# Why would split observed $x$ into two unobserved r.v $z_1,z_2$ consider a way to augmenting data in EM algorithm?

I am reading the materials on the EM algorithm, and I am a bit confused about the example provided on the material I am currently reading. The example is considered a classical missing data problem solving by EM algorithm. Here is the content of the material I am reading :

197 animals are distributed into four categories $$\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=(125,18,20,34)$$ and modeled with the multinomial distribution $$\text { multinomial }\left(n ; \frac{1}{2}+\frac{\theta}{4}, \frac{1}{4}(1-\theta), \frac{1}{4}(1-\theta), \frac{\theta}{4}\right)$$ Estimation is easier if the $$x_{1}$$ cell is split into two cells, so we create the augmented model $$\left(z_{1}, z_{2}, x_{2}, x_{3}, x_{4}\right) \sim \text { multinomial }\left(n ; \frac{1}{2}, \frac{\theta}{4}, \frac{1}{4}(1-\theta), \frac{1}{4}(1-\theta), \frac{\theta}{4}\right)$$ with $$x_{1}=z_{1}+z_{2}$$

In the introduction section of the EM algorithm, the setting is as follow:

Consider a sample of $$n$$ items, where $$n_{1}$$ of the items are observed while $$n_{2}=n-n_{1}$$ items are not observable Let $$Z_{j}$$ 's and $$X_{i}$$ 's be mutually independent and $$\begin{array}{ll} X_{1}, \ldots, X_{n_{1}} \sim g(x \mid \theta) & -\text { Observed Data } \\ Z_{1}, \ldots, Z_{n_{2}} \sim f(z \mid \theta) & -\text { Unobserved Data } \end{array}$$ $$(\boldsymbol{X}, \boldsymbol{Z})$$ is called the complete data, $$\boldsymbol{Z}$$ the augmented data, and $$\boldsymbol{X}$$ the incomplete data.

Here, there is a clear cut between augmented data and incomplete data. However, here I am rather confused: it seems $$(x_1,x_2,x_3,x_4)$$ will be the incomplete data while $$(z_1,z_2,x_2,x_3,x_4)$$ will be consider the complete data. I have no previous exposure to the concept of augmented data and I am very confused why simply introducing two other random variables would work.

Also, I know the pdf for multinomial distribution with four outcomes have expression $$f\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=\frac{n !}{x_{1} ! x_{2} ! x_{3} ! x_{4} !} \pi_{1}^{x_{1}} \pi_{2}^{x_{2}} \pi_{3}^{x_{3}} \pi_{4}^{x_{4}}$$

if we substitute $$x_1$$ with $$z_1+z_2$$, we have $$(\pi_1+c)^{z_1}(\pi_1+c)^{z_2}$$ but not $$\pi_1^{z_1}c^{z_2}$$, also, the factorial part of $$x_1!=(z_1+z_2)!\neq z_1!z_2!$$ How the substitution actually work here? What makes this substitution valid?

Here $$g(x_1,x_2,x_3,x_4)$$ is the distribution associated with observed data, whereas $$f(z_1,z_2,x_2,x_3,x_4)$$ is the distribution associated with the complete data. So, we are not substituting the variables $$z_1,z_2$$ into $$g$$. Rather, these are two different distributions and there is no reason why $$g(x_1,x_2,x_3,x_4) = f(z_1,z_2,x_2,x_3,x_4), \text{where,} \quad x_1 = z_1 + z_2$$ should hold in general.
The only relation between $$g$$ and $$f$$ is that $$g(x_1,x_2,x_3,x_4) = \sum_{z_1,z_2}f(z_1,z_2,x_2,x_3,x_4)$$ where the summation is over all possible $$z_1,z_2 \geq 0$$ such that $$x_1 = z_1 + z_2$$.