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.
Best Answer
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 $E(X|X\ge c) = c + (\sqrt{2/\pi}+o(1))c^{1/2}$.