# Is there a closed-form solution for the tail index of a GB2 distribution?

In the Generalized Beta distribution of the second kind (GB2), where a, p, and q are shape parameters and b is a scale parameter, the pdf is defined on $\mathbb{R}_+$ by: $$GB2(y;a,b,p,q) = \frac{|a|y^{ap-1}}{b^{ap}B(p,q)(1+(y/b)^a)^{p+q}}$$ In the Pareto distribution, the tail index is just the single shape parameter (usually denoted by α).

In the more complicated GB2, if the four parameters are known, is there a closed-form solution for the tail index? Do all four parameters occur in it?

The "tail index" of a distribution function $F$ describes the rate of decrease of the survival function $1-F(y)$. (Thus, since the $b$ is merely a scale parameter, it cannot possibly influence the tail index.)

The question gives the derivative $f(y)=F^\prime(y)$. We can determine the asymptotic rate of decrease of $f$ by inspection, since for very large $y$, $1$ is much smaller than $y/b$ and the rate doesn't depend on any multiplicative constants. Thus

$$f(y) = \frac{|a|y^{ap-1}}{b^{ap}B(p,q)(1+(y/b)^a)^{p+q}} \approx C \frac{y^{ap-1}}{(y^a)^{(p+q)}} = Cy^{-aq-1}$$

where $C$ collects all the multiplicative constants. In general, whenever a function proportional to a power $k$ of $y$ is integrated, the result is proportional to $y^{k+1}$ (provided $k\ne -1$, but that cannot be the case for otherwise $F$ would diverge instead of approaching $1$ as a limit). Therefore $1-F$ must behave asymptotically like a multiple of $y^{-aq}$. Such functions are said to have a "tail index" of $1/(aq)$--that is, the negative reciprocal of the power.

The foregoing can be made rigorous by expanding $f$ about $\infty$ to first order. The expansion is provided by the Binomial Theorem or Taylor's Theorem (you express $f$ in terms of $x=1/y$, expand around $x=0$, and then rewrite the result in terms of $y$) and is absolutely convergent for $|y| \gt b$, justifying integrating the expansion term-by-term when going from $f$ to $1-F$. You need to estimate the error in this expansion and show it is a "slowly varying function," but doing so is straightforward, requiring no new ideas or techniques.

• whuber, I find your mathematics both persuasive and enlightening, and consistent with the tail index definition you cite. But that definition is at odds with every statement of the tail index for the Pareto distribution I have ever seen. The tail index for the Pareto is always given as alpha -- the negative, but not the inverse, of the power. This would be consistent with the unfortunately deleted answer formerly above. – andrewH Oct 14 '17 at 0:13
• @andrewH I have no problem with such criticism: how one measures the tail index after all is a convention. But the convention in the reference I cite and link to is that the tail index is the negative reciprocal power, not the power (or its negative) itself. The same convention is used in the paper referenced in the deleted answer. – whuber Oct 14 '17 at 17:44
• @whuber I'm in the process of fixing my error(s) in my deleted answer in light of your correct answer and I agree it's all about being consistent in the convention for what a particular subject matter uses (although consistency across subject matters would be good, too). But Qi (2010) has the definition of the tail index as $1/\gamma$ (rather than $\gamma$) which corresponds to $a q$. Or am I misinterpreting Qi's paper again? – JimB Oct 14 '17 at 18:17
• @JimB It's just as likely that I have misread something. Therefore I would appreciate seeing you resurrect your previous answer. – whuber Oct 14 '17 at 19:41
• Wow, this is a terrific answer! It's true -- once pointed out, simplification implied by taking the limit in y just springs out at you! – andrewH Jan 30 '18 at 3:29

This is just an extended comment (and definitely not an answer) adding some details and checks to @whuber 's answer. (Mathematica is used when some code is needed.)

From Qi (2010) the tail index of the distribution function $F$ is $1/\gamma$ defined by

$$1-F(y)=y^{-1/\gamma} L(y)$$

for $y>0$ with the function $L$ satisfying

$$\lim_{t\rightarrow \infty} {{L(t y)}\over{L(t)}} = 1$$

We start with the density function for the generalized beta distribution of the second kind:

f[y_] := (Abs[a]/(b^(a p) Beta[p, q])) y^(a p - 1)/(1 + (y/b)^a)^(p + q)

As noted by @whuber the term $\left(\left(\frac{y}{b}\right)^a+1\right)^{-p-q}$ can be replaced by $\left(\left(\frac{y}{b}\right)^a\right)^{-p-q}$ when $y$ is large:

f4LargeY[y_] := FullSimplify[f[y] //. (1 + (y/b)^a)^(-p - q) -> y^(-a (p + q)) b^(a (p + q))]

which simplifies to

$$\frac{\left| a\right| b^{a q} y^{-a q-1}}{B(p,q)}$$

We should check on that assumption by taking the limit of the ratio of the two functions to see if that ratio approaches 1 as $y\rightarrow \infty$:

Limit[f[y]/f4LargeY[y], y -> \[Infinity], Assumptions -> {a > 0, b > 0, p > 0, q > 0}]
(* 1 *)

and the limit is 1.

We see that $1-F(y)$ is approximately $\int_y^{\infty } \text{f4LargeY}(t) \, dt$ for large enough values of $y$:

OneMinusF = Integrate[f4LargeY[t], {t, y, \[Infinity]},
Assumptions -> {a > 0, b > 0, p > 0, q > 0, y > 0}] /. (b/y)^(a q) -> y^(-a q) b^(a q)

We have $1-F(y) \approx \frac{b^{a q} y^{-a q}}{q B(p,q)}$ for large $y$. If we let $L(y)=\frac{b^{a q}}{q B(p,q)}$, then this function satisfies the requirement in Qi (2010) as $L(y)$ is just a constant.

L[y_] := b^(a q)/(q Beta[p, q])
Limit[L[t y]/L[y], t -> \[Infinity]]
(* 1 *)

So (and, yes, this is a bit of overkill in the use of Solve) we can solve for the tail index

FullSimplify[Solve[OneMinusF == y^(-tailIndex) L[y], tailIndex],
Assumptions -> {y > 0, a > 0, b > 0, p > 0, q > 0}][[1, 1]]
(* tailIndex -> a q *)