Series Representation of Jointly S$\alpha$S Distribution via Symmetric Covariations

Abstract

We introduce the notion of symmetric covariation, which is a new measure of dependence between two components of a symmetric $\alpha$-stable random vector, where the stability parameter $\alpha$ measures the heavy-tailedness of its distribution. Unlike covariation that exists only when $\alpha \in (1,2]$, symmetric covariation is well defined for all $\alpha \in (0,2]$. We show that symmetric covariation can be defined using the proposed generalized fractional derivative, which has broader usages than those involved in this work. Several properties of symmetric covariation have been derived. These are either similar to or more general than those of the covariance functions in the Gaussian case. The main contribution of this framework is the representation of the characteristic function of bivariate symmetric $\alpha$-stable distribution via convergent series based on a sequence of symmetric covariations. This series representation extends the one of bivariate Gaussian.

Appendix A. Proofs of Statements

1.1 A.1. Proof of Lemma 2.4

Proof

For $\beta \in {\mathbb {R}}_+\backslash {\mathbb {Z}}_+$, from (2.2) we write

$$\begin{aligned} {}_{a}^{m}\text {D}_x^\beta \left( |x-a|^{p}\right) =\mathbb {1}_{\{x\geqslant a\}}{}_a\widetilde{\text {D}}_x^\beta \left( |x-a|^{p}\right) +(-1)^{m}\mathbb {1}_{\{x<a\}}{}_x\widetilde{\text {D}}_a^\beta \left( |x-a|^{p}\right) .\nonumber \\ \end{aligned}$$

(A.1)

Let $n=\lfloor \beta \rfloor +1$. On the one hand, if $x\geqslant a$, taking the left Riemann–Liouville fractional derivative of the function $x\mapsto |x-a|^{p}$, it yields

$$\begin{aligned}&{}_a\widetilde{\text {D}}_x^\beta \left( |x-a|^{p}\right) \nonumber \\&\quad =\frac{1}{\Gamma (n-\beta )}\frac{\, \mathrm {d}^n}{\, \mathrm {d}x^n}\int _a^x(t-a)^p(x-t)^{n-\beta +1}\, \mathrm {d}t\nonumber \\&\quad =\frac{1}{\Gamma (n-\beta )}\frac{\, \mathrm {d}^n}{\, \mathrm {d}x^n}\int _0^1(x-a)^p(1-\tau )^p(x-a)^{n-\beta -1}\tau ^{n-\beta -1}(x-a)\, \mathrm {d}\tau \nonumber \\&\quad =\frac{1}{\Gamma (n-\beta )}\left( \frac{\, \mathrm {d}^n}{\, \mathrm {d}x^n}(x-a)^{p+n-\beta }\right) \left( \int _0^1(1-\tau )^p\tau ^{n-\beta -1}\, \mathrm {d}\tau \right) \nonumber \\&\quad =\frac{B(n-\beta ,p+1)}{\Gamma (n-\beta )}\left( \frac{\, \mathrm {d}^n}{\, \mathrm {d}x^n}(x-a)^{p+n-\beta }\right) \nonumber \\&\quad =\frac{\Gamma (p+1)}{\Gamma (n-\beta +p+1)}\left( \frac{\Gamma (n-\beta +p+1)}{\Gamma (p-\beta +1)}(x-a)^{p-\beta }\right) \nonumber \\&\quad =\frac{\Gamma (p+1)}{\Gamma (p-\beta +1)}(x-a)^{p-\beta }. \end{aligned}$$

(A.2)

Here we have used the change of variables $t=x-(x-a)\tau$ and the fact that the beta function B satisfies

$$\begin{aligned} B(x,y)=\frac{\Gamma (x)\Gamma (y)}{\Gamma (x+y)},~\text{ for } \text{ all }~x,y>0. \end{aligned}$$

On the other hand, if $x<a$, the right Riemann–Liouville fractional derivative yields

