Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Looking for a limiting value: $$\lim_{K\to \infty } \, -\frac{x \sum _{j=0}^K x (a+1)^{-3 j} \left(-(1-a)^{3 j-3 K}\right) \binom{K}{j} \exp \left(-\frac{1}{2} x^2 (a+1)^{-2 j} (1-a)^{2 j-2 K}\right)}{\sum _{j=0}^K (a+1)^{-j} (1-a)^{j-K} \binom{K}{j} \exp \left(-\frac{1}{2} x^2 (a+1)^{-2 j} (1-a)^{2 j-2 K}\right)}-1$$ with $a \in [0,1)$ and $x$ on the real line. The idea is to to take the second limit for $x \rightarrow \infty $. The aim is to compute the slope of the tail exponent of a distribution.

Making the variable continuous (binomial as ratio of gamma functions) makes it equivalent to: $$\lim_{N\to \infty } 1-\frac{x (1-a)^N \displaystyle\int_0^N -\frac{x (a+1)^{-3 y}\Gamma (N+1) (1-a)^{3 (y-N)} \exp \left(-\frac{1}{2} x^2 (a+1)^{-2 y} (1-a)^{2 y-2 N}\right)}{\Gamma (y+1) \Gamma (N-y+1)} \, \mathrm{d}y}{\sqrt{2} \displaystyle\int_0^N \frac{\left(\frac{2}{a+1}-1\right)^y \Gamma (N+1) \exp \left(-\frac{1}{2} x^2 (a+1)^{-2 y} (1-a)^{2 y-2 N}\right)}{\sqrt{2} \, \Gamma (y+1) \Gamma (N-y+1)} \, \mathrm{d}y}$$ Thank you in advance.

share|cite|improve this question
Someone must have a been a very, very bad boy/girl to deserve be given this nightmare...sigh. – DonAntonio Oct 25 '13 at 21:30
@DonAntonio That's true. – Felix Marin Nov 4 '13 at 22:12

The answer for the continuous version when the binomials are replaced with the $\Gamma$ function is $$\alpha = \frac{\log\left(-\frac{\log(1+a)}{\log(1-a)}\right)}{\log \frac{1-a}{1+a}}$$ and when the binomials are replaced with a normal distribution of mean $K/2$ and variance $K/4$. $$\alpha = -2\frac{\log \left(1-a^2\right)}{\log ^2\left(\frac{1-a}{a+1}\right)}$$

The discrete version does not converge. Consider $a=\phi^{-1}$ (the golden ratio conjugate) and $x=(2\phi+1)^n$ for some $n \in \mathbb{N}$, the variance will be equal to $x$ for $j=n+2K/3$. The tail exponent oscillates with a period of 3 and does not converge as $K\rightarrow \infty$

However, for all intent and purposes, the first $\alpha$ given is a good description of the asymptotic behavior of the tail, the tail just isn't smooth enough to see it at the infinitesimal scale.

share|cite|improve this answer
Can you walks us through the steps for f'/f ? – Nero Oct 30 '13 at 20:56
Doing the continuous approximation and change of variable $a=2 p -1$ (and using N instead of K), I get $$-\frac{x \int_0^N -\frac{2^{-4 N-\frac{1}{2}} x p^{-3 j}\, \Gamma (N+1) (1-p)^{3 (j-N)} \exp \left(-2^{-2 N-1} x^2 p^{-2 j} (1-p)^{2 j-2 N}\right)}{\sqrt{\pi } \Gamma (j+1) \Gamma (-j+N+1)} \, dj}{\int_0^N \frac{\left(\frac{1}{p}-1\right)^j (4-4 p)^{-N} \Gamma (N+1) \exp \left(-2^{-2 N-1} x^2 p^{-2 j} (1-p)^{2 j-2 N}\, \right)}{\sqrt{2 \pi } \Gamma (j+1) \Gamma (-j+N+1)} \, dj}-1$$ – Nero Oct 31 '13 at 18:58
I went the other route, to consider the Binomial sum as a ratio of gamma functions $$\binom{N}{j}=\frac{\Gamma (N+1)}{\Gamma (j+1) \Gamma (-j+N+1)}$$ and multiply the integral by $2^N$. – Nero Oct 31 '13 at 19:40
Looking at solution will try numerically then post the context/motivation when I have web connection. – Nero Nov 2 '13 at 19:58
A 3/2 convergence makes sense, with the 1< tail <2 so it could be an error. The background is in chapter 8 p 97.… – Nero Nov 4 '13 at 17:10

Your Answer


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.