2.5.1. Example 1

Assume that *<sup>U</sup>*(*x*) = exp *xβ* with *β* > 1. We have

$$\frac{\mathcal{U}'(x)}{\mathcal{U}(x)} = \beta x^{\beta - 1}$$

and

$$\frac{\mathcal{U}''(\mathbf{x})}{\mathcal{U}'(\mathbf{x})} = \beta \mathbf{x}^{\beta - 1} + (\beta - 1)\mathbf{x}^{-1}.$$

Using *g*(*x*) = *<sup>x</sup>*<sup>−</sup>*γ*, we find

$$h\_1(\mathbf{x}) = \mathbf{g}(\mathbf{x}) \frac{\mathcal{U}'(\mathbf{x})}{\mathcal{U}(\mathbf{x})} = \beta \mathbf{x}^{\beta - \gamma - 1}$$

and

$$h\_2(\mathbf{x}) = g(\mathbf{x}) \frac{\mathcal{U}^{\prime\prime}(\mathbf{x})}{\mathcal{U}^{\prime}(\mathbf{x})} = \beta \mathbf{x}^{\beta - \gamma - 1} + (\beta - 1)\mathbf{x}^{-\gamma - 1}.$$

If 0 < *β* − 1 < *γ*, we find that *h*1(*x*) → 0 and *h*2(*x*) → 0. The results of this section show that

$$\frac{\mathcal{U}(x+y\mathcal{g}(x))}{\mathcal{U}(x)} - 1 \sim h\_1(x)y\_\mathcal{H}$$

and Equation (1) holds with *r*(*x*) = <sup>1</sup>4*<sup>h</sup>*1(*x*) ∼ *<sup>x</sup>γ*+1−*<sup>β</sup>*4*β*.

2.5.2. Example 2

> Assume that *<sup>U</sup>*(*x*) = exp *x*<sup>−</sup>*β* with *β* > 0. Clearly, we have

$$\begin{array}{rcl} \frac{\mathcal{U}'(\mathbf{x})}{\mathcal{U}(\mathbf{x})} &=& -\beta \mathbf{x}^{-\beta - 1} \\ \frac{\mathcal{U}''(\mathbf{x})}{\mathcal{U}'(\mathbf{x})} &=& -\beta \mathbf{x}^{-\beta - 1} - (\beta + 1)\mathbf{x}^{-1} .\end{array}$$

We use *g*(*x*) = *xγ* and find

$$\begin{aligned} h\_1(\mathbf{x}) &= -\beta \mathbf{x}^{\gamma - \beta - 1} \\ h\_2(\mathbf{x}) &= -h\_1(\mathbf{x}) - (\beta + 1)\mathbf{x}^{\gamma - 1} \end{aligned}$$

If *γ* < *β* + 1, we have *h*1(*x*) → 0. If *γ* < 1, we have *h*1(*x*) → 0 and *h*2(*x*) → 0. The results of the previous section show that

$$\frac{\mathcal{U}(x+y\mathcal{g}(x))}{\mathcal{U}(x)} - 1 \sim h\_1(x)y\_{\mathcal{H}}$$

and Equation (1) holds with *r*(*x*) = 1/*h*1(*x*) ∼ <sup>−</sup>*<sup>x</sup>β*+1−*<sup>γ</sup>*/*β*.

2.5.3. Example 3

> Assume that *<sup>U</sup>*(*x*) = *xβ* where *β* -= 0. We have

$$h\_1(\mathbf{x}) = \operatorname{g}(\mathbf{x}) \frac{\mathcal{U}'(\mathbf{x})}{\mathcal{U}(\mathbf{x})} = \beta \frac{\operatorname{g}(\mathbf{x})}{\mathbf{x}}$$

and

$$h\_2(\mathbf{x}) = \operatorname{g}(\mathbf{x}) \frac{\mathcal{U}^{\prime\prime}(\mathbf{x})}{\mathcal{U}^{\prime}(\mathbf{x})} = (\beta - 1) \frac{\mathcal{g}(\mathbf{x})}{\mathbf{x}}.$$

