On the role of skewness and kurtosis in tempered stable (CGMY) Lévy models in finance

Abstract

We study the structure and properties of an infinite-activity CGMY Lévy process $X$ with given skewness $S$ and kurtosis $K$ of $X_{1}$, without a Brownian component, but allowing a drift component. The jump part of such a process is specified by the Lévy density which is $C\mathrm {e}^{-Mx}/x^{1+Y}$ for $x>0$ and $C\mathrm {e}^{-G|x|}/|x|^{1+Y}$ for $x<0$. A main finding is that the quantity $R=S^{2}/K$ plays a major role, and that the class of CGMY processes can be parametrised by the mean $\mathbb{E}[X_{1}]$, the variance $\mathrm {Var}[X_{1}]$, $S$, $K$ and $Y$, where $Y$ varies in $[0,Y_{\mathrm {max}}(R))$ with $Y_{\mathrm {max}}(R)=(2-3R)/(1-R)$. Limit theorems for $X$ are given in various settings, with particular attention to $X$ approaching a Brownian motion with drift, corresponding to the Black–Scholes model; for this, sufficient conditions in a general Lévy process setup are that $K\to 0$ or, in the spectrally positive case, that $S\to 0$. Implications for moment fitting of log-return data are discussed. The paper also exploits the structure of spectrally positive CGMY processes as exponential tiltings (Esscher transforms) of stable processes, with the purpose of providing simple formulas for the log-return density $f(x)$, short derivations of its asymptotic form, and quick algorithms for simulation and maximum likelihood estimation.

Appendices

Appendix A: Proofs

In this Appendix, we slightly change notations. The time index $t$ of a stochastic process is not written with a subscript as in $X_{t}$, but with an argument in brackets as in $X(t)$.

Various techniques for proving functional limit theorems have been outlined in Sect. 3. We add here that even if subtraction is not continuous in $D[0,\infty )$, it is so in ℝ, and it therefore suffices to establish a limit of $X(1)$ or $X^{*}(1)$ separately for the positive and negative parts and then subtract. We also repeatedly use the fact from analysis that for the convergence $x\to x_{\infty}$ in a metric space, say $D[0,\infty )$, it suffices that every subsequence $(x_{n})$ of $x$ has a further subsequence $(x_{n_{k}})$ such that $x_{n_{k}}\to x_{\infty}$ as $k\to \infty$.

Proof of Theorem 1.2

(a) Consider a sequence $(X_{n})$ of Lévy processes. We can write $X^{*}_{n}=(1-\sigma _{n}^{2})^{1/2}W+J^{*}_{n}$, where $J^{*}_{n}$ is the jump part with Lévy measure, say, $\nu _{n}$ and centered to mean 0 and $\mathrm {Var}[X_{n}^{*}(1)]=1$. Thus the cumulants are $\kappa _{1,n}=0$, $\kappa _{2,n}=1$ and

$$ \kappa _{k,n} = \int _{-\infty}^{\infty }x^{k}\nu _{n}(\,\mathrm {d}x), \qquad k=3,4,\ldots $$

Further, we have $\int x^{2}\nu _{n}(\,\mathrm {d}x)=\sigma _{n}^{2}$. The assumption is that $\kappa _{4,n}\to 0$, and we must prove that $X_{n}(1)\Rightarrow V$, where $V$ is standard normal.

Consider first the pure jump case $\sigma _{n}^{2}=\kappa _{2,n}=1$. If each $\nu _{n}$ is concentrated on $\{x:|x|\le 1\}$, we have for $k>4$ that $\kappa _{k,n}\le \kappa _{4,n}$ and hence $\kappa _{k,n}\to 0$. Also $\kappa _{3,n}\to 0$ since $\kappa _{3,n}^{2}\le \kappa _{2,n}\kappa _{4,n}\to 0$ by a general Lévy process inequality. Thus all cumulants and hence all moments converge, which is sufficient (e.g. Kallenberg [28, Chap. 5, Exercise 11]). In the case of a general support of the $\nu _{n}$, write $X_{n}(1)= X_{n}'(1)+ X_{n}''(1)$, where $X_{n}'(1)$ is the part of $X_{n}(1)$ coming from jumps of sizes $x$ with $|x|>1$ and $X_{n}''(1)$ the rest. Then

$$ \mathrm {Var}[X_{n}''(1) ] = \int _{\{|x|>1\}}x^{2}\nu _{n}(\,\mathrm {d}x) \le \int _{-\infty}^{\infty }x^{4}\nu _{n}(\,\mathrm {d}x) = \kappa _{4,n} \longrightarrow 0.$$

