A statistic t=T(X) is sufficient for the parameter \theta, if p(X=x | T(X) = t) does not depend on \theta.
This means that t has the same information against the parameter \theta.
Let's consider an example.
Let X \overset{i.i.d}{\sim} \mathrm{Bern}(\theta), the random variables X = \{X_1, \dots, X_N\} , and the realizations x = \{x_1, \dots, x_N\}.
\begin{align*}
p(X=x) &= p(X_1=x_1, \dots, X_N=x_N) \\
&= \prod_{n=1}^N p(X_n = x_n; \theta) \\
&= \theta^{\sum_n^N x_n} (1 - \theta)^{N - \sum_n^N x_n}
\end{align*}
In this case, let T(X) = \sum_n^N x_n then $T(X) \overset{i.i.d}{\sim} \mathrm{Bin}(N, \theta) $ .
\begin{align*}
p\left( T(X)=t; \theta \right) = \binom{N}{t} \theta^t (1 - \theta)^{N - t}
\end{align*}
\begin{align*}
p\left( X=x | T(x) = t \right) &= \frac{p\left( X=x, T(X) = t \right)}{p\left(T(x) = t\right)} \\
&= \frac{\theta^t (1 - \theta)^{N - t}}{\binom{N}{t} \theta^t (1 - \theta)^{N - t}} \\
&= \frac{1}{\binom{N}{t}} \\
\end{align*}
p\left( X=x | T(x) = t \right) does not depend on \theta, so T(X) = \sum_n^N x_n is a sufficient statistic.
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