Taking *g* ∈ *SN* and *r*(*x*) = *x*4*g*(*x*) (→ ∞) we find

$$r(x) \left( \frac{\mathcal{U}(x + y\mathcal{g}(x))}{\mathcal{U}(x)} - 1 \right) \to \beta y.$$

2.5.4. Example 4

Proposition 1 can be extended for some stable distributions. For instance, consider the density of an asymmetric stable distribution. The representation of such a stable density in the form of a convergen<sup>t</sup> series is, for 0 < *α* < 1 and for any *x* > 0 (see, e.g., [8]),

$$q(\mathbf{x}, \boldsymbol{\alpha}, \boldsymbol{\rho}) = \frac{1}{\pi} \sum\_{n=1}^{\infty} \frac{(-1)^{n-1} \Gamma(\boldsymbol{\alpha} n + 1)}{n!} \sin(n \boldsymbol{\rho} \pi) \boldsymbol{x}^{-\boldsymbol{\alpha} n - 1}.$$

Additionally, assume *xq*(*<sup>x</sup>*, *α*, *<sup>ρ</sup>*)4*q*(*<sup>x</sup>*, *α*, *ρ*) → *τ* (-= 0) as *x* → ∞.

Let *g*(*x*) and *r*(*x*) be positive functions satisfying *g*(*x*)4*x* → 0 and *r*(*x*)*g*(*x*)/*x* → *δ* > 0. Note that, for each *n* > 1 and for *x* large enough, we have, making use of *z* − 1 ∼ log *z* as *z* → 1,

$$\left(1+y\frac{\operatorname{g}(\mathbf{x})}{\mathbf{x}}\right)^{-an-1}-1 \sim -(an+1)\log\left(1+y\frac{\operatorname{g}(\mathbf{x})}{\mathbf{x}}\right) \sim -(an+1)y\frac{\operatorname{g}(\mathbf{x})}{\mathbf{x}}.$$

Then, we have for *x* large enough

$$\begin{split} &\frac{q(\mathbf{x} + \mathbf{y}\mathbf{g}(\mathbf{x}), \mathbf{a}, \boldsymbol{\rho})}{q(\mathbf{x}, \mathbf{a}, \boldsymbol{\rho})} - 1 \\ &= \quad \frac{1}{\pi q(\mathbf{x}, \mathbf{a}, \boldsymbol{\rho})} \sum\_{n=1}^{\infty} \frac{(-1)^{n-1} \Gamma(an + 1)}{n!} \sin(n\rho \pi) \mathbf{x}^{-an-1} \left( \left( 1 + y \frac{\mathbf{g}(\mathbf{x})}{\mathbf{x}} \right)^{-an-1} - 1 \right) \\ &\sim \quad yg(\mathbf{x}) \frac{-1}{\pi q(\mathbf{x}, \mathbf{a}, \boldsymbol{\rho})} \sum\_{n=1}^{\infty} (an+1) \frac{(-1)^{n-1} \Gamma(an+1)}{n!} \sin(n\rho \pi) \mathbf{x}^{-an-2} \\ &= \quad yg(\mathbf{x}) \frac{q'(\mathbf{x}, \mathbf{a}, \boldsymbol{\rho})}{q(\mathbf{x}, \mathbf{a}, \boldsymbol{\rho})}. \end{split}$$

Hence, we have

$$\lim\_{x \to \infty} r(x) \left( \frac{q(x + yg(x), \alpha, \rho)}{q(x, \alpha, \rho)} - 1 \right) = y \delta \tau. \frac{1}{2}$$

#### **3. The Case** *α -***= 0**

Now, suppose that *α* -= 0 and that

$$\lim\_{x \to \infty} r(x) \left( \frac{\mathcal{U}(x + y\mathcal{g}(x))}{\mathcal{U}(x)} - \varepsilon^{ay} \right) = \theta(y),$$

holds l.u. in *y*.

