Sufficient Statistics and Discrete Distributions Let $X_1, \ldots, X_n$ be a random sample of size $n$ from the following distribution:
$$f(x;\theta) = \left\{\begin{array}{ccc} \frac{1 - \theta}{6} & , & x = 1 \\ \frac{1 + \theta}{6} & , & x = 2 \\ \frac{2 - \theta}{6} & , & x = 3 \\ \frac{2 + \theta}{6} & , & x = 4\end{array}\right.$$
where $-1 < \theta < 1$. Find a minimal sufficient statistic for the parameter $\theta$.
Answer: I am attempting to use the Neyman Theorem:
$$f(x_1;\theta)\cdots f(x_n;\theta) = k_1\Big[u_1(x_1,\ldots, x_n); \theta\Big]k_1(x_1,\ldots, x_n)$$
So,
\begin{eqnarray*}
f(x_1;\theta)\cdots f(x_n;\theta) & = & \prod\limits_{i = 1}^n \left(\frac{1 - \theta}{6}\right)^{n_1}\left(\frac{1 + \theta}{6}\right)^{n_2}\left(\frac{2 + \theta}{6}\right)^{n_3}\left(\frac{2 - \theta}{6}\right)^{n_4}
\end{eqnarray*}
where $n = n_1 + n_2 + n_3 + n_4$.
However, I do not seem to be able to form $k_1$ and $k_2$ from this, neither am I able to obtain the sufficient statistic $u_1$. Do the $x$-values 1, 2, 3, 4 even play a role here?
 A: When looking at the joint density as [note the erroneous inclusion of the product symbol!]
\begin{eqnarray*}\require{cancel}
f(x_1;\theta)\cdots f(x_n;\theta) & = & \cancel{\prod\limits_{i = 1}^n} \left(\frac{1 - \theta}{6}\right)^{n_1}\left(\frac{1 + \theta}{6}\right)^{n_2}\left(\frac{2 + \theta}{6}\right)^{n_3}\left(\frac{2 - \theta}{6}\right)^{n_4}
\end{eqnarray*}
the factorisation is already achieved, starting from the joint density as
\begin{align*}
f(x_1;\theta)&\cdots f(x_n;\theta) = \prod\limits_{i = 1}^n \left(\frac{1 - \theta}{6}\right)^{\mathbb I_1(x_i)}\left(\frac{1 + \theta}{6}\right)^{\mathbb I_2(x_i)}\left(\frac{2 + \theta}{6}\right)^{\mathbb I_3(x_i)}\left(\frac{2 - \theta}{6}\right)^{\mathbb I_4(x_i)}\\
&= \left(\frac{1 - \theta}{6}\right)^{\sum_{i=1}^n\mathbb I_1(x_i)}\left(\frac{1 + \theta}{6}\right)^{\sum_{i=1}^n\mathbb I_2(x_i)}\left(\frac{2 + \theta}{6}\right)^{\sum_{i=1}^n\mathbb I_3(x_i)}\left(\frac{2 - \theta}{6}\right)^{\sum_{i=1}^n\mathbb I_4(x_i)}\\
&= \left(\frac{1 - \theta}{6}\right)^{n_1}\left(\frac{1 + \theta}{6}\right)^{n_2}\left(\frac{2 + \theta}{6}\right)^{n_3}\left(\frac{2 - \theta}{6}\right)^{n_4}
\end{align*}
which only depends on the four counters $n_1(\mathbf x),\ldots,n_4(\mathbf x)$, the factorisation exhibits the statistic $$S(X_1,\ldots,X_n)=\left(\sum_{i=1}^n\mathbb I_1(x_i),\sum_{i=1}^n\mathbb I_2(x_i),\sum_{i=1}^n\mathbb I_3(x_i),\sum_{i=1}^n\mathbb I_4(x_i)\right)$$
as being sufficient (if not minimal) since the product only depends on these four quantities. To find a further decomposition as $k_1(S(\mathbf X);\theta)k_2(\mathbf X)$ is somewhat moot (for instance $k_2(\mathbf x)=1$).
