Simulation of Student–Lévy processes using series representations

Abstract

Lévy processes have become very popular in many applications in finance, physics and beyond. The Student–Lévy process is one interesting special case where increments are heavy-tailed and, for 1-increments, Student t distributed. Although theoretically available, there is a lack of path simulation techniques in the literature due to its complicated form. In this paper we address this issue using series representations with the inverse Lévy measure method and the rejection method and prove upper bounds for the mean squared approximation error. In the numerical section we discuss a numerical inversion scheme to find the inverse Lévy measure efficiently. We extend the existing numerical inverse Lévy measure method to incorporate explosive Lévy tail measures. Monte Carlo studies verify the error bounds and the effectiveness of the simulation routine. As a side result we obtain series representations of the so called inverse gamma subordinator which are used to generate paths in this model.

A Appendix

In the “Appendix” we prove the statements of Sect. 3.

Proof of Lemma 2

Recall that

$$\begin{aligned} Q(\mathrm {d}u)=u^{-1}\int _0^{\infty }e^{-su}\nu 2\left[ \pi ^2 2\nu s\left( J_{\frac{\nu }{2}}^2(\sqrt{2\nu s})+Y_{\frac{\nu }{2}}^2(\sqrt{2\nu s})\right) \right] ^{-1} \mathrm {d}s\mathrm {d}u. \end{aligned}$$

We use the inequality

$$\begin{aligned} J_{\nu }^2(x)+Y_{\nu }^2(x)>\frac{2}{\pi x}, \end{aligned}$$

(32)

first derived by Schafheitlin (1906); an elegant proof can be found in Watson (1995). Hence,

$$\begin{aligned} \nu g_{\frac{\nu }{2}}(2\nu s)=\nu \left[ \pi ^2\nu s\left( J_{\frac{\nu }{2}}^2(\sqrt{2\nu s})+Y_{\frac{\nu }{2}}^2(\sqrt{2\nu s})\right) \right] ^{-1}<\sqrt{\frac{\nu }{2\pi ^2s}}. \end{aligned}$$

By standard integration,

$$\begin{aligned} Q(\mathrm {d}u)<u^{-1}\int _0^{\infty }e^{-su}\sqrt{\frac{\nu }{2\pi ^2s}}\mathrm {d}s\mathrm {d}u=\sqrt{\frac{\nu }{2\pi u^3}}\mathrm {d}u \end{aligned}$$

such that (16) follows. To derive (17) the tail mass function is bounded by

$$\begin{aligned} Q([z,\infty ))=\int _z^{\infty }Q(\mathrm {d}u)<\int _{z}^{\infty }\sqrt{\frac{\nu }{2\pi u^3}}\mathrm {d}u=\sqrt{\frac{2\nu }{\pi z}}. \end{aligned}$$

Since $Q([z,\infty ))$ is strictly decreasing and continuous, $Q^{\leftarrow }(y)$, is the true inverse and

$$\begin{aligned} Q^{\leftarrow }(y)&=\mathrm {inf}\{z>0:Q([z,\infty ))<y\}\\&<\mathrm {inf}\{z>0:\sqrt{\frac{2\nu }{\pi z}}<y\}\\&=\frac{2\nu }{\pi y^2}, \end{aligned}$$

which completes the proof. $\square$

Proof of Theorem 2

Note that

$$\begin{aligned} Y_t-Y_t^{(n)}&=\sum _{i=1}^{\infty }Q^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) {\mathbb {1}}_{\{U_i\le t\}}-\sum _{i=1}^nQ^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) {\mathbb {1}}_{\{U_i\le t\}}\\&=\sum _{i=n+1}^{\infty }Q^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) {\mathbb {1}}_{\{U_i\le t\}} \end{aligned}$$

Denote by $q(y):=\frac{2\nu }{\pi y^2}$ the bound for $Q^{\leftarrow }$ derived in Lemma 2. We start by proving (19). Taking expectations to obtain

$$\begin{aligned} E\left[ \sum _{i=n+1}^{\infty }Q^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) {\mathbb {1}}_{\{U_i\le t\}}\right]&=\sum _{i=n+1}^{\infty }E\left[ Q^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) \right] E\left[ {\mathbb {1}}_{\{U_i\le t\}}\right] \nonumber \\&< \frac{t}{T}\sum _{i=n+1}^{\infty }E\left[ q\left( \frac{\Gamma _i}{T}\right) \right] , \end{aligned}$$