> Equivalently, we have

$$\lim\_{x \to \infty} r(x) \left( \frac{\varepsilon^{-ay} \mathcal{U}(x + y\mathcal{g}(x))}{\mathcal{U}(x)} - 1 \right) = \varepsilon^{-ay} \theta(y).$$

and then (using log *z* ∼ *z* − 1)

$$\lim\_{x \to \infty} r(x)(\mathcal{W}(x + y\mathcal{g}(x)) - \mathcal{W}(x) - ay) = \Omega(y),\tag{4}$$

where *<sup>W</sup>*(*x*) = log *<sup>U</sup>*(*x*) and <sup>Ω</sup>(*y*) = *<sup>e</sup>*<sup>−</sup>*<sup>α</sup>yθ*(*y*).

*3.1. The Limit*

> In Equation (4), we replace *x* by *x* = *t* + *zg*(*t*) to find

$$\lim\_{x \to \infty} r(x) \left( \mathcal{W}(t + z\mathfrak{g}(t) + y\mathfrak{g}(t + z\mathfrak{g}(t))) - \mathcal{W}(t + z\mathfrak{g}(t)) - \mathfrak{a}y \right) = \Omega(y),$$

and

$$\begin{aligned} &\lim\_{x \to \infty} r(x) \\ &\quad \left( (\mathcal{W}(t + z\mathfrak{g}(t) + y\mathfrak{g}(t + z\mathfrak{g}(t))) - \mathcal{W}(t) - a(y + z) ) - (\mathcal{W}(t + z\mathfrak{g}(t)) - \mathcal{W}(t) - az) \right) \\ &= \Omega(y) .\end{aligned}$$

The second term converges to <sup>Ω</sup>(*z*) and thus we have

$$\begin{aligned} &\lim\_{x \to \infty} r(x) \\ &\left(\mathcal{W}\left(t + \left(z + y\frac{\mathcal{g}(t + z\mathcal{g}(t))}{\mathcal{g}(t)}\right)\mathbf{g}(t)\right) - \mathcal{W}(t) - \mathfrak{a}(y + z)\right) \\ &= \Omega(z) + \Omega(y) \end{aligned}$$

or

$$\begin{aligned} r(x) \left( \mathcal{W} \left( t + \left( z + y \frac{\mathcal{g}(t + z \mathcal{g}(t))}{\mathcal{g}(t)} \right) \mathcal{g}(t) \right) - \mathcal{W}(t) \right) \\ - a \left( z + y \frac{\mathcal{g}(t + z \mathcal{g}(t))}{\mathcal{g}(t)} \right) \right) &+ ay \left( \frac{\mathcal{g}(t + z \mathcal{g}(t))}{\mathcal{g}(t)} - 1 \right) \\ \to \quad \Omega(z) + \Omega(y). \end{aligned}$$

By l.u. convergence, the first part converges to <sup>Ω</sup>(*z* + *y*) and then we have

$$\sigma(x)ay\left(\frac{g(t+zg(t))}{g(t)}-1\right)\to\Omega(z)+\Omega(y)-\Omega(y+z).$$

Using the result of the previous subsection, we find that

$$
\alpha y \beta z = \Omega(z) + \Omega(y) - \Omega(y+z).
$$

We propose a solution of the form <sup>Ω</sup>(*z*) = *dx* + *cx*2. The previous equation gives

$$
\alpha \beta yz = cz^2 + cy^2 - c(y^2 + z^2 + 2yz),
$$

and hence *αβyz* = 2*cyz* so that *c* = *αβ*42. We conclude that <sup>Ω</sup>(*x*) = *dx* + *αβx*<sup>2</sup>4<sup>2</sup> and that *<sup>θ</sup>*(*x*) = (*dx* + *αβx*2/2)*eαx*.

We conclude:

**Theorem 3.** *Suppose that α* -= 0*. If*

$$\lim\_{x \to \infty} r(x) \left( \frac{\mathcal{U}(x + y\mathcal{g}(x))}{\mathcal{U}(x)} - \varepsilon^{ay} \right) = \theta(y)\_{\mathcal{A}}$$

*holds l.u. in y, or equivalently if*

$$\lim\_{x \to \infty} r(x) \left( \mathcal{W}(x + y\mathcal{g}(x)) - \mathcal{W}(x) - ay \right) = \Omega(y).$$

