Independence

Two events are independent when knowing that one happened tells you nothing about the other. The clean way to say this is multiplicative — the chance of both is the product of the chances:

\mathbb{P}(A \cap B) \;=\; \mathbb{P}(A)\,\mathbb{P}(B).

This is not a theorem; it is the definition. Whenever the joint probability factorises like this, the two events carry no information about one another.

A picture: the unit square

Let the whole sample space \Omega be the unit square, so area is probability. Make A a vertical strip of width \mathbb{P}(A) and B a horizontal strip of height \mathbb{P}(B). Their overlap A \cap B is a rectangle, and its area is the product of the side lengths:

\text{area}(A \cap B) \;=\; \mathbb{P}(A)\,\mathbb{P}(B).

Drag the sliders: however wide or tall you make the strips, the overlap's area is always the product — that is independence drawn in one rectangle.

Independent random variables

Random variables X, Y are independent when their joint distribution factorises into the marginals — for every x, y,

\mathbb{P}(X \le x,\, Y \le y) \;=\; F_X(x)\, F_Y(y), \qquad f_{X,Y}(x,y) = f_X(x)\, f_Y(y).

The headline consequence — the one that powers risk and portfolio calculations — is that independence lets expectations and variances split:

\mathbb{E}[XY] = \mathbb{E}[X]\,\mathbb{E}[Y], \qquad \operatorname{Var}(X + Y) = \operatorname{Var}(X) + \operatorname{Var}(Y).

Pairwise is not mutual

For three or more events, checking them two at a time is not enough. Pairwise independence (every pair factorises) is strictly weaker than mutual independence, which demands that every sub-collection factorises — including all three together:

\mathbb{P}(A \cap B \cap C) \;=\; \mathbb{P}(A)\,\mathbb{P}(B)\,\mathbb{P}(C).

It is entirely possible for each pair to be independent while the triple is not, so always say which you mean.