I have a vector $x$ of random variables and a finite set of classes $C$. Also $x$ has to belong to one class. In the context of classification problems I have the transformation
$P(x) = P(x|c_1)P(c_1) + P(x|c_2)P(c_2) \ \cdots P(x|c_n)P(c_n)= \sum_{c \in C}P(x|c)P(c)$
Why is this equality true?
Thank you very much!
On
It is due to the Law of Total Probability.
If the members of a set, $\{B_i\}$, are mutually exclusive and exhaustive events, then for any event $A$:
$$P(A) = \sum\limits_{B_i \in \{B_i\}} P(B_i)P(A\mid B_i)$$
Since each instance of $\vec X$ belongs to one and only one of the classes in $C$, we do have a set of mutually exclusive and exhaustive events.
$$\begin{align}\mathbb{\large P}(\vec X=\vec x) & = \mathbb{\large P}(\mathop{\large\cup}_{c\in C} (\vec X \in c\cap \vec X=\vec x)) \\ & = \sum\limits_{c \in C} \mathbb{\large P}(X\in c)\cdot \mathbb{\large P}( \vec X=\vec x \mid \vec X\in c) \\ & \color{grey}{= \mathbb{\large P}(\vec X\in c_1)\mathbb{\large P}(\vec X=\vec x\mid \vec X\in c_1)+\cdots+\mathbb{\large P}(\vec X\in c_n)\mathbb{\large P}(\vec X=\vec x\mid \vec X\in c_n) }\end{align}$$
In this answer $X$ stands for a random vector and $x$ is one of the values that it can take. I am not sure whether this meets your question (as my comment makes clear).
Maybe it will be of help anyhow. If not then let me know. In that case I will delete it.
Formally $P\left(X=x\mid c_{i}\right)P\left(c_{i}\right)=P\left(X=x\wedge c_{i}\right)$ where $c_{i}$ stands for the event that $X$ belongs to class $i$.
Then $P\left(X=x\mid c_{1}\right)P\left(c_{1}\right)+\cdots+P\left(X=x\mid c_{n}\right)P\left(c_{n}\right)=P\left(X=x\wedge c_{1}\right)+\cdots+P\left(X=x\wedge c_{n}\right)$
Also $P\left(X=x\right)=P\left(\left[X=x\wedge c_{1}\right]\vee\cdots\vee\left[X=x\wedge c_{n}\right]\right)=P\left(X=x\wedge c_{1}\right)+\cdots+P\left(X=x\wedge c_{n}\right)$
This because $X$ must belong to exactly one class.
So LHS of first equality equals LHS of second equality.