Sufficient Statistic: Fisher Theorem | Bernoulli(p)

[Julio1]

Let $\underline{X}=X_1,...,X_n$ an random sample of size $n$. Show that the statistic $T=\displaystyle\sum_{i=1}^n X_i$ is an sufficient statistic for $p$ if $X\sim Bernoulli(p)$ with $0<p<1.$

Hello :). I think that can use the Fisher Theorem (Factorization) for find the statistic and therefore show that is an sufficient statistic. Truth?

 

[Valkarie]

Yes that is correct. The Fisher Factorization theorem states that if a sample $\underline{X}$ is independent and identically distributed, then the joint probability distribution $f(\underline{X})$ can be written as a product of functions of the individual variables $X_i$ and functions of sufficient statistics $T$: $$f(\underline{X})=h(\underline{X})\cdot g(T)$$In this case, we have a Bernoulli random variable with parameter $p$ which has a probability mass function of $$f(x)=p^x (1-p)^{1-x}$$Since $X_i$ are independent and identically distributed, we can factorize the joint probability mass function as$$f(\underline{X})=\prod_{i=1}^n p^{X_i} (1-p)^{1-X_i} = p^{\sum_{i=1}^n X_i} (1-p)^{n-\sum_{i=1}^n X_i} = h(\underline{X})\cdot g(T)$$where $T = \sum_{i=1}^n X_i$ is the sufficient statistic for $p$.