*holds l.u. in y, then g*(*x*) *satisfies Equation (2),* <sup>Ω</sup>(*x*) = *dx* + *αβx*<sup>2</sup>4<sup>2</sup> *and <sup>θ</sup>*(*x*) = *dx* + *αβx*<sup>2</sup>42*e<sup>α</sup>x.*

## *3.2. Special Case*

We assume that *W* is differentiable and that *g*(*x*)*<sup>W</sup>*(*x*) → *α*. In this case, we have

$$\begin{aligned} \mathcal{W}(\mathbf{x} + y\mathbf{g}(\mathbf{x})) - \mathcal{W}(\mathbf{x}) &= \mathcal{G}(\mathbf{x}) \int\_0^y \mathcal{W}'(\mathbf{x} + z\mathbf{g}(\mathbf{x})) dz \\ &= \int\_0^y \frac{\mathcal{G}(\mathbf{x})}{\mathcal{g}(\mathbf{x} + z\mathbf{g}(\mathbf{x}))} \mathcal{g}(\mathbf{x} + z\mathbf{g}(\mathbf{x})) \mathcal{W}'(\mathbf{x} + z\mathbf{g}(\mathbf{x})) dz \\ &\to \quad \text{ay}. \end{aligned}$$

Now, suppose in addition that *r*(*x*)*g*(*x*)*<sup>W</sup>*(*x*) − *α* → *δ* and that

$$r(t)\left(\frac{g(\mathbf{x} + t\mathbf{g}(\mathbf{x}))}{g(\mathbf{x})} - 1\right) \to \beta t.$$

We have

$$\begin{split} & \quad \mathcal{W}(\mathbf{x} + y\mathbf{g}(\mathbf{x})) - \mathcal{W}(\mathbf{x}) - a\mathbf{y} \\ & \quad = \ & \mathbf{g}(\mathbf{x}) \int\_{0}^{y} \mathcal{W}'(\mathbf{x} + t\mathbf{g}(\mathbf{x})) dt - a\mathbf{y} \\ & \quad = & \int\_{0}^{y} \mathbf{g}(\mathbf{x} + t\mathbf{g}(\mathbf{x})) \mathcal{W}'(\mathbf{x} + t\mathbf{g}(\mathbf{x})) \frac{\mathbf{g}(\mathbf{x})}{\mathbf{g}(\mathbf{x} + t\mathbf{g}(\mathbf{x}))} dt - a\mathbf{y} \\ & \quad = & \int\_{0}^{y} \mathbf{g}(\mathbf{x} + t\mathbf{g}(\mathbf{x})) \mathcal{W}'(\mathbf{x} + t\mathbf{g}(\mathbf{x})) \left( \frac{\mathbf{g}(\mathbf{x})}{\mathbf{g}(\mathbf{x} + t\mathbf{g}(\mathbf{x}))} - 1 \right) dt \\ & \quad \quad + \int\_{0}^{y} (\mathbf{g}(\mathbf{x} + t\mathbf{g}(\mathbf{x})) \mathcal{W}'(\mathbf{x} + t\mathbf{g}(\mathbf{x})) - a) dt. \end{split}$$

For the first integral, by assumption, we have

$$r(t)\left(\frac{\mathbf{g}(\mathbf{x}+t\mathbf{g}(\mathbf{x}))}{\mathbf{g}(\mathbf{x})}-1\right)\to\beta t,$$

or

$$r(t)\frac{g(\mathbf{x}+t\mathbf{g}(\mathbf{x}))}{g(\mathbf{x})}\left(1-\frac{g(\mathbf{x})}{g(\mathbf{x}+t\mathbf{g}(\mathbf{x}))}\right)\to\beta t,\mathbf{y}$$

$$r(t)\left(\frac{g(\mathbf{x})}{g(\mathbf{x}+t\mathbf{g}(\mathbf{x}))}-1\right)\to-\beta t.$$