Thus $X_{n}''(1)\to 0$ and $\mathrm {Var}X_{n}'(1)\to 1$ so that we can neglect $X_{n}''(1)$ and use what was proved for the finite support case to get $X_{n}'(1)\Rightarrow V$ and hence $X_{n}(1)\Rightarrow V$.

In the presence of a Brownian component, it suffices to show that each subsequence has a further subsequence $(n_{k})$ with $X_{n_{k}}(1)\Rightarrow V$. This is clear if $\sigma _{n}^{2}\to 0$ along the subsequence. Otherwise, we can choose $n_{k}$ with $\sigma _{n_{k}}^{2}\to \sigma ^{2}$ for some $0<\sigma ^{2}\le 1$. Define $\widetilde{J}_{n}=J_{n}/\sigma _{n}$. Then the pure jump process $\widetilde{J}_{n}$ has mean 0, variance 1 and cumulants $\widetilde{\kappa}_{k,n}=\kappa _{k,n}/\sigma _{n}^{k}$, $k>2$. Thus $\widetilde{\kappa}_{4,n}\to 0$ because $\sigma ^{2}>0$, and we can use what was already proved to conclude that $\widetilde{J}_{n_{k}}(1)\Rightarrow \widetilde{V}$. Hence

$$\begin{aligned} X_{n_{k}}(1) = (1-\sigma _{n_{k}}^{2})^{1/2}W(1)+ \sigma _{n_{k}}\widetilde{J}_{n_{k}}(1) \Longrightarrow (1-\sigma ^{2})^{1/2}W(1)+ \sigma \widetilde{V}\ \stackrel{d}{=} \ V. \end{aligned}$$

The proof of (b) is almost the same, noting that by positivity we can replace the bounds in terms of $\kappa _{4,n}$ by bounds in terms of $\kappa _{3,n}$. □

Proof of Proposition 5.1

Adding a Brownian component increases $\kappa _{2}$, but leaves $\kappa _{3},\kappa _{4},\ldots$ unchanged, so that we may assume that $X$ has no Brownian component. Let $Z$ be a random variable with $\mathbb{P}[Z\in \mathrm {d}x]= x^{2}\nu (\mathrm {d}x)/\kappa _{2}$. Then $\mathbb{E}[Z^{k}]=\kappa _{k+2}/\kappa _{2}$, and thus the inequality $\mathbb{E}[Z^{2}] \ge (\mathbb{E}[Z])^{2}$ means that $\kappa _{3}^{2}/(\kappa _{2}\kappa _{4})\le 1$ as claimed in the first part. Equality holds if and only if $Z$ and then $\nu$ is degenerate at some $z_{0}$ (then $\lambda =\nu (\{z_{0}\})$.

For the second part, assume that we have $n(x)=\int _{0}^{\infty }\mathrm {e}^{-ax}\,V^{+}(\mathrm {d}a)$ for $x>0$ and $n(x)=\int _{0}^{\infty }\mathrm {e}^{-a|x|}\,V^{-}(\mathrm {d}a)$ for $x<0$, so that

$$ \kappa ^{\pm}_{k} = \int _{0}^{\infty }x^{k} \int _{0}^{\infty } \mathrm {e}^{-ax}\,V^{\pm}(\mathrm {d}a). $$

