If $X$ is binomial, $2X$ isn't binomial My question is really simple. I Don't understand why if a random variable $X \sim \text{Bin}(n,p)$, then $2X$ isn't binomial. I know every value of $2X$ is even, but I can't prove $2X$ isn't binomial using this fact.
 A: MGF uniquely defines the distribution of a random variable. That is if you can show that the moment generating function of $2X$ is the same as the moment-generating function of Binomial distribution, then $2X$ follows the same distribution.
$X\sim B(n,p) $
$M_X(t)=E(e^{tx})=(q+pe^t)^n$
$Y=2X \implies X=\dfrac{Y}{2}$
Substitute this and then compute MGF.
$M_\frac{Y}{2}(t)=M_{Y}(\frac{t}{2})=(q+pe^{\frac{t}{2}})^n\ne M_X(t)$
Thus not binomial.
A: It's because $X$ isn't just the value whose probability is being enumerated.  If it were, you could double those and enumerate the probabilities.
But in a binomial, the distribution is a set of pairs of frequencies and the associated probabilities, fitting a predetermined pattern. To be binomial, those frequencies must be exactly determined by the combinatorial rules that arise out of the number of different ways $TRUE$ and $FALSE$ can be picked from $m$ trials.  Doubling, or multiplying by any ratio $N$ populates the frequencies of success equivalent to $0\pmod N$ with the original probabilities rather than the ones in accordance with the new combinatorics, and populates every frequency in-between with probability zero.
