Given the Binomial distribution: $$\Pr(k;n,p) = \Pr(X = k) = \binom n k p^k (1-p)^{n-k}$$ It is clear that for choices of $n$ and $p$ we can shift the binomial distribution left or right since $\mu = np$. I want to consider the case for $p =0.5$, hence the mean and median are equivalent.
Does anyone know if there is an extension of this distribution which allows the distribution to be centered at any integer (including negative of course)? Where centered refers to centering with respect to the mean. This was somewhat addressed in the post but here they are looking for a shift to zero from a programming perspective not analytically.
Thanks.
As noted by @Chappers, you have not said what you mean by 'centered'. The mean of a binomial distribution is $np$, the median is (roughly) a nearby integer, and the mode is (rougly) an integer near $(n+1)p.$ See Wikipedia for details.
If you want a binomial distribution with a particular mean, then choose $n$ and $p$ appropriately. If you want an approximately binomial distribution with mean $\mu,$ then perhaps use the Poisson distribution with mean $\mu.$
If you have a particular shape of binomial distribution in mind (perhaps as determined by particular choices of $n$ and $p$) but it is not 'centered' to your liking, then perhaps add or subtract a constant to shift it however you please. However, I can't immediately think of an application where this kind of centering a binomial would be useful.
If you have a particular goal in mind, perhaps we could be more helpful with a clear description of that goal.
Addendum after revision: If $X \sim \mathsf{Binom}(2m, .5),$ then $E(X) = m.$ Let $Y = X - m$ so that $E(Y) = 0.$ Then $Z = Y + c$ has $E(Z) = c$ and $\operatorname{Var}(X) = \operatorname{Var}(Y) = \operatorname{Var}(Z) = 2m/4 = m/2.$ The 'centering' constant $c$ can be positive or negative.