Define $V=V^{+}+V^{-}$. Using that an $\text{exponential}(a)$ random variable $U_{a}$ has the $k$th moment $k!/a^{k}$, we get that for $k$ even,

$$\begin{aligned} \kappa _{k} = \kappa _{k}^{+} + \kappa _{k}^{-} &= \int _{0}^{\infty }x^{k} \int _{0}^{\infty }\mathrm {e}^{-ax}\,V(\mathrm {d}a)\,\mathrm {d}x \\ & = \int _{0}^{\infty }\int _{0}^{\infty }x^{k}\mathbb{P}[U_{a}\in \mathrm {d}x]\,a^{-1}V(\mathrm {d}a)\,\mathrm {d}x \\ & = k! \int _{0}^{\infty }a^{-k-1}V(\mathrm {d}a) = k! \, \kappa _{4} \mathbb{E}[Z^{4-k}/4!], \end{aligned}$$

where now $Z$ has distribution

$$ \mathbb{P}[Z\in \mathrm {d}a] = \frac{a^{-5}V(\mathrm {d}a)}{\int _{0}^{\infty }b^{-5}V(\mathrm {d}b)} = \frac{a^{-5}V(\mathrm {d}a)}{\kappa _{4}/4!}.$$

(A.1)

Similarly,

$$ |\kappa _{k}| \le \kappa _{k}^{+} +\kappa _{k}^{-} = k!\,\kappa _{4} \mathbb{E}[Z^{4-k}]/4! \qquad \text{for }k\text{ odd}.$$

(A.2)

Since $\mathbb{E}[Z^{0}]=1$, this gives

$$ \frac{S^{2}}{K} = \frac{\kappa _{3}^{2}}{\kappa _{2}\kappa _{4}} \le \frac{(3!\,\kappa _{4}\mathbb{E}[Z]/4!)^{2}}{(2!\,\kappa _{4}\mathbb{E}[Z^{2}]/4! )( 4!\,\kappa _{4}\mathbb{E}[Z^{0}]/4!)} \le \frac{6^{2}}{2\cdot 24} = \frac{3}{4}. $$

Equality holds if and only if $Z$ and then $V$ is degenerate at some $a_{0}$. However, the inequality in (A.2) is strict if $X$ is spectrally two-sided. Thus the necessary and sufficient condition for equality is that $X$ is either spectrally positive with $V=V^{+}$ degenerate at some $a_{0}$ or spectrally negative with $V=V^{-}$ degenerate at some $-a_{0}$. □

Remark A.1

The methods in the proof of Proposition 5.1 give similar bounds on other normalised cumulants. For example, for a spectrally positive finite-variation process, we get $\kappa _{3}\kappa _{4}\le \kappa _{2}\kappa _{5}$ and (use Hölder’s inequality with $\mathbb{P}[Z\in \mathrm {d}x]= x\nu (\mathrm {d}x)/\kappa _{1}$) also $\kappa _{2}^{2}\le \kappa _{1}\kappa _{3}$.

Proof of Proposition 5.2

For the first part, let $Z$ be as in the first part of the proof of Proposition 5.1. Then just use $(\mathbb{E}[Z^{2}])^{2}\le \mathbb{E}[Z^{4}]$, with equality if and only if $Z^{2}$ is degenerate at some $z_{0}^{2}$ (then $\lambda ^{+}=\nu (\{z_{0}\})$, $\lambda ^{-}=\nu (\{-z_{0}\})$). For the second part, use that $\mathbb{E}[Y^{-1}]\ge (\mathbb{E}[Y])^{-1}$ for $Y>0$ by Jensen’s inequality and the convexity of $y\mapsto 1/y$, $y>0$. Taking $Y=Z^{2}$ with now $Z$ as in (A.1), this gives

$$ \frac{\kappa _{4}^{2}}{\kappa _{2}\kappa _{6}} = \frac{\kappa _{4}^{2}}{(2!\,\kappa _{4}\mathbb{E}[Z^{2}]/4! )( 6!\,\kappa _{4}\mathbb{E}[Z^{-2}]/4!)} \le \frac{1}{(1/12)\cdot 30} = \frac{2}{5}. $$

Equality holds if and only if $Z^{2}$ is degenerate at some $a_{0}^{2}$, which is the same as the stated condition. □

