frac{p_\theta(x_1,x_2...x_n)}{p_\theta(T(x_1,x_2...x_n))}$ is independendent of $\theta$ then i can say $T(\bar X)$ is sufficient statistic. But if that particular form is not given, and if only distriubtion is given, then thinking of suitable form, i am finding difficulty. While i was refering to wiki, some distriubtions belong to exponential family. If i can express given distriubtion in exponential family form, then i feel i can find the sufficient statistic. But for what (ie., T(X) represent sufficient statistic for what function of $\theta$) i am not able to identify.

For ex.,

To find sufficient statistic for Binomial distribution, i did following

$ P_{\theta}(x_1,x_2...x_n) ={n \choose \sum_i x_i} \theta^{\sum_i x_i } (1-\theta)^{n-\sum_ix_i} \\ = {n \choose \sum_i x_i} e^{\sum_ix_i \ln \theta +(n-\sum_i x_i) \ln(1-\theta) } $

comparing this with exponential form $P_w(X) = \frac{e^{w^t\phi(X)}}{Z(w)}$

i get the following

Parameter $w= [\ln \theta, \ln (1-\theta)]^t$ and sufficient statistic as $\phi(\bar X) = (\sum_i x_i, -\sum_i x_i)$

But how to find what $\sum_i x_i $ is the sufficient statistic for what function of parameter $\theta - \sum_i x_i$ is sufficient statistic for $\theta$ or $\ln \theta$ or $-\ln (1-\theta)$

  • 2
    $\begingroup$ There is only one parameter and one sufficient statistic in this case. In the first line where you write $P_{\theta}(x_1,x_2,...,x_n)=...$, you already have it written so that it depends on $x_1,x_2,...,x_n$ only through the sum of those variables. Thus, the factorization theorem tells you the sum is a sufficient statistic. $\endgroup$ – John L Mar 22 at 15:07
  • $\begingroup$ sir i know it is sufficient statistic. But sum x_i represent sufficient statistic for what function to $\theta$. For some other distributions i am not able to identify. For ex., $f(x)=e^{-(x-\theta)}, \theta<x< \infty$. For this unless i saw the answer as min order statistic, i could not guess sufficient statistic. So i was thinking in this way. Further if u know any other way, kindly elaborate. I am a beginner. Request your patience $\endgroup$ – Nascimento de Cos Mar 22 at 19:09
  • $\begingroup$ If $T(X)$ is a sufficient statistic for a parameter $\theta$, then it is also sufficient for any function of $\theta$. See this question. $\endgroup$ – John L Mar 23 at 14:24

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