Slash distributions, generalized convolutions, and extremes

Abstract

An $\alpha$-slash distribution built upon a random variable X is a heavy tailed distribution corresponding to $Y=X/U^{1/\alpha }$, where U is standard uniform random variable, independent of X. We point out and explore a connection between $\alpha$-slash distributions, which are gaining popularity in statistical practice, and generalized convolutions, which come up in the probability theory as generalizations of the standard concept of the convolution of probability measures and allow for the operation between the measures to be random itself. The stochastic interpretation of Kendall convolution discussed in this work brings this theoretical concept closer to statistical practice, and leads to new results for $\alpha$-slash distributions connected with extremes. In particular, we show that the maximum of independent random variables with $\alpha$-slash distributions is also a random variable with an $\alpha$-slash distribution. Our theoretical results are illustrated by several examples involving standard and novel probability distributions and extremes.

Appendices

Appendix A: Basic facts on generalized convolutions

In this section we recall some basic facts on generalized convolutions and two crucial examples of Kendall and Kucharczak–Urbanik convolutions that serve as the main tools for analyzing extremes of random variables with $\alpha$-slash distributions. We refer to Urbanik (1964) for the background material on the theory of generalized convolutions and the corresponding characteristic functions as well as to more recent works such as Borowiecka-Olszewska et al. (2015) or Jasiulis-Gołdyn et al. (2020b, 2021) and the references therein for new interesting results and extensive examples.

We start with the definition of generalized convolution (see, e.g., Urbanik, 1964; Borowiecka-Olszewska et al., 2015, Section 2.1).

Definition 2

A commutative and associative operation $*: \mathcal {P}_+^2 \rightarrow \mathcal {P}_+$ is called generalized convolution if it satisfies the following conditions:

• (i) $\delta _0 *\lambda = \lambda$ for all $\lambda \in \mathcal {P}_+$;

• (ii) $(p\mu + (1-p)\nu )*\lambda = p(\mu *\lambda ) + (1-p)(\nu *\lambda )$ for all $p \in [0,1], \mu , \nu , \lambda \in \mathcal {P}_+$;

• (iii) $T_x\mu *T_x \nu = T_x(\mu *\nu )$, for all $x > 0, \mu , \nu \in \mathcal {P}_+$;

• (iv) If $\mu _n \Longrightarrow \mu$ for $\mu _n, \mu \in \mathcal {P}_+$, then $\mu _n*\lambda \Longrightarrow \mu *\lambda$ for all $\lambda \in \mathcal {P}_+$;

• (v) There exists a sequence $\{c_n\}_{n=1}^\infty , c_n > 0$ and $\mu \in \mathcal {P}_+, \mu \ne \delta _0$ such that $T_{c_n}\delta _1^{*n} \Longrightarrow \mu$, where $\delta _1^{*n}$ denotes the convolution of n identical measures $\delta _1$.

The pair $(\mathcal {P}_+, *)$ is called the generalized convolution algebra.

Remark 12

It is well known (see, e.g., Borowiecka-Olszewska et al., 2015, Section 2.1) that a generalized convolution $*$ is uniquely determined by the convolution of point-mass measures $\delta _x *\delta _y, x,y \ge 0$. That is, for every $\lambda _1, \lambda _2 \in \mathcal {P}_+$ and $A \in \mathcal {B}({\mathbb {R}}_+)$ we have

$$\begin{aligned} (\lambda _1 *\lambda _2)\! (A) = \int _{0}^{\infty } \int _{0}^{\infty } \left( \delta _x *\delta _y\right) \!(A)\ \lambda _1(\mathrm{d}x) \lambda _2(\mathrm{d}y). \end{aligned}$$

(20)

The main technical tool in the study of generalized convolutions, which plays an analogous role to that of the Laplace transform for ordinary convolutions, is a generalized characteristic function (see, e.g, Urbanik, 1964, Section 4; Borowiecka-Olszewska et al., 2015, Definition 2.2), defined below.

