Tail Conditional Expectation of a binomial random variable Let $X \sim B(n,c/n)$ be a binomially distributed random variable with
parameter $p = c/n$, and hence mean $c$. Here $c$ is some function of $n$ such that 
i)  $c \geq n^{2/3}$
ii) The function $c$ grows slower than any linear function of $n$ (i.e., in big-O notation, $c = o(n)$, or equivalently $\lim_{n \to \infty} c/n = 0$). 
For such a variable, I want a ball-park estimate of $E[X|X \geq c]$, i.e., tail conditional expectation (TCE), for large $n$. If the probability $c/n$ were a constant, then by central limit theorem the TCE is approximately $c + \sqrt{c}$.
However, $c/n$ is not a constant here. I am most interested in finding whether the following statement is true:
For all $c$ in the said range, the TCE is of the form $c + f(c)$ where $f(c) = o(c^{r})$ for some constant $r < 1$. 
The choice of lower-bound for $c$, namely $c \geq n^{2/3}$ has no significance. I would be happy with resolving the question for a much more restricted range of $c$ by placing a bigger lower bound on $c$. 
I tried writing the explicit expression for the TCE but I have not been able to get anything useful out of it. Also I saw a paper on TCE for binomial rv's, but it just gives the obvious formula obtained by using linearity of expectation and nothing more. 
 A: $X$ will "tend" to $N(c,c)$, even though the usual formulation of the CLT does not cover this case, and $f(c)$ will be of order $\sqrt{c}$. In fact, this is true for any $c=\omega(1)$. Notice that even with CLT, you do not get immediately an estimate for $f(c)$, but only convergence of the CDF.
Unfortunately, I don't have time to expand now, but you can calculate the exponential moments $\mathbb{E}(e^{\lambda X})$ with $\lambda=1/\sqrt{c}$ and then get a bound on the probability that $X>c+k\sqrt{c}$ which decays exponentially in $k$.
A: CLT (sufficiently powerful version such as Berry-Esseen inequality) says that $Pr(X\ge c)\to\frac12$, so any event that has tiny probability for $X$ also has tiny probability for $X_{\ge c}$.  So $E(X|X\ge c) = c + O(c^{1/2})$.  You can get a precise value for $E(X|X\ge c)$ by approximating the point probabilities of $X$ near $c$ using Stirling's formula, then using Euler-Maclaurin summation. I'm sure this has been done many times before though I don't recall a reference at the moment. I think the answer will be that $E(X|X\ge c) = c + (\sqrt{2/\pi}+o(1))c^{1/2}$.
