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)?

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.
