We consider the gamma uniform G distribution as specified by Torabi and Montazeri: $$f(x) = \frac{1}{\Gamma (a)}\frac{g(x)}{[1-G(x)]^2}\left[\frac{G(x)}{1-G(x)}\right]^{a-1}\exp\left[\frac{G(x)}{1-G(x)}\right]$$ where $G$ is a valid c.d.f., $g$ the corresponding p.d.f. and $a > 0$.

My question is this:

If $G$ belongs to the Gumbel domain of attraction, that is, if there exists $\gamma(t)>0$ such that $$\lim_{t \uparrow w(G)}\frac{1-G(t+x\gamma(t))}{1-G(t)}=e^{-x},$$ for all $x\in\mathbb{R}$, and if the upper end point $w(G) = w(F)$, then is it true that $F$ belongs to the same (Gumbel) domain of attraction as $G$?

The sticking point here is the exponential term, as we consider $$\lim_{t \uparrow w(F)}\frac{1-F(t+x\gamma(t))}{1-F(t)}=\lim_{t \uparrow w(F)}\frac{f(t+x\gamma(t))(1+x\gamma^\prime(t))}{f(t)},$$ where the equality follows from L'Hopital's rule. Now this gives rise to the term $$\lim_{t \uparrow w(G)}\frac{\exp\left[\frac{G(t+x\gamma(t))}{1-G(t+x\gamma(t))}\right]}{\exp\left[\frac{G(t)}{1-G(t)}\right]}.$$ How do I evaluate this limit?

Any help much appreciated.

  • $\begingroup$ I don't understand the question; "if there exists ... such that $lim...$ for all $x$" - what does this mean? Should that limit be equal to something, or must that limit exist for all $x$? $\endgroup$
    – adityar
    Nov 1, 2018 at 14:50
  • $\begingroup$ Fixed the mistake. $\endgroup$
    – Will
    Nov 1, 2018 at 14:56
  • $\begingroup$ What is $w(G)$? Is this the upper end point in the sense that $t \uparrow w(G)$ implies $G(t) \uparrow 1$? Or do you mean that $t \uparrow w(G)$ implies $G(t) \uparrow w$? Or do you mean something else? $\endgroup$
    – Ben
    Nov 6, 2018 at 4:42
  • $\begingroup$ In the definition of the gamma uniform G distribution, a minus sign should be found in the exponentiated bracket. $\endgroup$
    – Yves
    Dec 4, 2018 at 7:35

1 Answer 1


Preliminary interpretation: Your question does not clearly specify what you mean by $w(G)$, but you say that this is an "upper end point"of the distribution. I am going to assume that you mean that $t \uparrow w(G)$ implies $G(t) \uparrow 1$. My answer proceeds on this basis.

To simplify the notation in this problem we treat $\gamma$ as fixed and denote $G_x(t) \equiv G(t+x\gamma(t))$. We then define the functions:

$$E_x(t) \equiv \frac{1-G_x(t)}{1-G(t)} \quad \quad \quad \bar{E}_x(t) \equiv \frac{G_x(t)}{G(t)}.$$

These functions are related by:

$$\bar{E}_x(t) = \frac{1- E_x(t)(1-G(t))}{G(t)}.$$

By assumption, for all $x \in \mathbb{R}$ you have $\lim_{t \uparrow w(G)} E_x(t)= e^{-x}$. By the composition law of limits (and assuming the required continuity for it to apply) we also have:

$$\lim_{t \uparrow w(G)} \bar{E}_x(t) = \lim_{t \uparrow w(G)} \frac{1- e^{-x}(1-G(t))}{G(t)}.$$

We now define and simplify the function:

$$\begin{equation} \begin{aligned} H(t,x) &\equiv \frac{G_x(t)}{1-G_x(t)} - \frac{G(t)}{1-G(t)} \\[6pt] &= \frac{1-G(t)}{1-G_x(t)} \cdot \frac{G_x(t)}{G(t)} \cdot \frac{G(t)}{1-G(t)} - \frac{G(t)}{1-G(t)} \\[6pt] &= \frac{G(t)}{1-G(t)} \Bigg[ \frac{1-G(t)}{1-G_x(t)} \cdot \frac{G_x(t)}{G(t)} - 1 \Bigg] \\[6pt] &= \frac{G(t)}{1-G(t)} \Bigg[ \frac{\bar{E}_x(t)}{E_x(t)} - 1 \Bigg]. \\[6pt] \end{aligned} \end{equation}$$

Hence, applying the composition law again yields:

$$\begin{equation} \begin{aligned} \lim_{t \uparrow w(G)} (1-G(t)) \cdot H(t,x) &= \lim_{t \uparrow w(G)} G(t) \Bigg[ \frac{\bar{E}_x(t)}{E_x(t)} - 1 \Bigg] \\[6pt] &= \lim_{t \uparrow w(G)} G(t) \Bigg[ \frac{1- e^{-x}(1-G(t))}{e^{-x} G(t)} - 1 \Bigg] \\[6pt] &= \lim_{t \uparrow w(G)} G(t) \Bigg[ \frac{1- e^{-x} + e^{-x} G(t) - e^{-x} G(t)}{e^{-x} G(t)} \Bigg] \\[6pt] &= \lim_{t \uparrow w(G)} G(t) \cdot \frac{1- e^{-x}}{e^{-x} G(t)} \\[6pt] &= \frac{1- e^{-x}}{e^{-x}} \\[6pt] &= e^{x}-1. \\[6pt] \end{aligned} \end{equation}$$

Now, the limit you want to find is $\lim_{t \uparrow w(G)} \exp(H(t,x))$. This limit is infinity, but there is a related limit that may be useful:

$$\begin{equation} \begin{aligned} \lim_{t \uparrow w(G)} \exp(H(t,x))^{\exp(1-G(t))} &= \lim_{t \uparrow w(G)} \exp( (1-G(t)) \cdot H(t,x)) \\[6pt] &= \exp(e^x-1). \\[6pt] \end{aligned} \end{equation}$$

  • $\begingroup$ The next to last line doesn't make sense to me. Could you explain the jump from a limit of the form $e^{(a/b)}/e^{(c/d)}$ to $e^{(a/c)}/e^{(b/d)}$? $\endgroup$
    – Will
    Nov 8, 2018 at 22:59
  • $\begingroup$ Sorry, that's a mistake - I will edit to fix. $\endgroup$
    – Ben
    Nov 8, 2018 at 23:17
  • $\begingroup$ When can we expect the edit? $\endgroup$
    – Will
    Nov 9, 2018 at 0:34
  • $\begingroup$ @Will: I have now made a major edit to correct the previous misinterpretation of your question. It should be okay now. $\endgroup$
    – Ben
    Nov 9, 2018 at 1:18

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