Random vectors and sufficient statistic

Question

Let $X_1,\dots, X_n$ be a random sampling of a discrete random variable. Let $\theta$ a parameter. It is easy to see that $$ T(X_1,\dots,X_n) = (X_1,\dots,X_{n-1}) $$ is a statistic. My problem is when $T$ is sufficient? I don't know how to use the Neyman-Pearson theorem in that case or if it's a good idea to use it. Thank you for the help!

Answer 1

$T(\vec{X})$ is sufficient statistic if

$$ F (\vec{x}|T(\vec{X})=t, \theta) = F (\vec{x}|T(\vec{X})=t) $$

By definition

$$ \begin{aligned} F(\vec{x}|T(\vec{X})=(t_1, ..., t_{n-1}),\ \theta) &= \mathbb{P}(X_1 < x_1, ..., X_n < x_n |X_1 = t_1, ..., X_{n-1} = t_{n-1},\ \theta)\\ &= \begin{cases} 0, &\exists i:t_i > x_i\\ \mathbb{P}(X_n < x_n |\ \theta) &\text{if } \forall i: t_i < x_i \end{cases}\\ &= \begin{cases} 0, &\exists i:t_i > x_i\\ F(x_n|\theta) &\text{if } \forall i: t_i < x_i \end{cases} \end{aligned} $$

If $(X_1, ..., X_n)$ are independent, then $T(X_1, ..., X_n)$ is not sufficient in any case.