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? $$ f(x_1,\cdots,x_n) = f_{X_1}(x_1) f_{X_2|X_1}(x_2|x_1)f_{X_3|X_2,X_1}(x_3|x_2,x_1)\cdots f_{X_n|X_{n-1},\cdots ,X_{1}}(x_n|x_{n-1},\cdots,x_{1}) ? $$

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    $\begingroup$ I'm confused. What is your definition of $f_{X_i | X_{i-1},\dots,X_1}(x_i | x_{i-1},\dots,x_1)$? Your question is written so weirdly. Asking if some equation is true, and then asking what the stuff in the equation means. If we take the definition you gave in (2), then (1) definitely holds. $\endgroup$ – mathworker21 Dec 16 '19 at 4:44
  • $\begingroup$ Hehe. Alright. Please let me rewrite it more to your taste. $\endgroup$ – Mikkel Rev Dec 16 '19 at 13:55

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 $h=g(X)$.

$$\int_C hdP = \int_B g(x)d\mu_X = \int 1_B(x) g(x)d\mu_X$$ $$ = \int 1_B(x) \int yf_{Y|X}(y|x)dy d\mu_X = \int\int 1_B(x) yf_{Y|X}(y|x)dy f_X(x)dx$$ $$= \int\int 1_B(x) yf_{Y|X}(y|x)f_X(x)dxdy = \int\int 1_B(x) yf(x, y)dxdy$$ $$ = E(1_B(X)Y) = E(1_CY) = \int_C Y dP.$$

  • $\begingroup$ in the second line, by "we only need to", do you mean "it suffices to"? the direction of the logical implication of "need" goes opposite to that of "suffice". $\endgroup$ – mathworker21 Dec 17 '19 at 9:27
  • $\begingroup$ @mathworker21 - ah, but "ONLY need" = "need ONLY" = "it suffices"? ;) $\endgroup$ – antkam Dec 17 '19 at 16:59
  • $\begingroup$ Thank you both very much for the suggestions. I made the edits. $\endgroup$ – Xiaohai Zhang Dec 17 '19 at 18:26
  • $\begingroup$ @antkam I object to last equality. let's say I want to prove there are no solutions to $x^3+y^3 = z^3$ in positive integers. it suffices to prove fermat's last theorem, but i wouldn't say i "need only" prove it, since I don't need to prove it. The point is that "need only" still is saying I need to do something, which is very different from something sufficing. Also, ask a good question on this site please :). I want to ask one myself soon... $\endgroup$ – mathworker21 Dec 17 '19 at 20:11
  • $\begingroup$ @XiaohaiZhang +1 $\endgroup$ – mathworker21 Dec 17 '19 at 20:12

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