or Since *r*(*x*)*g*(*x*)*<sup>W</sup>*(*x*) − *α* → *δ*, we obtain

$$\begin{aligned} &r(\mathbf{x})\left(\mathcal{W}(\mathbf{x}+t\mathbf{g}(\mathbf{x}))-\mathcal{W}(\mathbf{x})-\alpha\mathbf{y}\right) \\ &=\int\_0^y \mathbf{g}(\mathbf{x}+t\mathbf{g}(\mathbf{x}))\mathcal{W}'(\mathbf{x}+t\mathbf{g}(\mathbf{x}))r(\mathbf{x})\left(\frac{\mathbf{g}(\mathbf{x})}{\mathbf{g}(\mathbf{x}+t\mathbf{g}(\mathbf{x}))}-1\right)dt \\ &+\int\_0^y r(\mathbf{x})(\mathbf{g}(\mathbf{x}+t\mathbf{g}(\mathbf{x}))\mathcal{W}'(\mathbf{x}+t\mathbf{g}(\mathbf{x}))-\alpha)dt \\ &\to \quad \alpha(-\beta)\frac{y^2}{2}+\delta y. \end{aligned}$$

*3.3. Representation Theorem*

> Now, consider *Q*(*x*) = *<sup>W</sup>*(*x*) + *<sup>α</sup><sup>A</sup>*(*x*), where *<sup>A</sup>*(*x*) = *xa*1/*g*(*t*)*dt* as before. We prove above that

*<sup>A</sup>*(*x* + *yg*(*x*)) − *<sup>A</sup>*(*x*) → *y*

l.u. in *y*. If *g* ∈ *SN* satisfies

$$\lim\_{\mathfrak{x}\to\infty} \mathfrak{x}\left(\frac{\mathfrak{g}(\mathfrak{x}+\mathfrak{y}\mathfrak{g}(\mathfrak{x}))}{\mathfrak{g}(\mathfrak{x})}-1\right) = \beta \mathfrak{y}\_{\mathfrak{x}}$$

then we also have

$$\begin{aligned} A(\mathbf{x} + y\mathbf{g}(\mathbf{x})) - A(\mathbf{x}) - y &= \quad \int\_0^y \left( \frac{\mathbf{g}(\mathbf{x})}{\mathbf{g}(\mathbf{x} + z\mathbf{g}(\mathbf{x}))} - 1 \right) dz \\ &= \quad - \int\_0^y \frac{\mathbf{g}(\mathbf{x})}{\mathbf{g}(\mathbf{x} + z\mathbf{g}(\mathbf{x}))} \left( \frac{\mathbf{g}(\mathbf{x} + z\mathbf{g}(\mathbf{x}))}{\mathbf{g}(\mathbf{x})} - 1 \right) dz, \end{aligned}$$

so that

$$r(\mathbf{x})(A(\mathbf{x} + y\mathbf{g}(\mathbf{x})) - A(\mathbf{x}) - y) \to -\beta \frac{y^2}{2}.$$

Using *Q*(*x*) = *<sup>W</sup>*(*x*) − *<sup>α</sup><sup>A</sup>*(*x*), we see that

$$\begin{aligned} &Q(\mathbf{x} + y\mathbf{g}(\mathbf{x})) - Q(\mathbf{x}) \\ &= \
 &W(\mathbf{x} + y\mathbf{g}(\mathbf{x})) - W(\mathbf{x}) - \alpha y \\ &- \alpha \left( A(\mathbf{x} + y\mathbf{g}(\mathbf{x})) - A(\mathbf{x}) - y \right) .\end{aligned}$$

Hence, using Equation (3),

$$\begin{aligned} &r(\mathbf{x})\left(Q(\mathbf{x}+y\mathbf{g}(\mathbf{x}))-Q(\mathbf{x})\right) \\ &=\quad r(\mathbf{x})\left(W(\mathbf{x}+y\mathbf{g}(\mathbf{x}))-W(\mathbf{x})-\alpha y\right) \\ &-\alpha r(\mathbf{x})\left(A(\mathbf{x}+y\mathbf{g}(\mathbf{x}))-A(\mathbf{x})-y\right) \\ &\to\quad \Omega(y)+\alpha\beta\frac{y^2}{2}=\Psi(y) \end{aligned}$$

l.u. in *y*. As in the previous subsection, we conclude that

$$r(\mathbf{x}) \left( Q(\mathbf{x} + y\mathbf{g}(\mathbf{x})) - Q(\mathbf{x}) \right) \to \Psi(y) = \lambda y$$

for some real number *λ*. The first representation of the previous subsection gives

$$Q(\mathbf{x}) = T(\mathbf{x}) + \frac{C(\mathbf{x})}{r(\mathbf{x})},$$

or

$$W(\mathbf{x}) = aA(\mathbf{x}) + T(\mathbf{x}) + \frac{C(\mathbf{x})}{r(\mathbf{x})}r$$

where *<sup>C</sup>*(*x*) → *C* and *r*(*x*)*g*(*x*)*<sup>T</sup>*(*x*) → *λ*.

