Independence means the joint law is the product of the marginals

Statement

Let $X$ and $Y$ be random variables on one probability space, with values in measurable spaces $(E_1, \mathcal{E}_1)$ and $(E_2, \mathcal{E}_2)$. Let $\mu_X$ and $\mu_Y$ be their laws, and let $\mu_{(X, Y)}$ be the joint law of the pair $(X, Y)$ on $E_1 \times E_2$. Then $X$ and $Y$ are independent if and only if $\mu_{(X, Y)} = \mu_X \otimes \mu_Y$, the product measure of the two marginal laws.

Remarks

The statement is a direct translation of the definition: $\mu_{(X, Y)}(A \times B) = P(X \in A, Y \in B)$ and $(\mu_X \otimes \mu_Y)(A \times B) = \mu_X(A) \, \mu_Y(B)$. Two probability measures on $E_1 \times E_2$ that agree on all rectangles $A \times B$ are equal, because the rectangles form a π-system that generates the product σ-algebra. For real random variables, it follows that $F_{(X, Y)}(x, y) = F_X(x) \, F_Y(y)$. If $X$ and $Y$ have densities, the joint density is $f_X(x) \, f_Y(y)$. The same result holds for $n$ random variables: they are independent if and only if their joint law is $\mu_{X_1} \otimes \dots \otimes \mu_{X_n}$.

Explanations

L1 human

For independent variables, the full joint picture is fixed by the two separate pictures. You find the chance of a pair of outcomes by multiplying the chances of the two outcomes.