Independence of $X$ and $2X$ Are these two random variables independent? Unfortunately, I don't know probability theory enough to answer this question. I know for a fact that if $X$ and $Y$ are independent random variables and $g$,$h$ are measurable functions, then $g(X)$ and $h(Y)$ are independent as well. However, I don't know if that can be used here.
I ran into this question in my work, where someone asked if the following moment generating function was valid:
$$M_{X}(t)M_{X}(2t) $$
and stated that this was, indeed, a valid moment generating function for a random variable $X+Y$ where $Y$ follows the same distribution of $2X$.
But is it true that you can write the moment generating function of $X+Y$ as such? (Basically, I doubt that $X$ and $2X$ are independent.)
My main question: Does there exist a random variable $X$ such that you can write the moment generating function of $X + 2X$ as stated above (assuming the MGFs exist)?

Think of independence as "Having informations on $X$ does not give me any information on $Y$". 
This is clearly not the case, if you know that $X = 2$, you also know that $2X = 4$, so they are not independent.
The formal definition for independence of $X$ and $Y$ is
$P(X \in A, Y \in B) = P(X \in A) \cdot P(Y \in B)$  for all $A, B$.
The choice of $A, B$ depends on the probability space on which $X$ is defined, but to make things easy, suppose $X$ is a real random variable (let's say a gaussian).
Then if you take $A = [0, 1], B = [3, 4]$, you know
$P(X \in A, 2X \in B) = 0$ (because if $X$ is in $A$, then $2X$ is not in $B$ and vice versa).
But $P(X \in A) \cdot P(X \in B) \neq 0$