Definition 3

We say that the generalized convolution algebra $(\mathcal {P}_+, *)$ admits a characteristic function if there exists one-to-one correspondence between probability measures $\lambda \in \mathcal {P}_+$ and functions $\Phi _\lambda (\cdot ): {\mathbb {R}}_+ \rightarrow {\mathbb {R}}$ such that

• (i) $\Phi _{p\mu + (1-p)\nu } = p\Phi _{\mu } + (1-p)\Phi _{\nu }$ for all $\mu , \nu \in \mathcal {P}_+, p \in [0,1]$;

• (ii) $\Phi _{\mu *\nu } = \Phi _\mu \Phi _\nu$ for all $\mu , \nu \in \mathcal {P}_+$;

• (iii) $\Phi _{T_x \mu }(t) = \Phi _\mu (xt)$ for all $x,t \ge 0, \mu \in \mathcal {P}_+$;

• (iv) The uniform convergence of $\Phi _{\lambda _n}$ on every bounded interval is equivalent to the weak convergence of $\lambda _n$.

The function $\Phi _\lambda (\cdot )$ is termed the $*$-generalized characteristic function of measure $\lambda$. Analogously, if a random variable X has distribution $\lambda$, then $\Phi _X(\cdot )$ denotes the $*$-generalized characteristic function of that $\lambda$.

If $(\mathcal {P}_+, *)$ admits a generalized characteristic function, then $\Phi _\lambda (t)$ is an integral transform of the form (see Urbanik, 1964, Theorem 6)

$$\begin{aligned} \Phi _\lambda (t) = \int _0^\infty h(tx) \lambda (\mathrm{d}x) \end{aligned}$$

(21)

for some kernel $h(\cdot )$. As shown in Kucharczak and Urbanik (1974), a function $h:{\mathbb {R}}_+ \rightarrow {\mathbb {R}}$ is a kernel of the characteristic function corresponding to a generalized convolution algebra if and only if it is quasi-stable (that is, h is a bounded, continuous function such that $\forall a,b \in {\mathbb {R}}_+$ the function $(T_a h)(T_b h)$ belongs to the closed convex hull of the set $\{T_x h: x\in {\mathbb {R}}_+\}$) and, in addition, $h(t) = 1 - t^q L(t)$ for some $q >0$ and a slowly varying (at the origin) function $L(\cdot )$. Further, there must exist a probability measure $\mu \in \mathcal {P}_+$ such that $\limsup _{t \rightarrow \infty } \int _0^\infty h(tx) \mu (\mathrm{d}x) < 1$.

A particular example of a function $h(\cdot )$ that satisfies these conditions, which plays a central role in our work, is the function (see, for example function $f_6$ in Kucharczak and Urbanik, 1974)

$$\begin{aligned} h(t) = (1-t^\alpha )_+^n, \quad t \in [0,1], \alpha > 0, n\in {\mathbb {N}}. \end{aligned}$$

(22)

We refer to Kucharczak and Urbanik (1974) for other examples and general discussion on the connections between quasi-stable functions and generalized convolutions.

Generalized convolution corresponding to the characteristic function with kernel (22) and its special case with parameter $n=1$ plays a crucial role in proving the results of Sect. 2.

Definition 4

Let $\alpha > 0, n \in {\mathbb {N}}$. A generalized convolution $*_{\alpha , n}$ with the characteristic function

$$\begin{aligned} \Phi ^{(\alpha , n)}_\lambda (t) = \int _0^\infty (1 - x^\alpha t^\alpha )^n_+ \lambda (\mathrm{d}x), \end{aligned}$$

(23)

that is with the kernel given by (22), is called Kucharczak–Urbanik convolution (see, e.g., Kucharczak and Urbanik, 1974, p. 268; Borowiecka-Olszewska et al., 2015, Example 2.8).

Definition 5

Let $\alpha > 0, n \in {\mathbb {N}}$. A generalized convolution $*_\alpha$ with the characteristic function

$$\begin{aligned} \Phi ^{(\alpha )}_\lambda (t) = \int _0^\infty (1 - x^\alpha t^\alpha )_+ \lambda (\mathrm{d}x) \end{aligned}$$

(24)

is called Kendall convolution (see, e.g., Urbanik, 1988; Borowiecka-Olszewska et al., 2015, Example 2.4).