**Theorem 4.** *We have Equation (3) if and only if <sup>W</sup>*(*x*) *is of the form*

$$W(\mathbf{x}) = \mathfrak{a}A(\mathbf{x}) + T(\mathbf{x}) + \frac{\mathbf{C}(\mathbf{x})}{r(\mathbf{x})},$$

*where <sup>C</sup>*(*x*) → *C and r*(*x*)*g*(*x*)*<sup>T</sup>*(*x*) → *λ.*

## *3.4. More Results*

In our next result, we consider the function *h*(*x*) = *f*(*x*)4*F*(*x*), where *f* is the density of *F*. We make the following assumptions about *h*:


**Lemma 1.** *If (a) and (b) hold, then*

$$r(\mathbf{x}) \left( \frac{h(\mathbf{x} + yg(\mathbf{x}))}{h(\mathbf{x})} - 1 \right) \to -\beta y.$$

**Proof.** We have

$$h(\mathbf{x} + y\mathbf{g}(\mathbf{x})) - h(\mathbf{x}) = \mathbf{g}(\mathbf{x}) \int\_0^y h'(\mathbf{x} + z\mathbf{g}(\mathbf{x})) dz.$$

Since *r*(*x*)*<sup>h</sup>*(*x*)4*h*<sup>2</sup>(*x*) → −*β* > 0, we have that *<sup>h</sup>*(*x*) ∈ <sup>Γ</sup>0(*g*) and, using *g*(*x*) = <sup>1</sup>4*h*(*x*), we obtain that 

$$r(\mathbf{x}) \left( \frac{h(\mathbf{x} + y\mathbf{g}(\mathbf{x}))}{h(\mathbf{x})} - 1 \right) = \frac{r(\mathbf{x})}{h^2(\mathbf{x})} \int\_0^y h'(\mathbf{x} + z\mathbf{g}(\mathbf{x})) dz \to -\beta y. \mathbf{x}$$

Now, we study the tail *<sup>F</sup>*(*x*).

**Lemma 2.** *If (a) and (b) hold, then*

$$r(\mathbf{x}) \left( \log \frac{\mathbb{F}(\mathbf{x} + y\mathbf{g}(\mathbf{x}))}{\mathbb{F}(\mathbf{x})} + y \right) \to \beta \frac{y^2}{2}.$$

**Proof.** Using *h*(*x*) = *f*(*x*)4*F*(*x*), we obtain that

$$\int\_{\infty}^{\mathbf{x} + y\mathbf{g}(\mathbf{x})} h(z)dz = \int\_{\mathbf{x}}^{\mathbf{x} + y\mathbf{g}(\mathbf{x})} \frac{f(z)}{\overline{F}(z)} dz.$$

so that

$$\lg(\mathbf{x}) \int\_0^y h(\mathbf{x} + z\mathbf{g}(\mathbf{x}))dz = -\log \frac{\mathbb{F}(\mathbf{x} + y\mathbf{g}(\mathbf{x}))}{\mathbb{F}(\mathbf{x})}.$$

It follows that (recall *g*(*x*) = 1/*h*(*x*))

$$r(\mathbf{x}) \int\_0^y \left( \frac{h(\mathbf{x} + z\mathbf{g}(\mathbf{x}))}{h(\mathbf{x})} - 1 \right) dz = -r(\mathbf{x}) \left( \log \frac{\mathsf{F}(\mathbf{x} + y\mathbf{g}(\mathbf{x}))}{\mathsf{F}(\mathbf{x})} + y \right), \quad \mathbf{x} \in \mathbb{R}^d$$

and using Lemma 1, it follows that

$$r(x) \left( \log \frac{\overline{F}(x + y\overline{g}(x))}{\overline{F}(x)} + y \right) \to \beta \frac{y^2}{2}.$$

This proves the result.