$$\begin{aligned}&{}_x\widetilde{\text {D}}_a^\beta \left( |x-a|^{p}\right) \nonumber \\&\quad =(-1)^n\frac{1}{\Gamma (n-\beta )}\frac{\, \mathrm {d}^n}{\, \mathrm {d}x^n}\int _x^a(a-t)^p(t-x)^{n-\beta +1}\, \mathrm {d}t\nonumber \\&\quad =(-1)^n\frac{1}{\Gamma (n-\beta )}\frac{\, \mathrm {d}^n}{\, \mathrm {d}x^n}\int _0^1(a-x)^p(1-\tau )^p(a-x)^{n-\beta -1}\tau ^{n-\beta -1}(a-x)\, \mathrm {d}\tau \nonumber \\&\quad =\frac{(-1)^n}{\Gamma (n-\beta )}\left( \frac{\, \mathrm {d}^n}{\, \mathrm {d}x^n}(a-x)^{p+n-\beta }\right) \left( \int _0^1(1-\tau )^p\tau ^{n-\beta -1}\, \mathrm {d}\tau \right) \nonumber \\&\quad =\frac{(-1)^nB(n-\beta ,p+1)}{\Gamma (n-\beta )}\left( \frac{\, \mathrm {d}^n}{\, \mathrm {d}x^n}(a-x)^{p+n-\beta }\right) \nonumber \\&\quad =\frac{(-1)^n\Gamma (p+1)}{\Gamma (n-\beta +p+1)}\left( (-1)^n\frac{\Gamma (n-\beta +p+1)}{\Gamma (p-\beta +1)}(a-x)^{p-\beta }\right) \nonumber \\&\quad =\frac{\Gamma (p+1)}{\Gamma (p-\beta +1)}(a-x)^{p-\beta }. \end{aligned}$$

(A.3)

Here we have taken $t=x+(a-x)\tau$. Therefore (2.4) holds for $\beta$ being non-integer, by using (A.1), (A.2) and (A.3).

For $\beta \in {\mathbb {Z}}_+$, by using (2.3) we write

$$\begin{aligned} {}_{a}^{m}\text {D}_x^\beta \left( |x-a|^{p}\right)= & {} \text {sign}^{m+n+1}(x-a)\frac{\, \mathrm {d}^\beta }{\, \mathrm {d}x^\beta }\left( |x-a|^{p}\right) \\= & {} \frac{\Gamma (p+1)}{\Gamma (p-\beta +1)}|x-a|^{p-\beta } \text {sign}^{m}(x-a), \end{aligned}$$

where $n=\lfloor \beta \rfloor +1$. Hence (2.4) is obtained for all real numbers $\beta \geqslant 0$. $\square$

1.2 A.2. Proof of Lemma 3.10

Proof

On the one hand, in view of Lemma 2.7.5 in [37], $\varvec{Y}=(Y_1,Y_2)$ is also an S$\alpha$S random vector. On the other hand, considering $\varvec{Y}=\left( \sum _{k=1}^{n}a_kX_k, \sum _{k=1}^{n}b_kX_k\right)$, we can rewrite the characteristic function as:

$$\begin{aligned}&{\mathbb {E}}\exp \{i(\theta _1Y_1+\theta _2Y_2)\}\nonumber \\&\quad ={\mathbb {E}}\exp \left\{ i\sum _{k=1}^{n}(\theta _1a_k +\theta _2b_k)X_k\right\} \nonumber \\&\quad =\exp \left\{ -\int _{S_n}\left| \theta _1\sum _{k=1}^{n}a_ks_k +\theta _2\sum _{k=1}^{n}b_ks_k\right| ^{\alpha }\varvec{\varGamma _X}(\, \mathrm {d}\varvec{s})\right\} . \end{aligned}$$

Now we want to show

$$\begin{aligned} \int _{S_n}\left| \theta _1\sum _{k=1}^{n}a_ks_k+\theta _2\sum _{k=1}^{n}b_ks_k \right| ^{\alpha }\varvec{\varGamma _X}(\, \mathrm {d}\varvec{s})=\int _{S_2}|\theta _1t_1+\theta _2t_2|^{\alpha }\varvec{\varGamma }(\, \mathrm {d}\varvec{t}), \end{aligned}$$

(A.4)

with $\varvec{\varGamma }=\widehat{\varvec{\varGamma _{X}}}\circ h^{-1}$. To verify the above equation, we first write

$$\begin{aligned} \int _{S_2}|\theta _1t_1+\theta _2t_2|^{\alpha }\varvec{\varGamma }(\, \mathrm {d}\varvec{t})= & {} \int _{S_2}|\theta _1t_1+\theta _2t_2|^{\alpha }\widehat{\varvec{\varGamma _{X}}} (\, \mathrm {d}(h^{-1}(\varvec{t}))). \end{aligned}$$

Then applying Lemma 3.9, we have

$$\begin{aligned}&\int _{S_2}|\theta _1t_1+\theta _2t_2|^{\alpha }\widehat{\varvec{\varGamma _{X}}}(\, \mathrm {d}(h^{-1}(\varvec{t})))\nonumber \\&\quad =\int _{S_n}\left| \frac{\theta _1\sum _{k=1}^{n}a_ks_k+\theta _2\sum _{k=1}^nb_ks_k}{\big (\big (\sum _{k=1}^{n}a_ks_k\big )^{2}+\big (\sum _{k=1}^{n}b_ks_k\big )^{2}\big )^{1/2}}\right| ^{\alpha }\widehat{\varvec{\varGamma _{X}}}(\, \mathrm {d}\varvec{s})\nonumber \\&\quad =\int _{S_n}\left| \theta _1\sum _{k=1}^{n}a_ks_k+\theta _2\sum _{k=1}^{n}b_ks_k\right| ^{\alpha }\varvec{\varGamma _X}(\, \mathrm {d}\varvec{s}). \end{aligned}$$

