TITLE: What is $\sum_{k=0}^n {n \choose k} p^k (1-p)^{n-k} e^{-s(k - np)^2}$?
QUESTION [5 upvotes]: Consider the following expression in three variables, $0 \leq p,s \leq 1$ and $n >0$
$$S_{n, p, s} = \sum_{k=0}^n {n \choose k} p^k (1-p)^{n-k} e^{-s(k - np)^2}$$
If $s = 0$ then $S_{n, p, 0}  = 1$.

Is there a closed form for the sum for $ s > 0$?  If not, can it be
  approximated if we assume that $n$ is large?

REPLY [2 votes]: Just an attempt in trying to simplify the sum (without any idea to reach a closed form):
$$S_{n, p, s} = \sum_{k=0}^n {n \choose k} p^k (1-p)^{n-k} e^{-s(k - np)^2}$$
$$ = \sum_{k=0}^n {n \choose k} p^k (1-p)^{n-k} \sum_{m=0}^\infty (-1)^m \frac{\left[s(k - np)^2\right]^m}{m!}$$
$$ = \sum_{k=0}^n {n \choose k} p^k (1-p)^{n-k} \sum_{m=0}^\infty \frac{(-1)^m}{m!} s^m \sum_{r=0}^{2m} k^r (-np)^{2m-r} $$
$$ = \sum_{k=0}^n {n \choose k} p^k (1-p)^{n-k} \sum_{m=0}^\infty \frac{(-1)^{m}}{m!} s^m \sum_{r=0}^{2m} (-k)^r (np)^{2m-r} $$
$$ = \sum_{k=0}^n {n \choose k} p^k (1-p)^{n-k} \sum_{m=0}^\infty \frac{(-1)^{m}}{m!} (sn^2p^2)^m \sum_{r=0}^{2m} \left(-\frac{k}{np}\right)^r $$
$$ = \sum_{k=0}^n {n \choose k} p^k (1-p)^{n-k} \sum_{m=0}^\infty \frac{(-1)^{m}}{m!} (sn^2p^2)^m \left[ \frac{\left(-\frac{k}{np}\right)^{2m+1}-1}{\left(-\frac{k}{np}\right)-1} \right] $$
$$ = \sum_{k=0}^n {n \choose k} p^k (1-p)^{n-k} \left[ \frac{\exp(-sk^2) \left(\frac{k}{np}\right)+\exp(-sn^2p^2)}{\left(\frac{k}{np}\right)+1} \right] $$
$$ = \sum_{k=0}^n {n \choose k} p^k (1-p)^{n-k} \left[ \frac{k\exp(-sk^2)+np\exp(-sn^2p^2)}{k+np} \right] $$
See if this helps in leading to a closed form at certain approximations of $n$, as well as any condition on $np$.