(33)

where we have used the monotonicity of the expected value. Since the $\Gamma _i$s are $\Gamma (i,1)$ distributed (with density function denoted by $\gamma _i(x)$), (33) is equal to

$$\begin{aligned} \frac{t}{T}\sum _{i=n+1}^{\infty }\int _0^{\infty }q(x/T)\gamma _i(x)\mathrm {d}x&=\frac{t}{T}\sum _{i=n+1}^{\infty }\frac{2\nu }{\pi }\frac{T^2}{(i-1)(i-2)}\nonumber \\&=\frac{2\nu }{\pi }\frac{T}{n-1}t \end{aligned}$$

(34)

It remains to prove (20). Using the monotone convergence theorem

$$\begin{aligned}&E\left[ \left( \sum _{i=n+1}^{\infty }Q^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) {\mathbb {1}}_{\{U_i\le t\}}\right) ^2\right] \\&\quad =\sum _{i=n+1}^{\infty }\sum _{j=n+1}^{\infty }E\left[ Q^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) {\mathbb {1}}_{\{U_i\le t\}}Q^{\leftarrow }\left( \frac{\Gamma _j}{T}\right) {\mathbb {1}}_{\{U_j\le t\}}\right] , \end{aligned}$$

since $Q^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) {\mathbb {1}}_{\{U_i\le t\}}\ge 0$ for all i. Next, by the Cauchy–Schwarz inequality

$$\begin{aligned}&\sum _{i=n+1}^{\infty }\sum _{j=n+1}^{\infty }E\left[ Q^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) {\mathbb {1}}_{\{U_i\le t\}}Q^{\leftarrow }\left( \frac{\Gamma _j}{T}\right) {\mathbb {1}}_{\{U_j\le t\}}\right] \nonumber \\&\quad \le \sum _{i=n+1}^{\infty }\sum _{j=n+1}^{\infty }\sqrt{E\left[ Q^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) ^2{\mathbb {1}}_{\{U_j\le t\}}\right] E\left[ Q^{\leftarrow }\left( \frac{\Gamma _j}{T}\right) ^2{\mathbb {1}}_{\{U_j\le t\}}\right] }\nonumber \\&\quad <\sum _{i=n+1}^{\infty }\sum _{j=n+1}^{\infty }\sqrt{\frac{t^2}{T^2}E\left[ q\left( \frac{\Gamma _i}{T}\right) ^2\right] E\left[ q\left( \frac{\Gamma _j}{T}\right) ^2\right] }\nonumber \\&\quad =\frac{t}{T}\sum _{i=n+1}^{\infty }\sum _{j=n+1}^{\infty }\sqrt{\int _0^{\infty }q(x/T)^2\gamma _i(x)\mathrm {d}x \times \int _0^{\infty }q(x/T)^2\gamma _j(x)\mathrm {d}x}\nonumber \\&\quad =\frac{t}{T}\frac{4\nu ^2}{\pi ^2}\sum _{i=n+1}^{\infty }\sum _{j=n+1}^{\infty }\sqrt{\frac{T^4}{(i-1)(i-2)(i-3)(i-4)}\frac{T^4}{(j-1)(j-2)(j-3)(j-4)}}, \end{aligned}$$

(35)

Since $i,j\ge n+1\ge 5$, we can bound (35) using $(i-1)(i-2)(i-3)(i-4)\ge (i-4)^4$ by

$$\begin{aligned} \frac{t}{T}\frac{4\nu ^2}{\pi ^2}\left( \sum _{i=n+1}^{\infty }\frac{T^2}{(i-4)^2}\right) ^2=\frac{4\nu ^2T^3}{\pi ^2}t\psi '(n-3)^2, \end{aligned}$$