Therefore, (A.4) holds and hence $\varvec{\varGamma }=\widehat{\varvec{\varGamma _{X}}}\circ h^{-1}$, denoted by $\varvec{\varGamma _Y}$, is a spectral measure of Y. $\square$

1.3 A.3. Proof of Proposition 3.16

Proof

We first prove (i). By using Remark 3.3, we have

$$\begin{aligned}&[aX_1,bX_2]_{\alpha ,1,1}+[aX_1,bX_2]_{\alpha ,\alpha -1,1}\\&=\int _{\{(s_1,s_2)\in S_2:~|as_1|\leqslant |bs_2|\}} \left( as_1\right) ^{\langle 1\rangle }\left( bs_2\right) ^{\langle \alpha -1\rangle }\varvec{\varGamma _X}(\, \mathrm {d}\varvec{s})\\&\qquad +\int _{\{(s_1,s_2)\in S_2:~|as_1|>|bs_2|\}} \left( as_1\right) ^{\langle \alpha -1\rangle }\left( bs_2\right) ^{\langle 1\rangle }\varvec{\varGamma _X}(\, \mathrm {d}\varvec{s})\\&\qquad +\int _{\{(s_1,s_2)\in S_2:~|as_1|\leqslant |bs_2|\}} \left( as_1\right) ^{\langle \alpha -1\rangle }\left( bs_2\right) ^{\langle 1\rangle }\varvec{\varGamma _X}(\, \mathrm {d}\varvec{s})\\&\qquad +\int _{\{(s_1,s_2)\in S_2:~|as_1|>|bs_2|\}} \left( as_1\right) ^{\langle 1\rangle }\left( bs_2\right) ^{\langle \alpha -1\rangle }\varvec{\varGamma _X}(\, \mathrm {d}\varvec{s})\\&\quad =\int _{S_2} \left( as_1\right) ^{\langle 1\rangle }\left( bs_2\right) ^{\langle \alpha -1\rangle }\varvec{\varGamma _X}(\, \mathrm {d}\varvec{s})+\int _{S_2} \left( as_1\right) ^{\langle \alpha -1\rangle }\left( bs_2\right) ^{\langle 1\rangle }\varvec{\varGamma _X}(\, \mathrm {d}\varvec{s})\\&\quad =ab^{\langle \alpha -1\rangle }[X_1,X_2]_{\alpha }+ba^{\langle \alpha -1\rangle }[X_2,X_1]_{\alpha }. \end{aligned}$$

Here in the last equality we have used the fact that $s_1^{\langle 1\rangle }=s_1$ for all $s_1\in {\mathbb {R}}$. Hence (3.8) is proved.

Next we prove (ii). Using (3.8), the following system of linear equations holds:

$$\begin{aligned} \left\{ \begin{array}{ll} &{}a_1b_1^{\langle \alpha -1\rangle }[X_1,X_2]_{\alpha }+b_1a_1^{\langle \alpha -1\rangle }[X_2,X_1]_{\alpha }=[a_1X_1,b_1X_2]_{\alpha ,1,1}+[a_1X_1,b_1X_2]_{\alpha ,\alpha -1,1}, \\ &{}a_2b_2^{\langle \alpha -1\rangle }[X_1,X_2]_{\alpha }+b_2a_2^{\langle \alpha -1\rangle }[X_2,X_1]_{\alpha }=[a_2X_1,b_2X_2]_{\alpha ,1,1}+[a_2X_1,b_2X_2]_{\alpha ,\alpha -1,1}. \end{array}\right. \end{aligned}$$

(A.5)

Since $a_1b_1^{\langle \alpha -1\rangle }b_2a_2^{\langle \alpha -1\rangle }\ne a_2b_2^{\langle \alpha -1\rangle }b_1a_1^{\langle \alpha -1\rangle }$, the solution of (A.5) is uniquely obtained as in (3.9). $\square$

1.4 A.4. Proof of Lemma 5.6

Proof

Assume $[X_1, X_2]_{\alpha , k, 1}=0$ for all $k\in {\mathbb {Z}}$. Applying this assumption to Theorem 4.3, we have

$$\begin{aligned} \sigma ^\alpha _{\varvec{X}}(1, 1)=\sum _{k \text{ is } \text{ even }}\frac{(\alpha )_k}{k!}[X_1, X_2]_{\alpha , k, 0}={\widetilde{T}}_1+{\widetilde{T}}_2, \end{aligned}$$

