Is the sum of random variables $X$ and $X$ $2X$? Suppose that $X$ is a random variable (say, a normal random variable with mean $a$ and variance $b$). Then is the sum $X + X$ equal to $2X$?
I am asking this because I know that $2X$ has mean $2a$ and variance $4b$. If we just apply $var(X + X) = var(X) + var(X) = 2b$, we get a different answer because $var$ cannot be applied this way to dependent random variables?
 A: It is because you missed the covariance in between.
$$ Var[X + X] = Var[X] + 2Cov[X, X] + Var[X] = b + 2b + b = 4b $$
A: With random variables, it is not true that $X + X = 2X$.  More formally, if $X$ and $Y$ are independent random variables, $X + Y$ and $2X$ don't have the same distribution.
For example let $X$ and $Y$ be the outcomes of two die rolls.  Then $X + Y$ is the sum of the numbers on the two dice and $2X$ is twice the number on the first die.  These don't have the same distribution - for example,$X + Y$ can be odd, and $2X$ is always even.
A: It is simpler. Formally, don't forget that $X$ is no more than a function wich satisfies certains properties: it has a domain, a codomain (which is a vector space in most of the cases), a correspondence rule and an extra propertie (measurability, but it is not important for your question).
If $\Omega$ and $V$ is your codomain, where for all $u,v\in V$ we have $u+v$ is well defined (for example $X:\Omega\to\mathbb{R}$ for obtain a way to sum $X(w)+X(w)$), then $X+X:\Omega\to\mathbb{R}$ is, by definition, $(X+X)(w)=X(w)+X(w)$ (remember that $X(w)$ is in certain form a number. Then $X(w)+X(w)$ is a number). 
Thus, all depend on whether the codomain has a sum or not, AND, if it is the case, how do you define the sum.
For example, take $U=\{A,B\}$ (letters $A$ and $B$) and $\sigma=2^U$. Now, set $\Omega=[0,1]$  with $\mathcal{F}=\mathcal{B}$, and $P:\mathcal{F}\to\mathbb{R}$ given by $P(A)=\lambda(A)$ (Lebesgue measure).
Let $X:\Omega\to U$ given by $X(w)=A$. Thus $X$ is a random variable and $(X+X)(w)=X(w)+X(w)=A+A$. But what is the value of $A+A$? $A$ is only a letter, and, at least I don't know how sum letters.