(36)

where $\psi '(x)$ denotes the first derivative of the digamma function $\psi (x):=\frac{\Gamma '(x)}{\Gamma (x)}$ (also called polygamma function of order 1). Guo et al. (2015) provide a sharp bound for polygamma functions. The inequality for $\psi '(x)$ is

$$\begin{aligned} |\psi '(x)|<\frac{1}{x+\frac{1}{2}}+\frac{1}{x^2}. \end{aligned}$$

(37)

Applying (37) to (36) we obtain the bound

$$\begin{aligned} E\left[ \left( \sum _{i=n+1}^{\infty }Q^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) {\mathbb {1}}_{\{U_i\le t\}}\right) ^2\right] <\frac{4\nu ^2T^3}{\pi ^2}t\left( \frac{1}{n-2.5}+\frac{1}{(n-3)^2}\right) ^2. \end{aligned}$$

(38)

$\square$

Proof of Theorem 3

Again,

$$\begin{aligned} X_t-X_t^{(n)}=\sum _{i=n+1}^{\infty }\sqrt{Q^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) }V_i{\mathbb {1}}_{\{U_i\le t\}}. \end{aligned}$$

Note that $E[V_i]=0$ and that $V_i$ is independent of $Q^{\leftarrow }\left( \frac{\Gamma _i}{T}\right)$ and $U_i$. Hence, by Fubini’s theorem,

$$\begin{aligned} E\left[ \sum _{i=n+1}^{\infty }\sqrt{Q^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) }V_i{\mathbb {1}}_{\{U_i\le t\}}\right] =0. \end{aligned}$$

Furthermore, analogously to Theorem 2,

$$\begin{aligned}&E\left[ \left( \sum _{i=n+1}^{\infty }\sqrt{Q^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) }V_i{\mathbb {1}}_{\{U_i\le t\}}\right) ^2\right] \nonumber \\ =&\sum _{i=n+1}^{\infty }\sum _{j=n+1}^{\infty }E\left[ \sqrt{Q^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) }V_i{\mathbb {1}}_{\{U_i\le t\}}\sqrt{Q^{\leftarrow }\left( \frac{\Gamma _j}{T}\right) }V_j{\mathbb {1}}_{\{U_j\le t\}}\right] \nonumber \\ =&\sum _{i=n+1}^{\infty }\sum _{j=n+1}^{\infty }E\left[ \sqrt{Q^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) }\sqrt{Q^{\leftarrow }\left( \frac{\Gamma _j}{T}\right) }{\mathbb {1}}_{\{U_i\le t\}}{\mathbb {1}}_{\{U_j\le t\}}\right] E\left[ V_iV_j\right] . \end{aligned}$$

(39)

Since $E[V_iV_j]=\delta _{i,j}$, (39) equals

$$\begin{aligned} \sum _{i=n+1}^{\infty }E\left[ Q^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) {\mathbb {1}}_{\{U_i\le t\}}\right] <\frac{t}{T}\sum _{i=n+1}^{\infty }E\left[ q\left( \frac{\Gamma _i}{T}\right) \right] =\frac{2\nu }{\pi }\frac{T}{n-1}t \end{aligned}$$

as in the proof of Theorem 2. $\square$

Proof of Corollary 2

Since $N_{\tau }=\#\{i\in {\mathbb {N}}:\Gamma _i\le \tau \}$ and the $\Gamma _i$ are unit Poisson arrival times, $N_{\tau }\sim Poi(\tau )$. We now use the law of iterated expectation.

$$\begin{aligned} E[(X_t-X_t^{\tau })^2|\Gamma _2\le \tau ]&=E\left[ E[(X_t-X_t^{\tau })^2|N_{\tau },\Gamma _2\le \tau ]|\Gamma _2\le \tau \right] \\&<E\left[ \frac{2\nu }{\pi }\frac{Tt}{N_{\tau }-1}\bigg |\Gamma _2\le \tau \right] , \end{aligned}$$

by Theorem 3. The conditional expected value $E\left[ \frac{1}{N_{\tau }-1}|\Gamma _2\le \tau \right]$ exists and $N_{\tau }|\Gamma _2\le \tau$ follows a truncated Poisson distributed with density function

$$\begin{aligned} P[N_{\tau }=k|N_{\tau }\ge 2]=\frac{e^{-\tau }\tau ^k}{k!(1-P[N_{\tau }\le 2])}=\frac{e^{-\tau }\tau ^k}{k!(1-\Gamma (3,\tau )/2)}. \end{aligned}$$

