Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 1 down vote accepted

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

share|improve this answer
    
Thanks Brendan. –  Balu Sep 18 '11 at 4:07
    
Incidentally, there is a little-known theorem that might help. E. Mailhot (Une propriété de la variance de certaines lois de probabilité réelles tronquées, C. R. Acad. Sci. Paris Sér. I Math. 301 (1985) 241–244) showed that truncating a log-concave distribution (either discrete or continuous) cannot increase its variance. That includes the binomial, Poisson and normal distributions and many others. So $var(X_{\ge c})<c$. This fact might seem obvious at first glance, but it isn't true in general, even for continuous unimodal distributions. –  Brendan McKay Sep 18 '11 at 5:51

$X$ 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 $k$.

share|improve this answer
    
Thanks Ori for the quick response. –  Balu Sep 18 '11 at 4:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.