X discrete random variable, then 2X cannot have a binomial distribution I saw the following question and answer: 
Q: True or false? Let X,Y be random discrete variables. 
If X has a binomial distribution with parameters n,p and Y binomial distribution with n_1,p, then X+Y has a binomial distribution with parameters n_1+n,p.  A: False, take Y=X, then X+Y=2X and 2X cant have a binomial distribution since its even. 
I was wondering does the argument of the answer holds (so the even -> no bin distr), do you guys have an idea? Thanks in advance
A: In general, you are right indeed. If we take $X$ and $Y$ to be independent however, the statement is true.
To see this, remember that a binomial random variable with parameters $n,p$ is just saying how many of $n$ independent experiments with probability of succeeding $p$ succeeded.
Hence looking at $X+Y$ is like doing $n$ of those experiments, then doing $n_1$ of them, and adding up how many succeeded. This is exactly the same as doing $n+n_1$ experiments, and then counting how many worked.
A: Let $ X $ and $Y $ are independent, it follows that for any integer $k$  between $ 0 $ and $ m +n $ 
$ P( X+ Y = k) \sum_{i=0}^{k} P( ( X =i)\cap (Y = k-i)) = \sum_{i=0}^{k}P(X=i) P(Y =k-i) = \sum_{i=0}^{k}{m\choose i}p^{i}(1-p)^{m-i}{n\choose k-i}p^{k-i}(1-p)^{n-(k-i)}= \sum_{i=0}^{k}{m\choose i}{n\choose k-i}p^{k}(1-p)^{m+n-k} $
By the Van der Monde formula 
$ P( X+ Y = k) = {m + n \choose k}p^{k}(1 -p)^{m+n -k} $
It means that $ X + Y$ has the binomial distribution $ B(m + n, p).$
$ if m = n $  we have $ B(2 n , p) $.
A: The pdf of $Y$ is given as follows $ P\left( Y=y \right)=P(2X=y)=P\left(X=\frac{y}{2} \right) = \begin{pmatrix}
 n \\ \frac{y}{2}
 \end{pmatrix} \left( p \right)^{\frac{y}{2}} \left( 1-p \right)^{n-\frac{y}{2}}, \ y= 0, \ 2, \ 4, \cdots, 2n $
(Robert V. Hogg, and Allen T. Craig, THE UNIVERSITY OF IOWA, Introduction to Mathematical
Statistics, Fourth Edition, page 128). As such, I have a feeling that this is still a binomial distribution.
Secondly, if $X \sim Bin(n, p)$, then $M_X (t)=(pe^t+1-p)^n$. Hence, $M_{2X}(t)=M_X(2t)=(pe^{2t}+1-p)^n$. If we choose $u=2t$, we have moment generating function of binomial distribution. I ask you colleagues to reconsider your answers. Good Luck,
