Sufficient statistic for p in a binom(1,p) I don't really understand why the statistic $T = \sum_{i=1}^{n} X_{i}$ is a sufficient statistic for p , for $X_{1}, ... X_{n}\sim^{iid} Binom(1,p)$.
Shouldn't it be $\bar{X}$, i.e. $\frac{T}{n}$?
 A: In general, 1-1 functions of sufficient statistics are still sufficient.
For some model $\mathcal{F}$, recall a statistic $T(y)$ is sufficient for $\theta$ if it takes the same value at two points $y,z$ only if $y$ and $z$ have equivalent likelihoods, i.e. if, for all $y,z\in\mathcal{Y}$ (the sample space),
$$
T(y)=T(z)\quad\Longrightarrow \quad L(\theta;y)\propto L(\theta;z)\qquad\text{for all }\theta\in\Theta
$$
Likelihoods are equivalent, denoted $L(\theta;y)\propto L(\theta;z)$ for all $\theta\in\Theta$, if the ratio $L(\theta;y)/L(\theta;z)$ is constant as a function of $\theta$. 
Define the partition of the sample space $\mathcal{Y}$ where the statistic $T(y)$ takes the value $t$:
$$
A_t=\{y:y\in\mathcal{Y},T(y)=t\}.
$$
For a sufficient statistic $T(y)$ consider an element of $A_t$. Then, for $y,z\in\mathcal{Y}$
$$
y,z\in A_t\quad\Longrightarrow\quad T(y)=T(z)=t\quad\Longrightarrow\quad L(\theta;y)\propto L(\theta;z)
$$
We can also look at induced partitions of one-to-one functions $U$ of $T(y)$,
$$y,z\in A_{U(t)}\Longrightarrow U(T(y))=U(T(z))=U(t)$$
This implies that one-to-one functions $U(T(y))$ of $T$ are also sufficient statistics, because 
$$
\begin{array}{ccccr}
y,z\in A_{U(t)}&\Longrightarrow&U(T(y))=U(T(z))=U(t)&\\[.25cm]
\Updownarrow&&\Updownarrow&\\[.25cm]
y,z\in A_{t}&\Longrightarrow&T(y)=T(z)=t&\Longrightarrow& L(\theta;y)\propto L(\theta;z)
\end{array}
$$
