Let $X_1, X_2,... X_n$ denote a random sample from a geometric distribution with the parameter $\theta$. Show that $\sum_{i=1}^n X_i$ is a sufficient statistic for $\theta$.

I know that a statistic is sufficient if the conditional distribution does not depend on $\theta$.

So I have $f(x;\theta)=(1-\theta)^{x-1}\theta$.

When i try to get the conditional distribution, I do not get anything that will cancel out theta when divided. Am I missing something or what is the conditional distribution of this geometric distribution?


Let the indicator function be. $\mathbb I(x)=\begin{cases}1&,\text{ if }x=1,2,3,\cdots\\0&,\text{ otherwise }\end{cases} $.

Now constructing the joint distribution $ \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}$

This implies $\displaystyle\sum_{i=1}^nx_i$ is sufficient statistic after comparing with the standard exponentially representation. Hope this helps.


Using Fisher–Neyman factorization Theorem we know that a statistic $T_n$ is sufficient if and only if we can write the likelihood $$L(x_1,\dots, x_n, \theta) =g(\theta,T_n)\cdot h(x_1,\dots, x_n)$$ as the product of a function $g$ which depends only on $\theta$ and $T_n$ and a function $h$ which depends only on the $x_i$ (it can be the Identity function as well of course).

In our example $$L(x_1,\dots, x_n, \theta) = \theta^n(1-\theta)^{\sum_{i=1}^n x_i-n}\prod_{i=1}^n\mathbb I(x_i) = \theta^n(1-\theta)^{T_n-n}\prod_{i=1}^n\mathbb I(x_i)$$

with $g(\theta,T_n)= \theta^n(1-\theta)^{T_n-n}$ and $h(x_1,\dots, x_n) = \prod_{i=1}^n\mathbb I(x_i)$.


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