Hence, the conditional expectation is

$$\begin{aligned} E\left[ \frac{1}{N_{\tau }-1}\big |\Gamma _2\le \tau \right]&=\sum _{k=2}^{\infty }\frac{1}{k-1}\frac{e^{-\tau }\tau ^k}{k!(1-\Gamma (3,\tau )/2)}\\&=\frac{e^{-\tau }}{(1-\Gamma (3,\tau )/2)}\left( \tau +1-e^\tau -\tau \gamma +\tau Ei(\tau )-\tau \log (\tau )\right) , \end{aligned}$$

which completes the proof. $\square$

Proof of Corollary 3

We only prove the claim for the deterministic truncation; the random truncation bound follows as in Corollary 2. Note that Lemma 2 implies that $\frac{\mathrm {d}Q}{\mathrm {d}Q_0}\le 1$ and that the tail inverse $Q_0^{\leftarrow }(y)=\frac{2\nu }{\pi y^2}$ exists in closed form. Let us start with the mean squared error

$$\begin{aligned}&E\left[ \left( \sum _{i=n+1}^{\infty }\sqrt{Q_0^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) }V_i{\mathbb {1}}\left( \left\{ \frac{\mathrm {d}Q}{\mathrm {d}Q_0}\left( Q_0^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) \right) \ge W_i\right\} \right) {\mathbb {1}}_{\{U_i\le t\}}\right) ^2\right] \\&\quad = \sum _{i=n+1}^{\infty }E\left[ Q_0^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) {\mathbb {1}}\left( \left\{ \frac{\mathrm {d}Q}{\mathrm {d}Q_0}\left( Q_0^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) \right) \ge W_i\right\} \right) {\mathbb {1}}_{\{U_i\le t\}}\right] , \end{aligned}$$

analogously as in the proof of Theorem 3, since $E[V_iV_j]=\delta _{i,j}$. By the law of iterated expectation, this is equal to

$$\begin{aligned}&\sum _{i=n+1}^{\infty }E\left[ E\left[ \left. Q_0^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) {\mathbb {1}}\left( \left\{ \frac{\mathrm {d}Q}{\mathrm {d}Q_0}\left( Q_0^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) \right) \ge W_i\right\} \right) \right| \Gamma _i\right] \right] E\left[ {\mathbb {1}}_{\{U_i\le t\}}\right] \\&\quad =\sum _{i=n+1}^{\infty }E\left[ Q_0^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) P\left[ \left. \frac{\mathrm {d}Q}{\mathrm {d}Q_0}\left( Q_0^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) \right) \ge W_i\right| \Gamma _i\right] \right] E\left[ {\mathbb {1}}_{\{U_i\le t\}}\right] \\&\quad =\sum _{i=n+1}^{\infty }E\left[ Q_0^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) \frac{\mathrm {d}Q}{\mathrm {d}Q_0}\left( Q_0^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) \right) \right] E\left[ {\mathbb {1}}_{\{U_i\le t\}}\right] \\&\quad \le \sum _{i=n+1}^{\infty }E\left[ Q_0^{\leftarrow }\left( \frac{\Gamma _i}{T}\right) \right] E\left[ {\mathbb {1}}_{\{U_i\le t\}}\right] , \end{aligned}$$

because $\frac{\mathrm {d}Q}{\mathrm {d}Q_0}\le 1$. The rest of the proof follows as in the proof of Theorem 3 since $Q_0^{\leftarrow }\equiv q$. $\square$

Proof of Proposition 3

Asmussen and Rosiński (2001) showed that the distributional convergence is implied by

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}\frac{\sigma _{\varepsilon }}{\varepsilon }=+\infty . \end{aligned}$$

(40)

We show that $\lim _{\varepsilon \rightarrow 0}\frac{\sigma _{\varepsilon }^2}{\varepsilon ^2}=+\infty$. Recall that

