University of Alberta Guide/STAT/222/Independence and Conditional Expectations

The basic idea behind independence is that if your random variables (or vectors) are independent then the combination of several of these random variable/vectors $\left(P(X_1\le x_1,...,X_d\le x_d)\right)$ can be multiplied together.

Given $T_1,T_2$ are independent $\lambda \text{-Exponential}$ random variables, then $E[T_1+T_2]=E\left[2T_1\right]=E\left[2T_2\right]$. But $\text{Var}(T_1+T_2)<\text{Var}\left(2T_1\right)$. This is because $\text{Var}(T_1+T_2)=\text{Var}\left(T_1\right)+\text{Var}\left(T_2\right)=2\text{Var}\left(T_1\right)$ by independence, whereas $\text{Var}\left(2T_1\right)=2^2\cdot \text{Var}\left(T_1\right)$.

See the equations section for some more examples.

Equations

$P(X_1\le x_1,...,X_d\le x_d)$	$=P(X_1\le x_1)\cdots P(X_d\le x_d)$
$F_{X_1,X_2,...X_d}(x_{1...d})$	$=F_{X_1}(x_1)\cdots F_{X_d}(x_d)$
$f_{X_1,X_2,...X_d}(x_{1...d})$	$=f_{X_1}(x_1)\cdots f_{X_d}(x_d)$
$E[X_1\cdot X_2\cdots X_d]$	$=E\left[X_1\right]\cdots E\left[X_d\right]$
$E[g_1\left(X_1\right)\cdot g_2\left(X_2\right)\cdots g_d\left(X_d\right)]$	$=E\left[g_1\left(X_1\right)\right]\cdots E\left[g_d\left(X_d\right)\right]$
$M_{X_1+...+X_d}(x)$	$=M_{X_1}(x)\cdots M_{X_1}(x)$
$F_{XY}(x,y)$	$=???$
$f_{XY}(x,y)$	$=???$
$F_{X|Y}(x|y)$	$=???$
$F_{X+Y}(x)$	$=P(X+Y\le x)$