Probability: Expected Value definition question Can anyone explain how the following equality is true?
a)
$$
E[X + Y] = E[X] + E[Y]
$$
I considered the following definitions:
b)
$$
E[X] = \sum_{i=1}^k c_i p_X(c_i)
$$
c)
$$
p_X(z) = P(\{s \in \Omega : X(s) = z\}) = P(X = z)
$$
d)
$$
E[X + Y] = \sum_{i=1}^k c'_i f_{X+Y}(c'_i) 
$$
e)
$$
p_{X+Y}(z) = P(\{s \in \Omega : X(s) + Y(s) = z\}) = P(X + Y = z)
$$
For example if form (c) has mutually exclusive sets of elementary events that satisfy the random variable equality.  $p_X(z) = P(s_1 \subset \Omega)$ and $p_Y(z) = P(s_2 \subset \Omega)$ and $s_1 \cap s_2 = 0$ then how can this be true, $E[X] + E[Y] = E[X+Y]$?
 A: $X$ and $Y$ don't have to have the same sample space/support, i.e. $X$ can have sample space $\Omega_1$ and $Y$ sample space $\Omega_2$. Further, if $z\in\Omega_1\cap\Omega_2$, then in the discrete case $f_X(z)=Pr(s_1\subset \Omega_1)=Pr(\{z\}\subset\Omega_1)$ and $f_Y(z)=Pr(s_2\subset \Omega_2)=Pr(\{z\}\subset\Omega_2)$ and so $s_1=s_2=\{z\}$ and so $s_1\cap s_2\ne 0$.
But it is not necessary that $\Omega_1\cap\Omega_2\ne0$. From the definitions above, $E(X)+E(Y)=E(X+Y)$ is $\sum_{i=1}^k c_i f_X(c_i)+\sum_{i=1}^{k''}c_i''f_Y(c_i'')=\sum_{i=1}^{k'}c_i'f_{X+Y}(c_i')$.
Example:
$X$ is $1$ with probability $1$, $Y$ is $2$ with probability $1$. Then $X+Y=3$ with probability 1. We have $E(X)+E(Y)=1\cdot 1+2\cdot 1=3$ and $E(X+Y)=3\cdot 1=3$. But $\left(\Omega_1=\{1\}\right)\cap\left(\Omega_2=\{2\}\right)=0$. Suppose instead that $Y=1$ with probability $1$. Then $X+Y=2$ with probability 1 and $f_X(1)=Pr(\{1\}\subset\Omega_1)$ and $f_Y(1)=Pr(\{1\}\subset\Omega_2)$ and so it is not possible that $s_1\cap s_2=0$ as $s_1=s_2$.
