The asymptotic behavior of the CDF of Binomial distribution I got stuck with the following problem which seemed not to be very complicated at the beginning!
I would like to compute the CDF of a Binomial distribution, 
\begin{equation*}
F(\ell;n,q) = \sum_{k=0}^{\ell} \binom{n}{k} q^k (1-q)^{n-k} 
\end{equation*}
where $\ell$ is the solution of
\begin{equation*}
\sum_{k=0}^{\ell} \binom{n}{k} = 2^{\alpha n}. \tag{*}
\end{equation*}
for some fixed $0 < \alpha \le 1$.
Of course the exact value of $F(\ell;n,q)$ is not important (and I guess it is not possible to compute). I would like to know whether $F(\ell;n,q)$ goes to $0$ as $n \to \infty$ or not? 
One way to see this is to check whether $\ell > n q$ or not. However, I cannot see how one can know this given (*). More precisely, I guess there should be a relatively simple criterion that tells us whether the CDF goes to $0$ or not (as $n$ gets large) by comparing $\alpha$ and $q$ (or perhaps $h_2(q)$, ...) 
Any ideas?
Thanks in advance!
 A: If $\alpha = 1$ then $\ell(n) = n$ so $F$ is constant and equal to $1$.
Otherwise, let $\eta$ be the unique real number such that $h_2(\eta) = \alpha$ and $0 < \eta < \frac{1}{2}$. One can show that $\ell(n) \sim \eta n $ by using the Stirling formula and the very crude bounds
$$
\log \binom{n}{\ell} \leq \log \sum_{k=0}^\ell \binom{n}{k} \leq \log n + \log \binom{n}{\ell}, \qquad \ell \leq \frac{n}{2}.
$$
By the weak law of large numbers, we now deduce that
$$
\lim_{n\to\infty}F(\ell(n);n,q) = \begin{cases}0 & \text{if } \eta < q\\1 & \text{if } \eta > q\end{cases}.
$$
The case $\eta = q$ requires to know the second term of the asymptotic expansion of $\ell(n)$.
A: Let $\gamma = (1 - \alpha)\log 2$ and $x = x(n) = \frac{n - 2l}{\sqrt{n}}$.
Approximating the binomial distribution with a normal one, we get
$$ \Phi(-x) \approx e^{-\gamma n}$$
and we wish to compute the limit of
$$
\Phi\left(\frac{l - qn}{\sqrt{nq(1-q)}}\right) =
\Phi\left(\frac{(1 - 2q)\sqrt{n} - x}{2\sqrt{q(1-q)}}\right).
$$
So, it all amounts to the limit of $$A = (1-2q)\sqrt{n} - x.$$
From section 5 of Dominici, we get an approximation for $x \approx -\Phi^{-1}(e^{-\gamma n})$:
$$ x \approx \sqrt{LW\left(\frac{1}{2\pi}e^{2\gamma n}\right)} $$
where $LW(y)$ is the Lambert W with asymptotic behaviour $LW(y) \approx \log(y) - \log(\log(y))$.
Now we just need to organise the computations.
To get rid of the square roots, we rewrite
$$ A = \frac{(1-2q)^2 n - x^2}{(1-2q)\sqrt{n} + x}$$
($q < 1/2$).
The numerator is then approximated by
$$ ((1-2q)^2 - 2\gamma)n + o(n)$$
while the denominator is $\sim \sqrt{n}$. We thus conclude:


*

*If $q < \frac{1}{2}(1 - \sqrt{2\gamma})$, then $A \rightarrow +\infty$ and the limit is 1.

*If $q > \frac{1}{2}(1 - \sqrt{2\gamma})$, then $A \rightarrow -\infty$ and the limit is 0 (the case $q \geq \frac{1}{2}$ is easy).


For the case $q = \frac{1}{2}(1 - \sqrt{2\gamma})$, the numerator is $\sim\log(n)$, $A \rightarrow 0$ and the limit is $1/2$.
