# Looking for a limit

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.

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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.
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