What is the theoretical definition in terms of $\omega \in \Omega$ for independence in probability?

Question

I am wondering if the independence of two random variables $X,Y$ can be stated as follows:

The joint distribution of $X$ and $Y$ is: $P(\omega\in\Omega: X(\omega)=x, Y(\omega)=y)$. Hence, the joint probability is calculating the probability of the set $\omega \in \Omega$ under the "constraint" that $\omega$, when "plugged into" $X(\omega),Y(\omega)$ will simulataneously, cause $X(\omega)=x$ and $Y(\omega) = y$.

Then, if the set of $\omega\in\Omega$ that cause $X(\omega)$ to be equal to $x$ is disjoint from the set of $\omega'\in\Omega$ that cause $Y(\omega') = y$, then it seems I can separate them into two different sets and so I can factor the probabilities apart, i.e.:

$$ P(\omega\in\Omega: X(\omega)=x, Y(\omega)=y) = P(\omega\in\Omega: X(\omega)=x)P(\omega'\in\Omega: Y(\omega)=y) $$

Is this an accurate portrayal of what is going on? Thanks.

Answer 1

No, it's completely wrong.

First of all, not all random variables are discrete. For a continuous random variable, $P(X = x) = 0$ for all $x$.

Second, except for trivial cases, those sets are not disjoint. If they were disjoint, you'd have $P(X=x, Y=y) = 0$, not $P(X=x) P(Y=y)$.

Answer 2

You want, for any intervals $\cal A, \beta$ that are within the respective supports of independent random variables $X,Y$, that: "the probability that the outcome is in the intersection of the relevant preimages of the intervals is equal to the product of the probabilities that the outcome is in the preimage of each interval."

$$\begin{align}\mathbb P\{\omega\in \Omega: X(\omega)\in\mathcal A, Y(\omega)\in\beta\}~&=~\mathbb P\{\omega\in\Omega:X(\omega)\in\mathcal A\}\cdot\mathbb P\{\omega\in\Omega:Y(\omega)\in\beta\} \\[1ex] \mathbb P\bigl({X}^{-1}(\mathcal A)\cap {Y}^{-1}(\beta)\bigr)~&=~\mathbb P\bigl({X}^{-1}(\mathcal A)\bigr)\cdot\mathbb P\bigl({Y}^{-1}(\beta)\bigr) \end{align}$$

Though more usually abbreviated as: $$\begin{align}\mathsf P(X\in\mathcal A, Y\in\beta)~&=~\mathsf P(X\in\mathcal A)\cdot\mathsf P(Y\in\beta)\end{align}$$

Or sometimes, $\mathsf P_{X,Y}(\mathcal A,\beta)~=~\mathsf P_X(\mathcal A)\cdot \mathsf P_Y(\beta)$

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Compare this to the definition of independence for events.   Events $A,B$ are independent iff $\Pr (A\cap B)=\Pr (A)\cdot\Pr(B)$ .

Note: The preimages are not disjoint sets, else the measure of their intersection would be zero rather than the product of their measures.

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The left semi-infinite intervals generate the cumulative distribution functions for real-valued random variables.

$$\begin{align}\mathbb P\{\omega\in \Omega: X(\omega)\leqslant x, Y(\omega)\leqslant y\}~&=~\mathbb P\{\omega\in\Omega:X(\omega)\leqslant x\}\cdot\mathbb P\{\omega\in\Omega:Y(\omega)\leqslant y\} \\[1ex] \mathbb P\bigl(X^{-1}(-\infty;x]\cap Y^{-1}(-\infty;y]\bigr)~&=~\mathbb P\bigl(X^{-1}(-\infty;x]\bigr)\cdot\mathbb P\bigl(Y^{-1}(-\infty;y]\bigr)\\[1ex] \mathsf P(X\leqslant x,Y\leqslant y)~&=~ \mathsf P(X\leqslant x)\cdot \mathsf P(Y\leqslant y)\\[1ex] F_{X,Y}(x,y)~&=~ F_X(x)\cdot F_Y(y) \end{align}$$

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When the random variables are discrete, this also applies to the atoms.   The probability measures of these are called the probability mass functions.$$\begin{align}\mathbb P\{\omega\in \Omega: X(\omega)=x, Y(\omega)=y\}~&=~\mathbb P\{\omega\in\Omega:X(\omega)=x\}\cdot\mathbb P\{\omega\in\Omega:Y(\omega)=y\} \\[1ex] \mathbb P\bigl(X^{-1}\{x\}\cap Y^{-1}\{y\}\bigr)~&=~\mathbb P\bigl(X^{-1}\{x\}\bigr)\cdot\mathbb P\bigl(Y^{-1}\{y\}\bigr)\\[1ex] \mathsf P(X=x,Y=y)~&=~ \mathsf P(X=x)\cdot \mathsf P(Y=y)\\[1ex] \mathsf P_{X,Y}(x,y)~&=~ \mathsf P_X(x)\cdot \mathsf P_Y(y) \end{align}$$

When the random variables are continuous, the atoms correspond to null sets (ie have zero probability measure), but we wish the property to carry onto the derivatives (called the probability density functions) when they exist.

$$\begin{align}\frac{\mathrm d^2~~}{\mathrm d y~\mathrm d x}\mathbb P\{\omega\in \Omega: X(\omega)\leqslant x, Y(\omega)\leqslant y\}~&=~\frac{\mathrm d~~}{\mathrm d x}\mathbb P\{\omega\in\Omega:X(\omega)\leqslant x\}\cdot\frac{\mathrm d~~}{\mathrm d y}\mathbb P\{\omega\in\Omega:Y(\omega)\leqslant y\} \\[2ex] F^{(1,1)}_{X,Y}(x,y) ~&=~ {F}'_X(x)\cdot {F}'_Y(y) \\[2ex] f_{X,Y}(x,y)~&=~ f_X(x)\cdot f_Y(y)\end{align}$$