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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.

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    $\begingroup$ 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. $\endgroup$ – kimchi lover Nov 26 '18 at 2:19
  • $\begingroup$ @kimchilover Thank you, that is the kind of suggestion I'm looking for. $\endgroup$ – eternalGoldenBraid Nov 26 '18 at 3:59
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    $\begingroup$ 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). $\endgroup$ – Song Nov 26 '18 at 7:47

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