Product rule of continuous random variables I have found loads of resources on the product rule, but none are on $n$ joint continuous random variables, but I suspect that it should still hold? Suppose $X_1,\cdots,X_n$ are continuous random variables with joint pdf $f(x_1,\cdots,x_n)$. If $f_{X_i|X_{i-1}\cdots,X_1}(x_i|x_{i-1},\cdots, x_{1})$ is the probability distribution of $X_i$ in the case that we have already observed that $X_{i-1} = x_{i-1},X_{i-2} = x_{i-2}, \cdots, X_{1} = x_{1}$, do we have the following decomposition?
$$
f(x_1,\cdots,x_n) = f_{X_1}(x_1) f_{X_2|X_1}(x_2|x_1)f_{X_3|X_2,X_1}(x_3|x_2,x_1)\cdots f_{X_n|X_{n-1},\cdots ,X_{1}}(x_n|x_{n-1},\cdots,x_{1}) ?
$$ 
 A: We first observe that the formula for $n$ variables can be easily derived iteratively from that of two variables, i.e. it suffices to show
$$ f(x, y)=f_X(x)f_{Y|X}(y|x). $$
In non-rigorous theory (classical theory), the above formula directly follows from the defintion of (pdf for) conditional distributions, i.e. $f_{Y|X}(y|x)\equiv \frac{f(x, y)}{f_X(x)}$ where $f_X(x)=\int f(x, y)dy $ is the marginal distribution. The conditional distribution makes sense because it is non-negative and integrates to 1, hence it is a pdf of certain random variable. It is interpreted as the distribution conditioned on $X=x$.
In rigorous measure-theoretical formulation, $f_{Y|X}(y|x)=\frac{f(x, y)}{f_X(x)}$ is still simply a definition of a version of pdf, and it is related to $E(Y|X)\equiv E(Y|\sigma(X))$ by
$$ E(Y|X) = g(X) \equiv \int yf_{Y|X}(y|X)dy.\quad (1) $$
Definition. Let $(\Omega, \mathcal{F}, P)$ be a probability space, and let
$\mathcal{C} \subset \mathcal{F}$ be a sub-$\sigma$-field. Let $Y$ be a random variable that is $\mathcal{F}/\mathcal{B}$ measurable and $E|Y| < \infty$.
We use the symbol $E(Y|\mathcal{C})$ to stand for any function $h : \Omega \rightarrow \mathcal{R}$ that is $\mathcal{C}/\mathcal{B}$ measurable
and that satisfies
$$\int_C hdP = \int_C YdP, \text{ for all } C \in \mathcal{C}.\quad (2)$$
We call such a function $h$, a version of the conditional expectation of $Y$ given $\mathcal{C}$.
There are theorems that state the aforementioned conditional expectation exists and is unique. In our problem at hand, we have $\mathcal{C} = \sigma(X)$ and $\mathcal{F} = \sigma(X, Y)$.
Proof of (1): Let $C \in \mathcal{C}\equiv\sigma(X)$. Then there exists $B \in \mathcal{B}$ so that $C=X^{-1}(B)$. To prove (1), it suffices to show (2) holds for $h=g(X)$.
$$\int_C hdP = \int_B g(x)d\mu_X = \int 1_B(x) g(x)d\mu_X$$
$$ = \int 1_B(x) \int yf_{Y|X}(y|x)dy d\mu_X  = \int\int 1_B(x) yf_{Y|X}(y|x)dy f_X(x)dx$$
$$= \int\int 1_B(x) yf_{Y|X}(y|x)f_X(x)dxdy = \int\int 1_B(x) yf(x, y)dxdy$$
$$ = E(1_B(X)Y) = E(1_CY)  = \int_C Y dP.$$