As discussed in Remark 12, the generalized convolution of measures $\lambda _1, \lambda _2 \in \mathcal {P}_{+}$ can be uniquely defined by the convolution of point mass measures, which in the case of Kendall convolution, has the following form (see, e.g., Example 2.4 in Borowiecka-Olszewska et al., 2015)

$$\begin{aligned} \delta _x *_{\alpha } \delta _y = T_{x\vee y} \left\{ \left( \frac{x\wedge y}{x\vee y}\right) ^\alpha \pi _{2\alpha } + \left( 1- \left( \frac{x\wedge y}{x\vee y}\right) ^\alpha \right) \delta _1 \right\} . \end{aligned}$$

(25)

In the above expression, the quantity $T_{x\vee y}$ is the shift operator, $\pi _{2\alpha }$ denotes the standard Pareto distribution (scale 1) with parameter $2\alpha$, and 0/0 is assumed to be 0.

Appendix B: Proofs

This section contains proofs of our main results.

1.1 Appendix B.1: Proof of Theorem 1

Let $t \ge 0$. Observe that, due to the independence of $X_i, U_i, i = 1, 2, \ldots , n$, we have

$$\begin{aligned} {\mathbb {P}}\left( \bigvee _{i=1}^n \frac{X_i}{(U_i)^\frac{1}{\alpha }} \le t\right) = \prod _{i = 1}^n{\mathbb {P}}\left( \frac{X_i}{(U_i)^\frac{1}{\alpha }} \le t\right) = \prod _{i=1}^n G_{X_i}^{(\alpha )}(t). \end{aligned}$$

By Proposition 1, we have

$$\begin{aligned} \prod _{i=1}^n G_{X_i}^{(\alpha )}(t) = \prod _{i=1}^n \Phi _{X_i}^{(\alpha )}\left( \frac{1}{t}\right) . \end{aligned}$$

Due to Definition 5, combined with Property (ii) in Definition 3 (see Appendix A), applied to the Kendall convolution $*_\alpha$, there exists a random variable X with distribution $\lambda _1 *_\alpha \lambda _2 *_\alpha \cdots *_\alpha \lambda _n$ such that

$$\begin{aligned} \prod _{i=1}^n \Phi _{X_i}^{(\alpha )}\left( \frac{1}{t}\right) = \Phi _X^{(\alpha )}\left( \frac{1}{t}\right) . \end{aligned}$$

Finally, due to Proposition 1, there exists a uniform random variable U, independent of X, such that

$$\begin{aligned} \Phi _X^{(\alpha )}\left( \frac{1}{t}\right) = {\mathbb {P}}\left( \frac{X}{U^\frac{1}{\alpha }} \le t\right) . \end{aligned}$$

This completes the proof. $\square$

1.2 Appendix B.2: Proof of Theorem 2

For $i=1, \ldots , n$, let $Y_1, \ldots , Y_n$ be independent $\alpha$-slash versions of the $\{X_i\}$, with the CDFs

$$\begin{aligned} F_{Y_i}(t) = G^{(\alpha )}_{X_i}(t), \end{aligned}$$

(26)

respectively. By the independence of the $\{Y_i\}$, their maximum $Y := \bigvee _{i=1}^n Y_i$ has the CDF of the form

$$\begin{aligned} F_Y(t) = \prod _{i=1}^n F_{Y_i}(t), \quad t\in {\mathbb {R}}_+, \end{aligned}$$

(27)

so that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d} t} F_Y(t) = F_Y(t) \sum _{i=1}^n \frac{\frac{\mathrm{d}}{\mathrm{d}t} F_{Y_i}(t)}{F_{Y_i}(t)}, \quad t\in {\mathbb {R}}_+. \end{aligned}$$

(28)

Theorem 1 guarantees that Y is a slash version of some nonnegative random variable X, where X and Y satisfy (1). Finally, an application of Lemma 1 combined with (28) shows that the CDF of $X = \bigoplus _{i=1}^n X_i$ is of the form

$$\begin{aligned} F_X(t) = F_{Y}(t) + \frac{t}{\alpha } \frac{\mathrm{d}}{\mathrm{d}t} F_{Y}(t) = F_Y(t)\left\{ 1 + \frac{t}{\alpha } \sum _{i=1}^n \frac{\frac{\mathrm{d}}{\mathrm{d}t}F_{Y_i}(t)}{F_{Y_i}(t)}\right\} ,\quad t\in {\mathbb {R}}_+. \end{aligned}$$

The proof is completed by substituting (27) and (26) in the above expression. $\square$

