Binomial distribution closed by linear operations? I understand that normal distributions are closed by linear operations, but is binomial distribution as well?
In other words, if $X$ is a binomial random variable, does $aX + b$ follow a binomial distribution for any real number $a$ and $b$?
 A: No.  Although if we zoom out on a binomial distribution it looks like a discretized normal distribution, and we even have the normal approximation to the binomial distribution to support this rigorously, the two distributions are not the same especially near the tails.
Translation is especially not obeyed by a binomial distribution: if $X\sim B(n,p)$ is a binomial random variable, $\min(X)=0$.  However, $\min(X+b)=b$, so $X+b$ cannot follow a binomial distribution unless $b=0$.
For scaling, remember the binomial distribution is discrete, so its support (the possible values a binomial random variable can have) is a finite set, in particular the integers from 0 to $n$ inclusive.  However, if we scale a binomial random variable $X$ by $a$, the gaps between the possible values stretch out by a factor of $a$.  If $a$ is not 0 or 1, the support of $aX$ will not be a set of integers from 0 to $an$ inclusive and so $aX$ cannot have binomial distribution.  This problem does not happen for continuous distributions.  (Although not all continuous distributions form a family closed under scaling, and similarly one could create a discrete distribution that was closed under scaling but it would have to either only be closed under scaling up by an integer or allow a discrete but non integer support, both of which are weird.)
A: Is Y=aX+b binomial for all real a,b? No, if a=2 then the support of Y is 0,2,4,...,2n which is not binomial.
Do the values of Y follow "binomial probabilities"? Yes. If $X\sim Bin(n,p), Y=aX+b$, then the support of Y is $aX+b, X=0,1,...,n$ and the pmf is $f(Y)={n\choose \frac{y-b}{a}}p^{\frac{y-b}{a}}(1-p)^{1-\frac{y-b}{a}}$.
Example
$X\sim Bin(3,.4)$, $Y=2X+1$
$X\begin{cases}0&.6^3\\
1&{3\choose 1}.4.6^2\\
2&{3\choose 2}.4^2.6\\
3&.4^3\end{cases}$
$Y\begin{cases}1&.6^3\\
3&{3\choose 1}.4.6^2\\
5&{3\choose 2}.4^2.6\\
7&.4^3\end{cases}$
