On a positivity preserving numerical scheme for jump-extended CIR process: the alpha-stable case

Abstract

We propose a positivity preserving implicit Euler–Maruyama scheme for a jump-extended Cox–Ingersoll–Ross (CIR) process where the jumps are governed by a compensated spectrally positive $\alpha$-stable process for $\alpha \in (1,2)$. Different to the existing positivity preserving numerical schemes for jump-extended CIR or constant elasticity variance process, the model considered here has infinite activity jumps. We calculate, in this specific model, the strong rate of convergence and give some numerical illustrations. Jump extended models of this type were initially studied in the context of branching processes and was recently introduced to the financial mathematics literature to model sovereign interest rates, power and energy markets.

Appendix

1.1 Moment estimate of X

In this subsection, we show that for $h(x) = |x|^\frac{1}{\alpha }$ the solution of (1.1) has $\beta$-th moment for $\beta \in [1,\alpha )$.

Lemma 5.1

For $\beta \in [1,\alpha )$, the $\beta$-th moment of X is finite or more explicitly, there exists a constant $C_0>0$ such that

$$\begin{aligned} \sup _{t\leqslant T} \mathbb {E}\big [|X_{t}|^{\beta }\big ] \leqslant C_0(\alpha -\beta )^{-1} e^{C_0(\alpha -\beta )^{-1}T}. \end{aligned}$$

(5.1)

Proof

In the following, let $(\tau _m)_{m\in \mathbb {N}^+}$ be a localizing sequence of stopping times so that when stopped at $\tau _m$, all local martingales are martingales. By apply the Itô formula to $X^{\beta }$, we obtain

$$\begin{aligned} (X_{t \wedge \tau _m})^{\beta } =x_0^{\beta } +M_{t \wedge \tau _m} +I_{t \wedge \tau _m} +J_{t \wedge \tau _m} +K_{t \wedge \tau _m}, \end{aligned}$$

(5.2)

where we have set

$$\begin{aligned} {M}_t&:= \sigma _1 \beta \int _{0}^{t} (X_{s})^{\beta -1/2} dW_{s}\\&\quad +\int _{0}^{t} \int _{0}^{\infty } \left\{ (X_{s-}+\sigma _2 h(X_{s-})z)^{\beta }-(X_{s-})^{\beta } \right\} {{\widetilde{N}}}(ds,dz),\\ {I}_t&:=\beta \int _{0}^{t} (X_{s})^{\beta -1} \{a-kX_{s}\}ds, \quad {J}_t := \sigma _1^2 \frac{\beta (\beta -1)}{2} \int _{0}^{t} (X_{s})^{\beta -1}ds,\\ {K}_t&:= \int _{0}^{t} \int _{0}^{\infty } \Big \{ (X_{s-}+\sigma _2 h(X_{s-})z)^{\beta }-(X_{s-})^{\beta }-\sigma _2 h(X_{s-})z (X_{s-})^{\beta -1} \Big \} \nu (dz)ds. \end{aligned}$$

The martingale term $M_{t\wedge \tau _m}$ can be removed after taking the expectation. Next we consider $K_{t \wedge \tau _m}$. For $z \in (0,1)$, by the second order Taylor’s expansion for the map $x \mapsto x^{\beta }$, we have

$$\begin{aligned} (y+xz)^{\beta }-y^{\beta }-\beta xz y^{\beta -1}&=\alpha (\beta -1) |xz|^2 \int _{0}^{1} \theta (y+\theta xz)^{\beta -2} d\theta \\&\leqslant \frac{\beta (\beta -1) |xz|^2 }{y^{2-\beta }}. \end{aligned}$$

For $z \in [1,\infty )$, by the first order Taylor’s expansion for the map $x \mapsto x^{\beta }$ and the Hölder continuity of the map $x \mapsto x^{\beta -1}$, we have

$$\begin{aligned} \left| (y+xz)^{\beta }-y^{\beta }-\beta xz y^{\beta -1}\right| \leqslant \alpha |xz| \int _{0}^{1} \left| (y+\theta xz)^{\beta -1}-y^{\beta -1} \right| d\theta \leqslant \alpha |xz|^{\beta }. \end{aligned}$$

Hence, the expectation of $|K_{t \wedge \tau _m}|$ is bounded by

$$\begin{aligned} \beta (\beta -1)\mathbb {E}\left[ \int _{0}^{t \wedge \tau _m} \int _{0}^{1} |X_s|^{\frac{2}{\beta }+\beta -2} z^2 \nu (dz)ds \right] +\beta \mathbb {E}\left[ \int _{0}^{t \wedge \tau _m} \int _{1}^{\infty } |X_s| z^{\beta } \nu (dz) ds \right] . \end{aligned}$$

