## Events

Events $A$ and $B$ are independent if

$P(A∩B)=P(A)⋅P(B)$Intuitively, it means that if we know the probability of event $A$, this does not tell us anything about event $B$, and whether it has occurred.

In addition, if $P(A)>0$ and $P(B)>0$ and $A$ and $B$ are independent, then the Conditional probability of $P(A∣B)=P(A)$ as well as $P(B∣A)=P(B)$.

## Random variables

Two Random variables $X,Y$ are independent if

$P(X≤x,Y≤y)=P(X≤x)⋅P(Y≤y)$for all $x,y∈R$. In terms of the CDF and joint CDF, we can also write

$F_{X,Y}(x,y)=F_{X}(x)⋅F_{Y}(y)$### Continuous

If $X$ and $Y$ are continuous with joint PDF $f_{X,Y}(x,y)$, then the following statements are all equivalent:

- $X$ and $Y$ are independent.
- $f_{X,Y}(x,y)=f_{X}(x)⋅f_{Y}(y)$ for all $x,y∈R$. That is, we can factor the joint PDF to be the product of two functions $g(x)$ and $h(y)$.
- $f_{Y∣X}(y∣x)=f_{Y}(y)$ for all $x$ such that $f_{X}(x)>0$.
- $f_{X∣Y}(x∣y)=f_{X}(x)$ for all $y$ such that $f_{Y}(y)>0$.

### Expectation of their product

If $X$ and $Y$ are independent r.v.s, then $E(XY)=E(X)⋅E(Y)$. Note that it is not always true the other way around (the equation being true doesn’t imply independence).

#### Proof

We first know that $f_{X,Y}(x,y)=f_{X}(x)⋅f_{Y}(y)$ is true. Then by 2D LOTUS, we know that $E(XY)=∫_{−∞}∫_{−∞}xy⋅f_{X}(x)⋅f_{Y}(y)dxdy$. The inner integral can be refactored to become $y⋅f_{Y}(y)∫_{−∞}x⋅f_{X}(x)dx$, and the integral is now simply the definition of the PDF, which integrates to 1.

The complete integral is now $∫_{−∞}y⋅f_{Y}(y)⋅E(X)dy$. In this context, $E(X)$ is constant because it does not depend on $y$, so we can take it outside the integral, leaving us the definition of the PDF again. Therefore, we get $E(X)⋅E(Y)$.