I'm having trouble getting my head around independence of random variables and distributions, mainly because I can't understand how Var(2X) = 4Var(X) but Var(X+Y) = Var(X) + Var(Y).
Wikipedia says that: "two events are independent, statistically independent, or stochastically independent[1] if the occurrence of one does not affect the probability of occurrence of the other. Similarly, two random variables are independent if the realization of one does not affect the probability distribution of the other."
Shouldn't, according to this definition, there be no difference between the first and second cases I presented above? How does the first distribution of X in any way influence the second? Any more than an instance of X influences an instance of Y?
A simple way to conceptualize the difference between $X+Y$ when $X$ and $Y$ are IID, versus $2X$, is that each of $X$ and $Y$ are allowed to vary independently of the other when computing the sum, but in $2X$, the randomness is only coming from a single variable, but the outcome is being scaled up by a factor of $2$.
For example, suppose you have two fair coins. Each coin for which you flip heads gets you a "point." If you flip both coins and add the total number of points, then this is like $X+Y$ where $X$ is the number of heads obtained with the first coin, and $Y$ is the number of heads obtained with the second coin. Then, $$\Pr[X + Y = 0] = 1/4, \quad \Pr[X + Y = 1] = 1/2, \quad \Pr[X + Y = 2] = 1/4.$$ On the other hand, if you tried to take a shortcut (flipping that second coin is such a hassle!) and you only flip one coin and score two points for heads, then that is like $2X$, and you cannot get $2X = 1$: instead, $$\Pr[2X = 0] = 1/2, \quad \Pr[2X = 2] = 1/2.$$ And this illustrates why $$\operatorname{Var}[X+Y] \ne \operatorname{Var}[2X];$$ moreover, it illustrates the qualitative difference in the behavior of the random variables $X+Y$ and $2X$.