Is the expected value of a random variable always constant? $
\DeclareMathOperator\Cov{Cov}
\DeclareMathOperator\E{E}
$Can I say that the expected value of a random variable is always a constant?
For some context, I am trying to wrap my head around the definition of covariance. So I have that: $$\Cov(X,Y) = \E[(X-\E[X])(Y-\E[Y])] $$ Expanding the right hand side we have $$\Cov(X,Y) = \E[XY]-\E[X\E[Y]]-\E[Y\E[X]]-\E[\E[X]\E[Y]] = \E[XY] - \E[X]\E[Y]$$
It all makes sense intuitively but I am looking at $\E[X\E[Y]]$ and thinking that for $X$ and $Y$, $$\E[X\E[Y]] = \E[X]\E[Y]$$
Is this correct? Why?
 A: This is due to the linearity of expectation: $E[cX]=cE[X]$. Now, just substitute $c=E[Y]$.
A: A real valued random variable is a measurable map $X\colon \Omega\to \mathbb R$, and its expected value is the real number
$$
\operatorname{E}[X] = \int_\Omega X\,\mathrm dP,
$$
where $P$ is the probability measure on the probability space $\Omega$. When we write something like $X-Y$ for random variables $X,Y\colon \Omega\to\mathbb R$, we mean the map $(X-Y)\colon \Omega \to\mathbb R$ defined by $\omega\mapsto X(\omega)-Y(\omega)$. When writing something like $X-c$, for a constant $c\in\mathbb R$, we consider $c$ as the constant random variable $\omega\mapsto c$ for all $\omega\in\Omega$, so $X-c$ is the random variable given by $\omega\mapsto X(\omega)-c$. Similarly $cX$ is given by $\omega\mapsto c X(\omega)$. Now you can of course apply $\operatorname{E}[cX] = c\operatorname{E}[X]$ for $c=\operatorname{E}[Y]$, since $\operatorname{E}[Y]$ is just a real number.
You also have $E[c] = c$ for $c\in\mathbb R$, where $c$ is again considered as a constant random variable. In your case, this is applied for $c=\operatorname{E}[X]\operatorname{E}[Y]$ to get the last term.