Now, we arrive at the main result here.

**Theorem 5.** *If (a) and (b) hold, then*

$$r(x)\left(\frac{\overline{F}(x+y\mathfrak{g}(x))}{\overline{F}(x)}-e^{-y}\right)\to\beta\frac{y^2}{2}e^{-y}.$$

**Proof.** Using Lemma 2, we have

$$r(\mathfrak{x}) \log e^{\mathfrak{y}} \frac{\overline{F}(\mathfrak{x} + y\mathfrak{g}(\mathfrak{x}))}{\overline{F}(\mathfrak{x})} \to \beta \frac{\mathfrak{y}^2}{2}.$$

Using log *z* ∼ *z* − 1, it follows that

$$r(x) \left( e^y \frac{\mathbb{F}(x + yg(x))}{\mathbb{F}(x)} - 1 \right) \to \beta \frac{y^2}{2},$$

or

$$r(\mathbf{x}) \left( \frac{\overline{F}(\mathbf{x} + y\mathbf{g}(\mathbf{x}))}{\overline{F}(\mathbf{x})} - \varepsilon^{-y} \right) \to \beta \frac{y^2}{2} \varepsilon^{-y}.$$

The previous theorem can be useful in extreme value theory as follows.

We assume that (a) and (b) hold and that *F* is strictly increasing. We define *an* by the equality *nF*(*an*) = 1. It is clear that *an* ↑ ∞. In the result of Theorem 5, we replace *x* by *an* to see that

$$r(a\_{\mathcal{U}}) \left( n\overline{F}(a\_{\mathcal{U}} + \mathcal{yg}(a\_{\mathcal{U}})) - e^{-y} \right) \to \beta \frac{y^2}{2} e^{-y} \cdot \frac{1}{2}$$

Now, we use log(*z*)+(<sup>1</sup> − *z*) = *O*(1)(1 − *z*)<sup>2</sup> and write

$$\begin{aligned} n\mathbb{F}(a\_n + y\mathbf{g}(a\_n)) &= \, \_n\mathbb{F}(a\_n + y\mathbf{g}(a\_n)) + n\log F(a\_n + y\mathbf{g}(a\_n)) - n\log F(a\_n + y\mathbf{g}(a\_n)) \\ &= \, \_0\mathrm{O}(1)n\overline{\mathbb{F}}^2(a\_n + y\mathbf{g}(a\_n)) - \log F'(a\_n + y\mathbf{g}(a\_n)). \end{aligned}$$

Now, notice that

$$r(a\_n)n\mathbb{P}^2(a\_n + y\mathcal{g}(a\_n)) = O(1) \\ r(a\_n)n\mathbb{P}^2(a\_n) = O(1)\frac{r(a\_n)}{n}.$$

.

If *r*(*an*)4*n* → 0, we obtain that

$$r(a\_n) \left( \log \mathbb{F}^{\mathbb{N}}(a\_n + yg(a\_n)) + e^{-y} \right) \to -\beta \frac{y^2}{2} e^{-y} \rho$$

and hence also that

$$r(a\_{ll})\log e^{\exp\left(-y\overline{\mathcal{F}}^{\eta}\left(a\_{n}+y\mathcal{G}(a\_{n})\right)\right)} \rightarrow -\beta \frac{y^{2}}{2} e^{-y} \omega$$

and

$$\sigma(a\_n) \left( e^{\exp^{-y}\mathcal{J}^{\text{gr}}} (a\_n + y\mathcal{g}(a\_n)) - 1 \right) \to -\beta \frac{y^2}{2} e^{-y} \nu$$

or

$$r(a\_{\hbar})\left(\overline{F}^{\eta}(a\_{\hbar} + y\mathcal{g}(a\_{\hbar})) - \exp - \varepsilon^{-y}\right) \to -\beta \frac{y^2}{2} \varepsilon^{-y} \exp - \varepsilon^{-y}.$$