Proof of Proposition 5.3, range of $R_{4}$ for CGMY

Assume without loss of generality that $G<\infty$ and let $\rho =M/G$. Then

$$\begin{aligned} R_{4}\ &= \frac{\Gamma (4-Y)^{2}}{\Gamma (2-Y)\Gamma (6-Y)} \frac{(M^{Y-4}+G^{Y-4})^{2}}{(M^{Y-2}+G^{Y-2})(M^{Y-6}+G^{Y-6})} \\ &= \frac{(2-Y)(3-Y)}{(4-Y)(5-Y)} \frac{\rho ^{2Y-8}+1+2\rho ^{Y-4}}{\rho ^{2Y-8}+1+\rho ^{Y-2}+\rho ^{Y-6}} \\ & \le \frac{(2-Y)(3-Y)}{(4-Y)(5-Y)}, \end{aligned}$$

(A.3)

where the last step follows from $(2\rho ^{Y-4})/(\rho ^{Y-2}+\rho ^{Y-6})=(2\rho ^{2})/(\rho ^{4}+1)$ and the inequality $2z\le 1+z^{2}$ with $z=\rho ^{2}$. Here the right-hand side of (A.3) decreases monotonically from $3/5$ to 0 as $Y$ increases from 0 to 2, with $R_{4}=3/10$ if and only if $Y=0$, $\rho =1$. □

Proof of Lemma 8.1

(a) Continuity of $\varphi _{Y}$ is clear, as well as $\varphi _{Y}(1/2)=0$, $\varphi _{Y}(1)=1$. Also, by elementary calculus, one gets $f'_{Y}=h_{Y}k_{Y}/g_{Y}^{2}$, where

$$ k_{Y}(\pi ) = 2g_{Y}(\pi )h'_{Y}(\pi )-h_{Y}(\pi )g'_{Y}(\pi ) = \pi ^{P}+(1- \pi )^{P}+(3Q+1)\pi ^{P}(1-\pi )^{P}$$

with $P=(3Q-1)/2$. Here $g_{Y}(\pi )$ and $k_{Y}(\pi )$ are obviously strictly positive for $\pi \in [1/2,1]$, and $h_{Y}(\pi )$ is equally so for $\pi \in (1/2,1)$. This implies $f'_{Y}(\pi )>0$ for $\pi \in (1/2,1)$ and the strictly increasing property. That $\varphi _{Y}(\pi )\sim 1-\pi$ as $\pi \uparrow 1$ follows since then $g_{Y}(\pi )\sim 1$, $h_{Y}(\pi )\sim 1$, $k_{Y}(\pi )\sim 1$ and hence $f'_{Y}(\pi )\sim 1$.

(b) Let $L=L(\pi )=\ell (1-\pi )=(1-\pi )/\pi$ and $T=1/(2-Y)$, so that $L<1$ and $Q=1+2T$, $Q_{1}=1+T$. We can write

$$\begin{aligned} \varphi _{Y}(\pi ) = \frac{ (\pi \pi ^{T}-(1-\pi )(1-\pi )^{T} )^{2}}{\pi \pi ^{2T}+(1-\pi )(1-\pi )^{2T}} = \pi \frac{(1-L^{1+T})^{2}}{1+L^{1+2T}} \end{aligned}$$

(for the second expression, multiply both the numerator and denominator by $\pi ^{-2T}$). Since $L<1$, $L^{T}$ is a strictly decreasing function of $T$ and hence of $Y$, implying $\varphi _{Y}(\pi )$ to be strictly increasing in $Y$. Since $L<1$ and $T\to \infty$, we have $L^{T}\to 0$ for such a fixed $\pi$, giving $\varphi _{Y}(\pi )\to \pi$. □

Proof of Proposition 6.7

Parts (a) and (b) are direct corollaries of parts (a) resp. (b) of Theorem 1.2. For part (c), we use characteristic functions together with

$$\begin{aligned} M^{*} &= \sqrt{\frac{(2-Y)(3-Y)}{K}} = \sqrt{ \frac{R(3-Y)^{2}}{K(1-R)}}, \end{aligned}$$

(A.4)

$$\begin{aligned} \kappa ^{*}(s) &= \frac{{M^{*}}^{2}}{Y(Y-1)}\bigl((1-s/M^{*})^{Y}-1 \bigr)+s\frac{M^{*}}{Y-1}, \end{aligned}$$

