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)?
 A: The formula for the mgf of a sum $X+Y$ is correct when $X$ and $Y$ are independent.
Apart from  a few degenerate examples, $X$ and $2X$ are not independent. 
For instance, toss a fair coin, and let $X=1$ if we get a head, and $X=0$ otherwise. 
Then $\Pr(X=1\cap 2X=0)=0$. But $\Pr(X=1)\nee 0$, and $\Pr(2X=0)\ne 0$, so 
$$\Pr(X=1\cap 2X=0)\ne \Pr(X=1)\Pr(2X=0).$$
The above was a formal showing of non-independence. At the more informal level, if we know that $X=1$, we know a great deal, indeed everything, about $2X$. 
Edit: For the question about whether the mgf of $X+2X$ is ever the product of the mgf, let $X=0$ with probability $1$. 
A: You've written your question in a rather confused and confusing way.  In your question you refer to "a random variable $X+Y$ where $Y$ follows the same distribution of $2X$."  But elsewhere in your question, you mention the question of whether $X$ and $2X$ are independent, which is quite a different matter.
So:


*

*$X$ and $2X$ are not independent except when $X$ is constant;


but


*

*That has nothing to do with questions about $X+Y$ where $X$ and $Y$ are independent and $Y$ has the same distribution as $2X$.


Let's address the second question:
\begin{align}
M_{X+Y}(t) = \mathbb E(e^{t(X+Y)}) & = \mathbb E(e^{tX} e^{tY}) \\[10pt]
& = \mathbb E(e^{tX})\mathbb E(e^{tY}) \text{ by independence} \\[10pt]
& = M_X(t) \mathbb E(e^{t(2X)}) \tag 1 \\[10pt]
& = M_X(t)\mathbb E(e^{(2t)X}) \\[10pt]
& = M_X(t) M_X(2t).
\end{align}
The equality in $(1)$ is true because expected values depend only on the distribution.
A: 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 $P(X \in A) \cdot P(X \in B) \neq 0$