1.3 Appendix B.3: Proof of Proposition 2

In view of (20), in order to prove Proposition 2 it is enough to prove (6). Upon applying Theorem 1 to deterministic $\{X_i\}$, i.e., $X_i = x_i$, $i = 1,2,\ldots , n$, we obtain

$$\begin{aligned} G^{(\alpha )}_{\bigoplus _{i=1}^n x_i}(t) = \prod _{i=1}^n \left[ 1 - \left( \frac{x_i}{t}\right) ^\alpha \right] \pmb {1}(t \ge x_i) . \end{aligned}$$

Hence, by Lemma 1, we have

$$\begin{aligned} F_{\bigoplus _{i=1}^n x_i}(t)&= \prod _{i=1}^n \left[ 1 - \left( \frac{x_i}{t}\right) ^\alpha \right] \pmb {1}(t \ge x_i) + \frac{t}{\alpha } \frac{\mathrm{d}}{{\mathrm{d}}t} \prod _{i=1}^n \left[ 1 - \left( \frac{x_i}{t}\right) ^\alpha \right] \pmb {1}(t \ge x_i) \\&= \left[ 1 + \sum _{i=1}^n (-1)^i \frac{\sum _{U \in \mathcal {U}_{i}^{(n)}}\prod _{j \in U}x_j}{t^{i\alpha }} + \sum _{i=1}^n (-1)^{i+1} i \frac{\sum _{U \in \mathcal {U}_{i}^{(n)}}\prod _{j \in U}x_j}{t^{i\alpha }}\right] \\&\quad \times \pmb {1}\left( t \ge \bigvee _{i=1}^n x_i\right) \\&= \left[ 1 + \sum _{i=2}^{n} (-1)^{i+1} (i-1) \frac{\sum _{U \in \mathcal {U}_{i}^{(n)}}\prod _{j \in U}x_j}{t^{i\alpha }}\right] \pmb {1}\left( t \ge \bigvee _{i=1}^n x_i\right) . \end{aligned}$$

The proof is completed by replacing $i-1$ with i in the above summation. $\square$

1.4 Appendix B.4: Direct proof of property (8)

First, we show that for any $x,y, t> 0$, we have $G^{(\alpha )}_x(t)G^{(\alpha )}_y(t) = G^{(\alpha )}_{x\oplus y}(t)$, that is

$$\begin{aligned} \left( 1 - \left( \frac{x}{t}\right) ^\alpha \right) _+ \left( 1 - \left( \frac{y}{t}\right) ^\alpha \right) _+ = \int _0^t \left( 1 - \left( \frac{s}{t}\right) ^\alpha \right) \mathrm{d}F_{x \oplus y}(s). \end{aligned}$$

(29)

Upon integrating by parts, we find the right-hand side of the above equation to be

$$\begin{aligned} \int _0^t \left( 1 - \left( \frac{s}{t}\right) ^\alpha \right) \mathrm{d}F_{x \oplus y}(s)= & {} \frac{\alpha }{t^\alpha } \int _0^t s^{\alpha - 1} F_{x\oplus y}(s) \mathrm{d}s \end{aligned}$$

(30)

$$\begin{aligned}= & {} \frac{\alpha }{t^\alpha } \int _0^t s^{\alpha - 1}\left( 1 - \frac{(xy)^\alpha }{s^{2\alpha }}\right) \pmb {1}\!\left( {s \ge x \vee y}\right) \mathrm{d}s\nonumber \\= & {} I_1 - I_2, \end{aligned}$$

(31)

where we have

$$\begin{aligned} I_1 = \frac{\alpha }{t^\alpha } \int _{x\vee y}^t s^{\alpha - 1} \mathrm{d}s = 1 - \frac{(x \vee y)^{\alpha }}{t^\alpha } \end{aligned}$$

(32)

and

$$\begin{aligned} I_2 = \frac{\alpha (xy)^\alpha }{t^\alpha } \int _{x\vee y}^t s^{-\alpha - 1} \mathrm{d}s = - \frac{(xy)^\alpha }{t^{2\alpha }} + \frac{(x\wedge y)^\alpha }{t^\alpha }. \end{aligned}$$

(33)