(A.5)

as follows from Proposition 6.1 and $\kappa ^{*}_{2}=1$. Here (A.5) holds for $Y\ne 0,1$ and the final term is needed to ensure ${\kappa ^{*}}'(0)=0$. Now $2-Y=R/(1-R)$ and so (A.4) gives first that $Y\to 2$, $M^{*}\to 0$ and next that

$$\begin{aligned} & |(1-\mathrm {i}s/M^{*})^{Y-2} | \\ &= {M^{*}}^{2-Y}\bigl(1+{\mathrm {o}}(1)\bigr) = \exp \big((2-Y)\log M^{*} \big)\bigl(1+{\mathrm {o}}(1)\bigr) \\ &= \exp \bigg(\frac{R}{1-R}\bigl(\log R/2-\log K/2-\log (1-R)/2+ {\mathrm {o}}(1)\big)\bigg)\bigl(1+{\mathrm {o}}(1)\bigr) \\ & = \exp \big({\mathrm {o}}(1)\big)\bigl(1+{\mathrm {o}}(1)\bigr) = 1+{\mathrm {o}}(1). \end{aligned}$$

Here we used that $R\log K\to 0$ and $K\to \infty$ implies $R\to 0$ and hence $R\log R\to 0$. Hence by (A.5),

$$\begin{aligned} \kappa ^{*}(\mathrm {i}s) &= \frac{{M^{*}}^{2}}{Y(Y-1)}\Big((1-\mathrm {i}s/M^{*})^{2} \big(1+{\mathrm {o}}(1)\big)-1\Big)+\mathrm {i}s\frac{{M^{*}}}{Y-1} \\ &= \frac{{M^{*}}^{2}}{Y(Y-1)}(1-\mathrm {i}s/M^{*})^{2}{\mathrm {o}}(1)+ \frac{{M^{*}}^{2}}{Y(Y-1)}\bigl((1-\mathrm {i}s/M^{*})^{2}-1\bigr) + \mathrm {i}s\frac{{M^{*}}}{Y-1} \\ &= {\mathrm {o}}(1)+ \frac{{M^{*}}^{2}}{Y(Y-1)} (-s^{2}/{M^{*}}^{2}-2 \mathrm {i}s/M^{*} ) +\mathrm {i}s\frac{{M^{*}}}{Y-1} \\ &= {\mathrm {o}}(1)-\frac{s^{2}}{2+{\mathrm {o}}(1)}+\mathrm {i}s \frac{{M^{*}}}{Y-1}\bigg(1-\frac{2}{Y}\bigg) = -\frac{s^{2}}{2}+ {\mathrm {o}}(1). \end{aligned}$$

 □

Proof of Proposition 6.8

The gamma distribution of $X(1)$ corresponding to the asserted limit has cumulant generating function $-\kappa _{2}\gamma ^{2}\log (1-s/\gamma )$ for $s< M$; so it suffices to prove that $\kappa (s)$ has this asymptotic form. But $R\uparrow 2/3$ implies $Y\downarrow 0$ and $M\to \sqrt{6\kappa _{2}/\kappa _{4}}= \gamma$, and so by (6.3f),

$$\begin{aligned} \kappa (s) &= \frac{\kappa _{2}}{Y(Y-1)}M^{2-Y}\bigl((M-s)^{Y}-M^{Y} \bigr) \\ & = \frac{\kappa _{2}}{Y(Y-1)}M^{2}\Big(\exp \big(Y\log (1-s/M)\big)-1 \Big) \\ \vspace{-5mm} &\sim \frac{\kappa _{2}}{Y-1}\gamma ^{2}\log (1-s/M)\ \sim -\sigma ^{2} \gamma ^{2}\log (1-s/M). \end{aligned}$$

 □

Proof of Proposition 7.2

This is almost immediate from Proposition 6.7 and the positive and negative parts of a non-skewed CGMY process being identical. One just needs to notice that the $R$ of these is given by $R=(2-Y)/(3-Y)$, cf. (6.4), so that $(2-Y)/K\to \infty$ is equivalent to $R/K\to \infty$. Further, in $C$, one needs to replace $\kappa _{2}$ by $\kappa ^{+}_{2}=\kappa _{2}/2$. □

