# how to show sufficient statistic

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 $$h(x_1,\dots, x_n) = \prod_{i=1}^n\mathbb I(x_i)$$.