It means that, if *Xi* are independent and identically distributed random variables with distribution function *F*, then

$$r(a\_n) \left( P\left(\frac{M\_n - a\_n}{\mathcal{g}(a\_n)} \le y\right) - \Lambda(y) \right) \to \Phi(y),$$

where *Mn* = max(*<sup>X</sup>*1, *X*2, ..., *Xn*), <sup>Λ</sup>(*y*) = exp −*e*<sup>−</sup>*y* and <sup>Φ</sup>(*y*) = −*β y*22 *e*<sup>−</sup>*y* exp <sup>−</sup>*e*<sup>−</sup>*y*.

It means that *F* is in the max-domain of attraction of the double exponential and the convergence rate is determined by *<sup>r</sup>*(*an*).

*3.5. Examples*

3.5.1. Example 1

> The following example is related to Theorem 5.

Let *<sup>U</sup>*(*x*) = exp −*x*<sup>2</sup> for *x* > 0. Using *g*(*x*) = 1/(<sup>2</sup>*x*), we have *<sup>U</sup>*(*x*) ∈ <sup>Γ</sup>−<sup>1</sup>(*g*). Now, we consider the difference

$$\frac{\mathcal{U}(x + y\mathcal{g}(x))}{\mathcal{U}(x)} - e^{-y\mathcal{g}}$$

.

We have

$$\begin{aligned} \frac{\mathcal{U}(\mathbf{x} + y\mathbf{g}(\mathbf{x}))}{\mathcal{U}(\mathbf{x})} - \mathfrak{e}^{-y} &=& \mathfrak{e}^{-y - y^2 \mathfrak{g}^2(\mathbf{x})} - \mathfrak{e}^{-y} \\ &=& \mathfrak{e}^{-y} (\mathfrak{e}^{-y^2 \mathfrak{g}^2(\mathbf{x})} - 1) \\ &\sim& -\mathfrak{e}^{-y} y^2 \mathfrak{g}^2(\mathbf{x}) \end{aligned}$$

and

$$x^2 \left( \frac{\mathcal{U}(x + y\mathcal{g}(x))}{\mathcal{U}(x)} - e^{-y} \right) \to -\frac{1}{4} y^2 e^{-y}.$$

3.5.2. Example 2

Let *<sup>U</sup>*(*x*) = exp *<sup>x</sup>β*, *β* > 1. We have *<sup>W</sup>*(*x*) = log *<sup>U</sup>*(*x*) = *xβ* and *<sup>W</sup>*(*x*) = *βxβ*−1. Taking *g*(*x*) = *<sup>x</sup>*1−*β*, we have

$$
\lg(\mathbf{x})\mathsf{W}'(\mathbf{x}) = \boldsymbol{\beta}.
$$

As for *g*(*x*), we have *g*(*x*)4*x* → 0 and

$$\frac{\operatorname{g}(x+y\operatorname{g}(x))}{\operatorname{g}(x)} - 1 = \left(1 + y\frac{\operatorname{g}(x)}{x}\right)^{1-\beta} - 1 \sim (1-\beta)y\frac{\operatorname{g}(x)}{x}.$$

Taking *r*(*x*) = *x*4*g*(*x*) = *<sup>x</sup>β*, we have

$$r(\mathbf{x}) \left( \frac{\mathbf{g}(\mathbf{x} + \mathbf{y}\mathbf{g}(\mathbf{x}))}{\mathbf{g}(\mathbf{x})} - 1 \right) \to (1 - \beta)\mathbf{y}.$$

The result of Section 3.2 shows that

$$r(x)\left(\mathcal{W}(x+y\mathcal{g}(x))-\mathcal{W}(x)-\beta y\right)\to\beta(\beta-1)\frac{y^2}{2},$$

and then

$$r(\mathbf{x}) \left( \frac{\mathsf{U}(\mathbf{x} + \mathsf{y}\mathbf{g}(\mathbf{x}))}{\mathsf{U}(\mathbf{x})} - \mathsf{e}^{\beta y} \right) \to \beta(\beta - 1) \frac{y^2}{2} \mathsf{e}^{\beta y} \mathsf{y}$$