(A.6)

where

$$\begin{aligned} {\widetilde{T}}_1=\sum _{k \text{ is } \text{ even }}\frac{(\alpha )_k}{k!}\int _{\{(s_1,s_2)\in S_2:~|s_1|\leqslant |s_2|\}}|s_1|^k|s_2|^{\alpha -k}\varvec{\varGamma _X}(\, \mathrm {d}\varvec{s}) \end{aligned}$$

(A.7)

and

$$\begin{aligned} {\widetilde{T}}_2=\sum _{k \text{ is } \text{ even }}\frac{(\alpha )_k}{k!}\int _{\{(s_1,s_2)\in S_2:~|s_1|>|s_2|\}}|\lambda s_1|^{\alpha -k}|s_2|^k\varvec{\varGamma _X}(\, \mathrm {d}\varvec{s}). \end{aligned}$$

(A.8)

From Proposition 4.2, we have that for all $b\ne 0$ and all $x\in [-|b|,|b|]$,

$$\begin{aligned} |x+b|^\alpha +|x-b|^\alpha= & {} \sum _{k=0}^{+\infty }\frac{(a)_k}{k!}|b|^{\alpha -k}\text {sign}^k(b)x^k+\sum _{k=0}^{+\infty }\frac{(a)_k}{k!}|b|^{\alpha -k}\text {sign}^k(-b)x^k\nonumber \\= & {} \sum _{k=0}^{+\infty }\frac{(a)_k}{k!}|b|^{\alpha -k}\text {sign}^k(b)x^k\left( 1+(-1)^k\right) \nonumber \\= & {} 2\sum _{k \text{ is } \text{ even }}\frac{(a)_k}{k!}|b|^{\alpha -k}\text {sign}^k(b)x^k\nonumber \\= & {} 2\sum _{k \text{ is } \text{ even }}\frac{(a)_k}{k!}|x|^k|b|^{\alpha -k}. \end{aligned}$$

(A.9)

It follows from (A.9), (A.7) and (A.8) that

$$\begin{aligned} {\widetilde{T}}_1=\frac{1}{2}\int _{\{(s_1,s_2)\in S_2:~|s_1|\leqslant |s_2|\}}\left( |s_1+s_2|^\alpha +|s_1-s_2|^{\alpha }\right) \varvec{\varGamma _X}(\, \mathrm {d}\varvec{s}) \end{aligned}$$

and

$$\begin{aligned} {\widetilde{T}}_2=\frac{1}{2}\int _{\{(s_1,s_2)\in S_2:~|s_1|>|s_2|\}}\left( |s_1+s_2|^\alpha +|s_1-s_2|^{\alpha }\right) \varvec{\varGamma _X}(\, \mathrm {d}\varvec{s}). \end{aligned}$$

(A.6) then becomes

$$\begin{aligned} \int _{ S_2}|s_1+s_2|^\alpha \varvec{\varGamma _X}(\, \mathrm {d}\varvec{s})=\frac{1}{2}\int _{ S_2}\left( |s_1+s_2|^\alpha +|s_1-s_2|^{\alpha }\right) \varvec{\varGamma _X}(\, \mathrm {d}\varvec{s}). \end{aligned}$$

(A.10)

Now recall the following inequality (see, e.g., Lemma 2.7.13 in [37]): for $x,y\in {\mathbb {R}}$ and $p\geqslant 0$,

$$\begin{aligned} |x+y|^p+|x-y|^p\geqslant \min \{2^{p},2\}\max \{|x|^p,|y|^p\}. \end{aligned}$$

(A.11)

It then results from (A.10) and (A.11) that

$$\begin{aligned} \int _{ S_2}|s_1+s_2|^\alpha \varvec{\varGamma _X}(\, \mathrm {d}\varvec{s})\geqslant \min \{2^{\alpha -1},1\}\max \left\{ \int _{ S_2}|s_1|^{\alpha }\varvec{\varGamma _X}(\, \mathrm {d}\varvec{s}), \int _{ S_2}|s_2|^{\alpha }\varvec{\varGamma _X}(\, \mathrm {d}\varvec{s})\right\} , \end{aligned}$$

i.e.,

$$\begin{aligned} \Vert X_1+X_2\Vert _\alpha ^\alpha \geqslant \min \left\{ 2^{\alpha -1},1\right\} \max \left\{ \Vert X_1\Vert _\alpha ^\alpha ,\Vert X_2\Vert _\alpha ^\alpha \right\} . \end{aligned}$$

This proves Lemma 5.6. $\square$