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I am trying to find a sufficient and complete statistics function for $0<\theta<1$ of a sample $X = X_1, \dots, X_n$ from the Geometric Distribution.

We have $f(x;\theta) = (1-\theta)^{x-1}\theta $. Thus :

$$p(x;\theta)=\prod_{i=1}^n(1-\theta)^{x_i-1}\theta=\theta^n\prod_{i=1}^n (1-\theta)^{x_i-1}$$

$$=$$

$$\theta^n(1-\theta)^{\sum_i^n (x_i-1)} = \theta^n\exp\bigg\{(1-\theta)\ln\bigg(\sum_{i=1}^n(x_i-1)\bigg)\bigg\}$$

$$=$$

$$\theta^n\exp\bigg\{(1-\theta)\sum_{i=1}^n\ln(x_i-1)\bigg\}$$

This is an expression of the form of the Exponential Distribution Family and since the support does not depend on $\theta$, we can conclude that it belongs in the exponential distribution family. Thus, a sufficient and complete statistics function for $\theta$, is :

$$\sum_{i=1}^n\ln(x_i-1) \longrightarrow T(x) = \sum_{i=1}^n\ln x_i$$

Question : Is my approach and my conclusion correct ? Thanks in advance for your time.

marked as duplicate by Xi'an, kjetil b halvorsen, Michael Chernick, Robert Long, jpmuc Jun 17 at 18:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • @Xi'an I am asking about a sufficient and complete, not for MVUE. Thus not duplicate. – Rebellos Jun 7 at 9:08
  • @Rebellos: Sufficiency is dealt with in the post referenced by Xi'an. – F. Tusell Jun 12 at 10:02

You have made an error while writing the exponent of $e$.

Define $\mathbb I(x)=\begin{cases}1&,\text{ if }x=1,2,3,\cdots\\0&,\text{ otherwise }\end{cases}$

Then joint density of $(X_1,X_2,\cdots,X_n)$ is \begin{align}f_{\theta}(x_1,x_2,\cdots,x_n)&=\prod_{i=1}^n\theta(1-\theta)^{x_i-1}\mathbb I(x_i)\\&=\theta^n(1-\theta)^{\sum_{i=1}^n x_i-n}\prod_{i=1}^n\mathbb I(x_i)\\&=\exp\left[n\ln \theta+\left(\sum_{i=1}^nx_i-n\right)\ln(1-\theta)+\sum_{i=1}^n\ln \mathbb I(x_i)\right]\\&=\exp\left[\ln(1-\theta)\sum_{i=1}^nx_i+n\ln\theta-n\ln(1-\theta)+\sum_{i=1}^n\ln \mathbb I(x_i)\right]\\&=\exp\left[\ln(1-\theta)\sum_{i=1}^nx_i+n\ln\left(\frac{\theta}{1-\theta}\right)+\sum_{i=1}^n\ln \mathbb I(x_i)\right]\end{align}

So we have expressed the joint density in the form$$f_{\theta}(x_1,\cdots,x_n)=\exp\left[a(\theta)\sum_{i=1}^nu(x_i)+b(\theta)+c(x_1,x_2,\cdots,x_n)\right]$$ This implies $\displaystyle\sum_{i=1}^nu(x_i)$ is our complete sufficient statistic for $\theta$, where $u(x)=x$ in this case.

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