Proof of Theorem 8.3

(a) That $Y\uparrow Y_{\mathrm {max}}(R)$ is equivalent to $A\uparrow 1$ and hence by an easy continuity argument, we then have $\pi _{2}\uparrow 1$. This implies

$$\begin{aligned} \pi _{4}&=\frac{\pi _{2}^{Q}}{(1-\pi _{2})^{Q}+\pi _{2}^{Q}}\sim \frac{1}{0+1}=1, \\ \ 1-\pi _{4}&=\frac{(1-\pi _{2})^{Q}}{(1-\pi _{2})^{Q}+\pi _{2}^{Q}} \sim (1-\pi _{2})^{Q}={\mathrm {o}}(1-\pi _{2}). \end{aligned}$$

The expressions for $C,G,M$ then show that $G\to \infty$ and that $M$ and $C$ have the limits given by Proposition 6.1. This proves (a).

(b) First, $Y\to 0$ is equivalent to $A\to 3R/2$. Since $3R/2$ is an interior point of $(0,1)$, the limit $\pi _{2}$ of the solution to $\varphi _{Y}(\pi _{2})=A$ is indeed as asserted. The rest is then easy. □

Proof of Theorem 8.4

We establish (a) and (b) by showing that (d) $K(2-Y)\to 0$ implies $S^{+}\to 0$ and $S^{-}\to 0$. By Theorem 1.2(b), this gives that both $({X^{+}})^{*}$ and $({X^{-} })^{*}$ have Brownian limits. Hence so has $X^{*}$ as desired. Now

$$ K^{+}=\frac{\kappa _{4}^{+}}{({\kappa _{2}^{+}})^{2}}= \frac{\pi _{4}\kappa _{4}}{\pi _{2}^{2}\kappa _{2}^{2}}= \frac{\pi _{4}}{\pi _{2}^{2}}K,\qquad \text{and similarly} \qquad K^{-}= \frac{1-\pi _{4}}{(1-\pi _{2})^{2}}K. $$

(A.6)

Since $\pi _{2},\pi _{4}\in [1/2,1)$, this gives that $(S^{+})^{2}=K^{+}R^{+}=K^{+}(2-Y)/(3-Y)\to 0$ as desired. Similarly, $({S^{-}})^{2}=K^{-}(2-Y)/(3-Y)\to 0$ except when possibly $\pi _{2}\to 1$ along some subsequence. However, then ${X^{-}}$ can be ignored in the limit.

The proof of (c) is a similar application of (A.6), only using Proposition 6.7(c) instead of Theorem 1.2(b). □

Appendix B: NIG

The density of $X_{1}$ is

$$\frac{\alpha \delta }{\pi }\exp \bigg(\delta \sqrt{\alpha ^{2}-\beta ^{2}} +\beta (x-\mu )\bigg) \frac{K_{1} (\alpha \sqrt{\delta ^{2}+(x-\mu )^{2}} )}{\sqrt{\delta ^{2}+(x-\mu )^{2}}}, $$

which is called the $\text{NIG}(\alpha ,\beta ,\mu ,\delta )$ density; $\alpha$ is the tail heaviness of steepness, $\beta$ is the skewness, $\delta$ is the scale, and $\mu$ is the location. For the density of $X_{t}$, just replace $\delta$ by $\delta t$ and $\mu$ by $\mu t$. Numerically, we just used Matlab’s besselk routine for $K_{1}$. Scaling by $v$ changes the parameters to $\alpha v$, $\beta v$, $\delta /v$.

The cumulants are

$$ \kappa _{k} = \delta \frac{q_{k}(\beta )}{(\alpha ^{2}-\beta ^{2})^{k-1/2}},\qquad k\ge 2,$$

(B.1)

where the polynomials $q_{k}$ are given by the recurrence relation

$$ q_{k}(s) = (\alpha ^{2}-s^{2})q'_{k-1}(s)+(2k-3)sq_{k-1}(s)$$

starting from $q_{2}(s)=\alpha ^{2}$. In particular,

$$\begin{aligned} q_{3}(\beta ) &= 3\alpha ^{2}\beta ,\qquad q_{4}(\beta )=3\alpha ^{4}+12 \alpha ^{2}\beta ^{2}, \qquad q_{5}(\beta ) = 45\alpha ^{4}s+60 \alpha ^{2}\beta ^{3}, \\ q_{6}(\beta ) &= 45\alpha ^{6}+540\alpha ^{4}\beta ^{2}+360\alpha ^{2} \beta ^{4}. \end{aligned}$$

