TITLE: Refining the central limit theorem on discrete random vars
QUESTION [7 upvotes]: Let $x_i$ be iid nonnegative discrete random variables $E[x_i]=N/M$ for some integers $N, M$, variance $\sigma^2$ and higher moments known (finite). 
Then, the sum $\displaystyle S = \sum_{i=1}^M x_i$ will have $E[S]=N$. 
I'm interested in the probability that 
$S$ takes that precise value: $A=P\left(S=E[S]\right)$.
Applying the central limit theorem, I can write
$\displaystyle A \approx \frac{1}{\sqrt{ 2 \pi M \sigma^2}}$
My question is: can this approximation be refined?
ADDED: To add some example-context-motivation:
Lets consider $X$ as a sum of $N$ Bernoullis (0/1) with prob=$p$, such that $E(X)=E(N p)$ is an integer. We can compute exactly the probability that $X$ attains its expected value, it's a Binomial:
$\displaystyle P = P(X= N p) = {N \choose N p} p^{N p} q^{N q} \hspace{2cm}$  [1a]
We might also get an approximate value of that probability using the CTL (Central Limit Theorem)  
$\displaystyle P \approx \frac{1}{\sqrt{2 \pi N p q}} \hspace{2cm} $  [2a]
If we take [1a] and use the Stirling approximation, with $K \approx (K/e)^K \sqrt{2 \pi K}$, we get the same value. Fine.
Now, we may try to refine the approximation, both from [1a] and [2a].
Plugging the next orden Stirling approximation in [1a], we get (I am not mistaken)
$\displaystyle P \approx \frac{1}{\sqrt{2 \pi N p q}} \left(1 - \frac{1- p q}{12 N p q} \right) \hspace{2cm} $  [1b]
To refine the CTL, one can think of 

use some "continuity correction" to evaluate more precisely the (hipothetical) gaussian integral
add some terms from the Edgeworth expansions
do nothing of the above - because the CLT does not justify those procedures in this scenario (just one value of a discrete variable)

I'm not sure which is the correct way.
But let's try the first one:  the next order approximation of the integral gives me (again, if I'm not mistaken)
$\displaystyle P \approx \frac{1}{\sqrt{2 \pi N p q}} \left(1 - \frac{1}{24 N p q} \right) \hspace{2cm} $  [2b]
This is not the same as [1b], but it's close.
Is this just casual? Was it a reasonable thing to do? Should I look (also/instead) after the Edgeworth expansions?

REPLY [9 votes]: For a discrete random variable $X$ with support $\mathbb{Z}$, the Fourier transform of the probability distribution $P_x \equiv P[X=x]$ is given by
$$
\tilde{P}(k) = \sum_{x=-\infty}^{\infty} e^{ikx} P_x = E\left[e^{ikx}\right] = e^{h(k)},
$$
where
$$
h(k) = \sum_{n=1}^{\infty} \kappa_{n} \frac {(ik)^{n}}{n!}
$$
is the natural logarithm of the characteristic function of $X$, and $\kappa_{n}$ is the $n$th cumulant of $X$.  Recall that $\kappa_{1} = \mu$ is the mean and $\kappa_{2} = \sigma^2$ is the variance.  The probability that a sum of $M$ independent variables $X_i$ with the same distribution is exactly $x \in {\mathbb{Z}}$ is then
$$
\begin{eqnarray}
P\left[\sum_{i=1}^{M} X_i = x\right] &=& \int_{-\pi}^{\pi}\frac{dk}{2\pi} e^{-ikx}\tilde{P}(k)^M \\
&=& \int_{-\pi}^{\pi}\frac{dk}{2\pi} e^{Mh(k)-ikx} \\
&=& \int_{-\pi}^{\pi}\frac{dk}{2\pi} e^{ik(M\mu - x) - \frac{1}{2}M\sigma^2 k^2} \exp\left(\sum_{n=3}^{\infty}M\kappa_{n}\frac{(ik)^{n}}{n!}\right).
\end{eqnarray}
$$
Considering the desired case where $x = M\mu \in {\mathbb{Z}}$, and making the change of variable $k \rightarrow k/(\sigma\sqrt{M})$, we have
$$
P\left[\sum_{i=1}^{M} X_i = M\mu\right] = \frac{1}{\sigma\sqrt{2\pi M}}\int_{-\pi\sigma\sqrt{M}}^{\pi\sigma\sqrt{M}} d\Phi(k) \exp\left(\sum_{n=3}^{\infty} \sigma^{-n}M^{1-\frac{1}{2}n}\kappa_{n}\frac{(ik)^{n}}{n!}\right),
$$
where $d\Phi(k) = \phi(k) dk$ is the standard normal distribution (with mean $0$ and variance $1$).  Here we assume that exponential decays rapidly away from $k=0$, so we may replace the limits of integration by $\pm\infty$.  Then, expanding the exponential in inverse powers of $M$, and using the fact that the $n$th central moment of the standard normal distribution vanishes for odd $n$ and is equal to $(n-1)!!$ for even $n$, we obtain the following:
$$
P\left[\sum_{i=1}^{M} X_i = M\mu\right] = \frac{1}{\sigma\sqrt{2\pi M}}\left(1 + \frac{\kappa_4}{8M\sigma^4} - \frac{5\kappa_3^2}{24M\sigma^6} + O(M^{-2})\right).
$$
This is essentially the Edgeworth expansion.  If $X$ is the Bernoulli distribution with probability of success $p = \frac{1}{2}(1+a)$ (and of failure $q=\frac{1}{2}(1-a)$), then it is straightforward to verify that
$$
\begin{eqnarray}
\kappa_2 &=& \sigma^2 = pq = \frac{1}{4}(1-a^2) \\
\kappa_3 &=& \frac{1}{4}(1-a^2)(-a) = -\frac{1}{4}a(1-a^2) \\
\kappa_4 &=& \frac{1}{8}(1-a^2)(3a^2-1),
\end{eqnarray}
$$
and hence
$$
\begin{eqnarray}
\frac{5\kappa_3^2}{24\sigma^6} &=& \frac{5a^2}{6(1-a^2)} \\
\frac{\kappa_4}{8\sigma^4} &=& \frac{3a^2 - 1}{4(1-a^2)},
\end{eqnarray}
$$
for a total correction term proportional to
$$
-\frac{5\kappa_3^2}{24M\sigma^6} + \frac{\kappa_4}{8M\sigma^4} = \frac{9a^2-3-10a^2}{12M(1-a^2)} = -\frac{3+a^2}{12M(1-a^2)} = -\frac{1-pq}{12Mpq},
$$
which agrees with the Stirling approximation to the exact result.