By combining (32) and (33) with (31), we obtain

$$\begin{aligned} \int _0^t \left( 1 - \left( \frac{s}{t}\right) ^\alpha \right) \mathrm{d}F_{x \oplus y}(s)&= 1 - \frac{(x \vee y)^{\alpha }}{t^\alpha } - \frac{(x\wedge y)^\alpha }{t^\alpha } + \frac{(xy)^\alpha }{t^{2\alpha }}\\&= \left( 1 - \left( \frac{x}{t}\right) ^\alpha \right) _+ \left( 1 - \left( \frac{y}{t}\right) ^\alpha \right) _+, \end{aligned}$$

producing (29). For general X with distribution $\lambda _1 \in \mathcal {P}_{+}$ and Y with distribution $\lambda _2 \in \mathcal {P}_{+}$, integration by parts leads to

$$\begin{aligned} G^{(\alpha )}_{X \oplus Y}(t) = \int _0^t \left( 1 - \left( \frac{s}{t}\right) ^\alpha \right) \mathrm{d}F_{X \oplus Y}(s) = \frac{\alpha }{t^\alpha }\int _0^t s^{\alpha - 1} F_{X \oplus Y}(s) \mathrm{d}s. \end{aligned}$$

Due to (20), the last expression is equivalent to

$$\begin{aligned} \frac{\alpha }{t^\alpha }\int _0^t s^{\alpha - 1} \int _0^\infty \int _0^\infty F_{x \oplus y}(s) \mathrm{d}F_X(x) \mathrm{d}F_Y(y)\mathrm{d}s, \end{aligned}$$

which, by Fubini’s theorem, is equivalent to

$$\begin{aligned} \int _0^\infty \int _0^\infty \left\{ \frac{\alpha }{t^\alpha }\int _0^t s^{\alpha - 1} F_{x \oplus y}(s) \mathrm{d}s \right\} \mathrm{d}F_X(x) \mathrm{d}F_Y(y)= & {} \int _0^\infty \int _0^\infty G^{(\alpha )}_{x\oplus y}(t) \mathrm{d}F_X(x) \mathrm{d}F_Y(y) \end{aligned}$$

(34)

$$\begin{aligned}= & {} \int _0^\infty \int _0^\infty G^{(\alpha )}_{x}(t) G^{(\alpha )}_{y}(t) \mathrm{d}F_X(x) \mathrm{d}F_Y(y)\\= & {} \int _0^\infty G^{(\alpha )}_{x}(t) \mathrm{d}F_X(x) \int _0^\infty G^{(\alpha )}_{y}(t) \mathrm{d}F_Y(y),\nonumber \end{aligned}$$

(35)

where (34) is by (30) and (35) is by (29). This completes the proof. $\square$

1.5 Appendix B.5: Proof of Theorem 3

We provide only a sketch of the proof of Theorem 3, as it is similar to the proof of Theorem 1. Let $\lambda _1, \lambda _2, \ldots , \lambda _n \in \mathcal {P}_{+}$ be distributions of random variables $X_1, X_2, \ldots , X_n$, respectively. First, observe that for any probability measure $\mu$ we have $\limsup _{x \rightarrow \infty } \int _0^\infty h(xt) \mu (dx) < 1$. Thus, due to Theorem 2 in Kucharczak and Urbanik (1974), the function h(u) satisfies all the conditions for being a kernel of a generalized ChF for some generalized convolution $*$. Hence, $\Phi _{X_i}\left( \frac{1}{t}\right) := {\mathbb {P}}\left( \frac{X_i}{T_i} \le t \right) = \int _0^\infty h\left( \frac{x}{t}\right) \lambda _i (dx)$ is a proper generalized characteristic function at $\frac{1}{t}$ for the generalized convolution $*$. Then, there exists a random variable X with distribution $\lambda _1 *\lambda _2 *\ldots *\lambda _n$ such that $\prod _{i = 1}^n \Phi _{X_i} \left( \frac{1}{t}\right) = \Phi _X\left( \frac{1}{t}\right)$. Thus, there exists a random variable T, independent of X and with survival function h(u), such that

$$\begin{aligned} {\mathbb {P}}\left( \bigvee _{i = 1}^n \frac{X_i}{T_i} \le t \right) = \prod _{i = 1}^n \Phi _{X_i} \left( \frac{1}{t}\right) = \Phi _X\left( \frac{1}{t}\right) = {\mathbb {P}}\left( \frac{X}{T} \le t \right) . \end{aligned}$$

This completes the proof. $\square$