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}\frac{\sigma _{\varepsilon }^2}{\varepsilon ^2}=\lim _{\varepsilon \rightarrow 0}\frac{\int _0^{\varepsilon }u^2\int _0^{\infty }u^{-1}e^{-su}\nu g_{\frac{\nu }{2}}(2\nu s)\mathrm {d}s\mathrm {d}u}{\varepsilon ^2}. \end{aligned}$$

(41)

Using l’Hôspital’s rule, (41) is equal to

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}\frac{\varepsilon ^2\int _0^{\infty }\varepsilon ^{-1}e^{-s\varepsilon }\nu g_{\frac{\nu }{2}}(2\nu s)\mathrm {d}s}{2\varepsilon }=\lim _{\varepsilon \rightarrow 0}\frac{1}{2}\int _0^{\infty }e^{-s\varepsilon }g_{\frac{\nu }{2}}(2\nu s)\mathrm {d}s. \end{aligned}$$

(42)

The monotone convergence theorem can be applied to (41) and thus

$$\begin{aligned} \frac{1}{2}\int _0^{\infty }\lim _{\varepsilon \rightarrow 0}e^{-s\varepsilon }\nu g_{\frac{\nu }{2}}(2\nu s)\mathrm {d}s=\frac{1}{2}\int _0^{\infty }\nu g_{\frac{\nu }{2}}(2\nu s)\mathrm {d}s=\infty . \end{aligned}$$

$\square$

Proof of Proposition 4

Note that in the non-finite variation case $\mu _{\varepsilon }$ has to be zero (Sato 1999). Recall the Lévy measure for the univariate Student–Lévy process (with no drift and standard scaling) is given by

$$\begin{aligned} \Pi (\mathrm {d}x)=\frac{\nu 2^{\frac{3}{4}}|x|^{-\frac{1}{2}}}{\pi ^{\frac{1}{2}}}\int _{0}^{\infty }s^{\frac{1}{4}}K_{\frac{1}{2}}\left( \sqrt{2s}|x|\right) g_{\frac{\nu }{2}}(2\nu s)\mathrm {d}s\mathrm {d}x. \end{aligned}$$

(43)

In the following we use the identity

$$\begin{aligned} K_{\frac{1}{2}}(z)=\sqrt{\frac{\pi }{2}}e^{-z}z^{-\frac{1}{2}} \end{aligned}$$

(44)

for $z>0$. As $\Pi$ is a symmetric measure,

$$\begin{aligned} \sigma _{\varepsilon }^2=\int _{-\varepsilon }^{\varepsilon }x^2\Pi (\mathrm {d}x)=2\int _{0}^{\varepsilon }x^2\Pi (\mathrm {d}x). \end{aligned}$$

Again using l’Hôspital’s rule, for some constant $C>0$ that may change from line to line

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}\frac{\sigma _{\varepsilon }^2}{\varepsilon ^2}&=\lim _{\varepsilon \rightarrow 0}\frac{C\int _0^{\varepsilon }x^2|x|^{-\frac{1}{2}}\int _{0}^{\infty }s^{\frac{1}{4}}K_{\frac{1}{2}}\left( \sqrt{2s}|x|\right) g_{\frac{\nu }{2}}(2\nu s)\mathrm {d}s\mathrm {d}x}{\varepsilon ^2}\\&=\lim _{\varepsilon \rightarrow 0}C\frac{\varepsilon ^2\varepsilon ^{-\frac{1}{2}}\int _{0}^{\infty }s^{\frac{1}{4}}e^{-\sqrt{2s}\varepsilon }\left( \sqrt{2s}\right) ^{-\frac{1}{2}}\varepsilon ^{-\frac{1}{2}}g_{\frac{\nu }{2}}(2\nu s)\mathrm {d}s}{\varepsilon }\\&=\lim _{\varepsilon \rightarrow 0}C\int _0^{\infty }e^{-\sqrt{2s}\varepsilon }g_{\frac{\nu }{2}}(2\nu s)\mathrm {d}s\\&=C\int _0^{\infty }g_{\frac{\nu }{2}}(2\nu s)\mathrm {d}s\\&=\infty . \end{aligned}$$

The second-to-last last step uses the monotone convergence theorem.