The proof of linearity for expectation given random variables are independent is intuitive. What is the proof given there they are dependent?
Formally,$$ E(X+Y)=E(X)+E(Y)$$where $X$ and $Y$ are dependent random variables.
The proof below assumes that $X$ and $Y$ belong to the sample space. That is, they map from the sample space to a real number line. Is that also a condition for linearity of expectation?
Proof:$$E\left(X+Y\right) =\sum\limits_{s}\left(X+Y\right)\left(s\right) P\left({s}\right) $$$$E\left(X+Y\right) =\sum\limits_{s}\left(X\left(s\right)+Y\left(s\right)\right) P\left({s}\right) $$$$E\left(X+Y\right) =\sum\limits_{s} X\left(s\right)P\left({s}\right) + \sum\limits_{s} Y\left(s\right)P\left({s}\right) $$$$E\left(X+Y\right) =E\left(X\right)+E\left(Y\right)$$Here $S$ is the sample space and $s$ is an event in the sample space.
Reference Lecture for proof.
Also, more reasoning for step 2 would be helpful. I don't understand it completely.
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$\begingroup$The proof below assumes that $X$ and $Y$ belong to the sample space. That is, they map from the sample space to a real number line. Is that also a condition for linearity of expectation?
No. It's the definition of a random variable.
Basically any random variable $X$ is a function that maps the sample space to the reals (or a subset there of, called the support). $$X: \Omega \mapsto \Bbb R$$
If $X$ and $Y$ are both random variables of the same sample space, then so is their sum. $X+Y$. (That is not defined if they are not of the same sample space.)
$$ X:\Omega\mapsto\Bbb R~\wedge~ Y:\Omega\mapsto \Bbb R ~~\implies~~ X+Y:\Omega\mapsto\Bbb R\\\forall s\in\Omega,\quad(X+Y)(s) := X(s)+Y(s)$$
Linearity of Expectation then follows from its definition.
$\begin{align} \mathsf E(X+Y) =&~ \sum_{\omega\in\Omega} (X+Y)(\omega)~\mathsf P(\omega) \\[1ex] =&~ \sum_{\omega\in \Omega} X(\omega)~\mathsf P(\omega)+\sum_{\omega\in \Omega} Y(\omega)~\mathsf P(\omega) \\[1ex] =&~ \mathsf E(X)+\mathsf E(Y) \end{align}$
Of course, this is for discrete random variables. For continuous random variables we use integration , but everything is analogous by no coincidence.
$\begin{align} \mathsf E(X+Y) =&~ \int_{\Omega} (X+Y)(\omega)~\mathsf P(\mathrm d \omega) \\[1ex] =&~ \int_{\Omega} X(\omega)~\mathsf P(\mathrm d \omega)+\int_{\Omega} Y(\omega)~\mathsf P(\mathrm d \omega) \\[1ex] =&~ \mathsf E(X)+\mathsf E(Y) \end{align}$
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