How to prove something is a sufficient statistic? If you have $n$ random variables that are iid with density $\frac{1}{p}e^{-x/p}$, how do you show that the sum of the $x_i$'s is a sufficient statistic?
Attempt: Take likelihood function and express in terms of $g(p)h(x)$ and use factorization theorem to show that it is a sufficient statistic.
So likelihood = $\frac{1}{p^n} \exp(\sum \frac{-x_i}{p}) = \frac{1}{p^n} \exp(\frac{1}{p}  \sum (-x_i))$.
 A: These are exponential random variables. Consider the joint probability density. This would be 
$$
\prod_{i=1}^{n} \frac{1}{p}e^{-\frac{x_i}{p}}=p^{-n}e^{-\frac{1}{p}T(x)},\qquad T(x)= \sum_{i=1}^{n} x_i$$
Let $h(x) =1$ and $g(p,t) = p^{-n}e^{-t/p}$. Then the result follows from the factorization theorem.
A: Let $y$ represent observations and let $x$ be the latent variable of interest, then we have the following equivalent statements that can be helpful in proving that $t=t(y)$ is sufficient statistic wrt to our model $p_y(y;x)$
Non-Bayesian Viewpoint


*

*Definition: $p_{y|t}(y|t; x)$ is not a function of x  

*Proportional likelihoods (same ML estimate): $L_y(x) \propto L_t(x)$   

*Neyman factorization theorem: $p_y(y;x) = a(t(y),x)\times b(y)$
Bayesian Viewpoint  


*

*Conditioning on $t$ removes dependence on $x$: $p_{y|t,x}(y|t(y),x)
    = p_{y|t}(y|t(y))$  

*Same posteriors: $p_{x|y}(x|y) = p_{x|t}(x|t(y))$

*Data processing inequality $I(X;Y) \geq I(X;Z)$ for a Markov Chain
$X\rightarrow Y\rightarrow Z$ is satisfied with equality: $I(X;Y) =
I(X;T(Y))$ if $Z = T(Y)$, i.e. sufficient statistic of data $Y$. 


For example for an exponential family  
$p_y(y;x) = \exp \{\lambda(x) t(y) + \beta(y) - \alpha(x)\}$ 
where $\lambda(x)$ are natural parameters, $t(y)$ is sufficient statistic, $\beta(y)$ is log-base function and $\alpha(x)$ is log-normalizer, we can show that $t(y)$ is sufficient statistic by applying Neyman factorization theorem:
$p_y(y;x) = \exp \{\lambda(x) t(y) - \alpha(x)\} \times \exp\{\beta(y)\} = a(t(y), x) \times b(y)$