These expressions for the $\kappa _{k}$ with $k\le 4$ are classical for $k\le 4$, while for $k>4$, we refer to Asmussen and Bladt [4]. With $\rho =\beta ^{2}/\alpha ^{2}$, we get

$$R = \frac{q_{3}(\beta )^{2}}{q_{2}(\beta )q_{4}(\beta )} = \frac{3\rho}{1+4\rho}, \qquad R_{4} = \frac{q_{4}(\beta )^{2}}{q_{2}(\beta )q_{6}(\beta )} = \frac{(1+4\rho )^{2}}{5+60\rho +40\rho ^{2}}. $$

Letting $\rho$ vary from 0 to 1 shows that $R$ ranges from 0 to $3/5=0.6$. For $R_{4}$, the range is $[0.171,5/21)=[0.171,0.238)$.

Proposition B.1

Given values $\mathring {{\kappa }}_{1},\mathring {{\kappa }}_{2},\mathring {{\kappa }}_{3},\mathring {{\kappa }}_{4}$, an NIG process satisfying $\kappa _{k}=\mathring {{\kappa }}_{k}$ for $k=1,2,3,4$ exists if and only if $\mathring {{R}}=\mathring {{S}}^{2}/\mathring {{K}}\le 3/5$. In that case, the fitted parameters can be computed by first letting $\rho = \mathring {{R}}/(3-5\mathring {{R}})$ and then

$$\begin{aligned} \alpha = \frac{1}{1- \widehat {{\rho }}}\sqrt{ \frac{3\mathring {{\kappa }}_{2}(1+4\rho )}{\mathring {{\kappa }}_{4}}}, \qquad \beta ={\mathrm{sign}}( \mathring {{\kappa }}_{3}) \alpha \sqrt{\rho}, \\ \delta =\frac{1}{\alpha ^{2}}\mathring {{\kappa }}_{2}(\alpha ^{2}-\beta ^{2})^{3/2}, \qquad m=\mathring {{\kappa }}_{1}- \frac{\delta \beta}{\sqrt{\alpha ^{2}-\beta ^{2}}}. \end{aligned}$$

Note that related results are given in Eriksson et al. [18] and Ghysels and Wang [21].

Proof of Proposition B.1

With $\rho =\beta ^{2}/\alpha ^{2}$, we have

$$ R=\frac{3\rho /4}{\rho +1/4} = \frac{3}{4} \bigg(1-\frac{1}{4\rho +1} \bigg) $$

which increases monotonically from 0 to $3/5$ as $\rho$ varies from 0 to 1. We also get $\rho =R/(3-5R)$. The stated expression for $\alpha$ then follows from

$$ \frac{\kappa _{4}}{\kappa _{2}} = \frac{3(\alpha ^{2}+4\beta ^{2})}{(\alpha ^{2}-\beta ^{2})^{2}} = \frac{3(1+4\rho )}{\alpha ^{2}(1-\rho )^{2}}, $$

and those for $\beta$, $\delta$ from $\rho =\beta ^{2}/\alpha ^{2}$ resp. (B.1) with $k=2$. □

Appendix C: Meixner

Some basic references are Schoutens [49], [50, Chap. 5] and Mozzola and Muliere [36]. It is the special case $b^{+}=b^{-}$ of the generalised $z$-process in Grigelionis [23], for which $b$ is allowed to take different values $b^{+},b^{-}$ for $x>0$ resp. $x<0$. The cumulant function and the mean/variance/skewness/kurtosis are

$$\begin{aligned} \kappa (s) &= 2d\log \bigl(\cos (b/2)\bigr)-2d\log \big(\cos (as+b)/2 \big)+ms, \\ \kappa _{1}&=m+da\tan (b/2),\qquad \kappa _{2}=d \frac{a^{2}}{2\cos ^{2}(b/2)}, \\ S&=\frac{\sin b}{\sqrt{d(\cos b+1)}}=\frac{1}{\sqrt{d}}\sqrt{2}\sin (b/2), \qquad K=\frac{1}{d}\bigl(3-2\cos ^{2}(b/2)\bigr). \end{aligned}$$

