# Sufficient statistic of a discrete distribution that can take $n$ values.

Use factorization criterion to determine a sufficient statistic bassed on a sample of size N, where each observation come frome a family of distributions of a random variable that can take a finite list of values $$x_1,\cdots,x_n$$ with probability $$p_1,\cdots,p_n,$$ respectively.

My attempt

The factorization theorem establish that for a random sample $$\vec{x}=(x_1,\cdots,x_n)$$, is possible write: $$f(\vec{x}|\theta)=g(T(\vec{x})|\theta)h(\vec{x}),\ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$

where $$f(\vec{x}|\theta)$$ is the sample density, $$g(T(\vec{x})|\theta)$$ is a function that depends on theta and $$h(\vec{x})$$ does not depend on $$\theta.$$

One way to write the discrete density of given random variable is via indicator functions as follows

$$f(X=y_1|p_1,\cdots,p_n)=p_1{\bf 1}_{x_1}(y_1)+\cdots p_n{\bf 1}_{x_n}(y_1).$$

so I'm trying to write the sample distribution of an arbitrary sample $$y_1,y_2,\cdots,y_N$$:

$$\begin{eqnarray} f(y_1,\cdots,y_N|p_1,\cdots,p_n)&=&\displaystyle\Pi_{i=1}^Nf(y_i|p_1,\cdots,p_n)\\ &=&\displaystyle\Pi_{i=1}^N(\sum_{i=1}^n p_i{\bf{1}}_{x_i}(y_i)). \end{eqnarray}$$

It's possible to write the previous equation in the form of (1)?.

There is not suitable to write density in the form of a sum. Rewrite it as a product: $$f(X=y_1|p_1,\cdots,p_n)=p_1^{{\bf 1}_{x_1}(y_1)}\cdots p_n^{{\bf 1}_{x_n}(y_1)}.$$ And then $$\begin{eqnarray} f(y_1,\cdots,y_N|p_1,\cdots,p_n)&=&\prod_{i=1}^Nf(y_i|p_1,\cdots,p_n)\\ &=&\prod_{i=1}^N\prod_{j=1}^n p_j^{{\bf{1}}_{x_j}(y_i)} \\ &=& \prod_{j=1}^n p_j^{\sum_{i=1}^N{\bf{1}}_{x_j}(y_i)}. \tag{2} \end{eqnarray}$$
This is a function that depends on $$p_1,\ldots,p_n$$ and on a sufficient statistics $$T(y_1,\ldots,y_n) = \left(\sum_{i=1}^N{\bf{1}}_{x_1}(y_i),\ldots,\sum_{i=1}^N{\bf{1}}_{x_{n-1}}(y_i)\right).$$ Note that really there are $$n-1$$ unknown parameters since $$p_n=1-p_1-\ldots-p_{n-1}$$. Also $$T$$ is $$(n-1)$$-dimensional since the missed last sum of indicators can be found from the others: $$\sum_{i=1}^N{\bf{1}}_{x_n}(y_i) = N-\sum_{j=1}^{n-1}\sum_{i=1}^N{\bf{1}}_{x_j}(y_i).$$ The expression (2) can be rewritten as $$f(y_1,\cdots,y_N|p_1,\cdots,p_n) = \prod_{j=1}^{n-1} p_j^{\sum_{i=1}^N{\bf{1}}_{x_j}(y_i)}\times (1-p_1-\ldots-p_{n-1})^{N-\sum_{j=1}^{n-1}\sum_{i=1}^N{\bf{1}}_{x_j}(y_i)}$$ $$=g(T,p_1,\ldots,p_{n-1})\cdot \underbrace{h(y_1,\ldots,y_n)}_1.$$