# Clarifying the statement $f(x,y) = f_X(x) f_Y(y)$ for all $x, y$ (when $X$ and $Y$ are independent)

The textbook A First Course in Probability by Ross defines random variables $$X$$ and $$Y$$ to be independent when the following condition is satisfied: if $$A$$ and $$B$$ are subsets of $$\mathbb R$$, then $$\tag{0}P(X \in A, Y \in B) = P(X \in A) P(Y \in B).$$ (I think technically we should assume that $$X$$ and $$Y$$ are Borel, but Ross glosses over that issue.)

The textbook goes on to state that if $$X$$ and $$Y$$ are jointly continuous random variables with PDFs $$f_X$$ and $$f_Y$$ and joint PDF $$f$$, then $$\tag{1} f(x,y) = f_X(x) f_Y(y) \quad \text{for all } x,y.$$ However, I think this statement is not correct because $$f_X$$ and $$f_Y$$ are only defined up to sets of measure $$0$$. (For example, I think you could redefine $$f_X$$ arbitrarily on a set of measure $$0$$, and you would still have a perfectly valid PDF for $$X$$.)

Question: So how can we state equation (1) in a way that is correct (so that we are not speaking nonsense to students) but which is also understandable to undergrads?

Further details: Let $$F$$ be the joint CDF for $$X$$ and $$Y$$, and let $$F_X$$ and $$F_Y$$ be the CDFs for $$X$$ and $$Y$$ respectively. It can be shown that equation (0) is equivalent to $$F(a,b) = F_X(a) F_Y(b)$$ for all $$a, b \in \mathbb R$$. In other words, $$F(a,b) = \int_{-\infty}^a f_X(x) \, dx \int_{-\infty}^b f_Y(y) \, dy$$ for all $$a,b \in \mathbb R$$. If $$f_X$$ is continuous at $$a$$, then differentiating both sides with respect to $$a$$ yields $$\frac{\partial F}{\partial a} = f_X(a) \int_{-\infty}^b f_Y(y) \, dy.$$ If $$f_Y$$ is continuous at $$b$$, then differentiating with respect to $$b$$ yields $$\frac{\partial^2 F}{\partial a \partial b} = f_X(a) f_Y(b).$$ If we knew that $$\tag{2} \frac{\partial^2 F}{\partial a \partial b} = f(a,b)$$ then we would have established that $$f(a,b) = f_X(a) f_Y(b)$$.

So let's see what assumptions we need in order to conclude that $$\frac{\partial^2 F}{\partial a \partial b} = f(a,b)$$. We know that $$\tag{3} F(a,b) = \int_{-\infty}^a \int_{-\infty}^b f(x,y) \, dy \, dx.$$ If the function $$x \mapsto \int_{-\infty}^b f(x,y) \, dy$$ is continuous at $$a$$, then by the fundamental theorem of calculus differentiating both sides of (3) with respect to $$a$$ yields $$\frac{\partial F}{\partial a} = \int_{-\infty}^b f(a,y) \, dy.$$ If the function $$y \mapsto f(a,y)$$ is continuous at $$b$$, then differentiating with respect to $$b$$ yields $$\frac{\partial^2 F}{\partial a \partial b} F(a,b) = f(a,b).$$

So, in order to establish (2), we needed the following two assumptions:

1. The function $$x \mapsto \int_{-\infty}^b f(x,y) \, dy$$ is continuous at $$a$$.
2. The function $$y \mapsto f(a,y)$$ is continuous at $$b$$.

Question: What is a nice, simple assumption I can state which will guarantee that these two conditions are met? If $$f$$ is continuous at $$(a,b)$$, then the second assumption is satisfied. But I don't see that the continuity of $$f$$ at $$(a,b)$$ would guarantee that the first condition is satisfied.

• You could say that (1) holds for all $(x,y)\in U{\times} V$, for any open $U$ and $V$ for which $f_X$, $f_Y$, and $f$ are continuous on $U$, $V$, and $U{\times} V$ respectively. Most density functions seen in undergraduate courses are continuous, so you are not losing many examples this way. – kimchi lover Nov 26 '18 at 2:19
• @kimchilover Thank you, that is the kind of suggestion I'm looking for. – eternalGoldenBraid Nov 26 '18 at 3:59
• The correct statement is: $f(x, y) = f_X(x) f_Y(y)$ holds for almost every $x,y$ (with respect to 2-dimensional Lebesgue measure). – Song Nov 26 '18 at 7:47