Letting $t(s)=\tan ((as+b)/2) )$ and using that the derivative of $\tan x$ is $1+\tan ^{2} x$, one gets from $\kappa ^{(1)}(s)=dat(s)$ that $\kappa ^{(k)}(s)$ is a polynomial $p_{k}(t(s))$ of degree $k$ in $t(s)$, given by the recursion

$$ p_{k}(x) = \frac{a}{2}p'_{k-1}(x)(1+x^{2}), \qquad k\ge 2.$$

Thus the cumulants are $\kappa _{k}=p_{k}(\tan (b/2))$. In particular,

$$\begin{aligned} p_{2}(x)&= \frac{da^{2}}{2}(1+x^{2}), \qquad p_{3}(x)= \frac{da^{3}}{4}(2x+2x^{3}), \\ p_{4}(x)&= \frac{da^{4}}{8} (2+8x^{2}+6x^{4} ), \qquad p_{5}(x) = \frac{da^{5}}{16} (16x+40x^{3}+24x^{5} ), \\ p_{6}(x)&= \frac{da^{6}}{32} (16+136x^{2}+120x^{4}+120x^{6} ). \end{aligned}$$

From these formulas, we found numerically (letting $x^{2}=\tan ^{2}(b/2)$ vary in $[0,\infty )$) that the range of $R_{4}$ is $[0.240,0.340]$.

The density of $X_{1}$ is

$$\frac{(2\cos (b/2))^{2d}}{3a\pi \Gamma (2d)}\mathrm {e}^{b(x-m)/a} \big| \Gamma \big(d+\mathrm {i}(x-m)/a\big) \big|^{2}. $$

Numerically, the most obvious possibility is to write the absolute values of the gamma function $|\Gamma (d+\mathrm {i}y)|^{2}$ of a complex argument as $I_{1}^{2}+I_{2}^{2}$, where

$$ I_{1} = \int _{0}^{\infty }z^{d-1}\cos (y\log z)\mathrm {e}^{-z}\, \mathrm {d}z, \qquad I_{2} = \int _{0}^{\infty }z^{d-1}\sin (y\log z) \mathrm {e}^{-z}\,\mathrm {d}z, $$

and evaluate $I_{1},I_{2}$ by numerical integration. However, in some cases, Matlab’s integration routine failed to perform this task, possibly because of the highly oscillatory nature of the integrands. Instead, we then used simulation based on the representations $I_{1}=\Gamma (d)\mathbb{E}[\cos (y\log Z)]$, $I_{2}=\Gamma (d)\mathbb{E}[\sin (y\log Z)]$ with $Z$ gamma-distributed with form parameter $d$ and rate parameter 1.

For the density of $X_{t}$, replace $d$ by $dt$. Scaling by $v$ changes $a$ to $a/v$.

Proposition C.1

Given values $\mathring {{\kappa }}_{1},\ldots ,\mathring {{\kappa }}_{4}$, a Meixner process satisfying $\kappa _{k}=\mathring {{\kappa }}_{k}$ for $k=1,2,3,4$ exists if and only if $\mathring {{R}}<2/3$. In that case, the fitted parameters can be computed by first letting $q^{2}=(2-3\mathring {{R}})/(2(1-\mathring {{R}}))$ and then recursively defining

$$\begin{aligned} d&=2(1-q)^{2}/\mathring {{S}}^{2},\qquad b={\mathrm{sign}}(\mathring {{\kappa }}_{3})2\arcsin \Big(\sqrt{1-q^{2}}\Big), \\ a&=\sqrt{2\mathring {{\kappa }}_{2} q^{2}/d}, \qquad m=\mathring {{\kappa }}_{1}-da\tan (b/2). \end{aligned}$$

Proof

In terms of $q=\cos (b/2)$, we have $R=S^{2}/K=2(1-q^{2})/(3-q^{2})$, so we must have $q^{2}=(2-3R)/(2(1-R))$. The rest is then easy. □