Since $\beta \in [1,\alpha )$, then $0< \frac{2}{\beta } + \beta -2 \leqslant \beta$ and there exists a constant $C>0$ such that $|x|^{\beta -1}\vee |x| \vee |x|^{\frac{2}{\beta }+\beta -2} \leqslant C(1+|x|^{\beta })$. Then by taking expectation on (5.2), and using the fact that $\nu (dz) \,\,\propto \,\, z^{-(1+\alpha )} dz$, we see that there exists $C_0>0$ such that

$$\begin{aligned} \mathbb {E}\big [|X_{t \wedge \tau _m}|^{\beta }\big ] \leqslant C_0(\alpha -\beta )^{-1} +C_0 (\alpha -\beta )^{-1}\int _{0}^{t} \mathbb {E}\big [|X_{s \wedge \tau _m}|^{\beta }\big ] ds. \end{aligned}$$

By Gronwall’s inequality, we obtain

$$\begin{aligned} \mathbb {E}\big [|X_{t \wedge \tau _m}|^{\beta }\big ] \leqslant C_0(\alpha -\beta )^{-1} e^{C_0(\alpha -\beta )^{-1}T}. \end{aligned}$$

Finally, we conclude by using Fatou’s lemma. $\square$

1.2 Yamada–Watanabe approximation technique

We introduce below the Yamada and Watanabe approximation technique, see also [27]. For each $\delta \in (1,\infty )$ and $\varepsilon \in (0,1)$, we select a continuous function $\psi _{\delta , \varepsilon }: \mathbb {R}\rightarrow \mathbb {R}^+$ with support of $\psi _{\delta , \varepsilon }$ belongs to $[\varepsilon /\delta , \varepsilon ]$ and is such that

$$\begin{aligned} \int _{\varepsilon /\delta }^{\varepsilon } \psi _{\delta , \varepsilon }(z) dz = 1 \quad \text { and } \quad 0 \leqslant \psi _{\delta , \varepsilon }(z) \leqslant \frac{2}{z \log \delta }, \,\,\,z > 0. \end{aligned}$$

We define a function $\phi _{\delta , \varepsilon } \in C^2(\mathbb {R};\mathbb {R})$ by setting

$$\begin{aligned} \phi _{\delta , \varepsilon }(x)&:=\int _0^{|x|}\int _0^y \psi _{\delta , \varepsilon }(z)dzdy. \end{aligned}$$

(5.3)

It is straight forward to verify that $\phi _{\delta , \varepsilon }$ has the following useful properties:

$$\begin{aligned}&|x| \leqslant \varepsilon + \phi _{\delta , \varepsilon }(x), \hbox { for any}\ x \in \mathbb {R}, \\&0 \leqslant |\phi '_{\delta , \varepsilon }(x)| \leqslant 1, \hbox { for any}\ x \in \mathbb {R}, \\&\phi '_{\delta , \varepsilon }(x) \geqslant 0, \text { for } x\geqslant 0 \text { and } \phi '_{\delta , \varepsilon }(x)< 0, \text { for } x< 0, \\&\phi ''_{\delta , \varepsilon }(\pm |x|)=\psi _{\delta , \varepsilon }(|x|) \leqslant \frac{2}{|x|\log \delta }\mathbf{1}_{[\varepsilon /\delta , \varepsilon ]}(|x|) \leqslant \frac{2\delta }{\varepsilon \log \delta }, \text { for any }x \in \mathbb {R}\setminus \{0\}. \end{aligned}$$

Lemma 5.2

(Lemma 1.3 in [22]) Suppose that the Lévy measure $\nu$ satisfies $\int _0^{\infty } \{z \wedge z^2\} \nu (dz) < \infty$. Let $\varepsilon \in (0,1)$ and $\delta \in (1,\infty )$. Then for any $x \in \mathbb {R}$, $y \in \mathbb {R}\setminus \{0\}$ with $xy \geqslant 0$ and $u>0$, it holds that

$$\begin{aligned}&\int _{0}^{\infty } \{\phi _{\delta ,\varepsilon }(y+xz)-\phi _{\delta ,\varepsilon }(y)-xz\phi _{\delta ,\varepsilon }'(y)\} \nu (dz)\nonumber \\&\quad \leqslant 2 \cdot \mathbf{1}_{(0,\varepsilon ]}(|y|) \left\{ \frac{|x|^2}{\log \delta } \left( \frac{1}{|y|} \wedge \frac{\delta }{\varepsilon }\right) \int _{0}^{u} z^2 \nu (dz) + |x| \int _{u}^{\infty } z \nu (dz) \right\} . \end{aligned}$$

Lemma 5.3

(Lemma 1.4 in [22]) Suppose that the Lévy measure $\nu$ satisfies $\int _0^{\infty } \{z \wedge z^2\} \nu (dz) < \infty$. Let $\varepsilon \in (0,1)$ and $\delta \in (1,\infty )$. Then for any $x,x' \in \mathbb {R}$, $y \in \mathbb {R}$ and $u \in (0,\infty ]$, it holds that

$$\begin{aligned}&\int _{0}^{\infty }\left| \phi _{\delta ,\varepsilon }(y+xz)-\phi _{\delta ,\varepsilon }(y+x'z)-(x-x')z\phi _{\delta ,\varepsilon }'(y) \right| \nu (dz) \nonumber \\&\quad \leqslant 2 \left\{ \frac{\delta ( |x-x'|^2+|x'||x-x'|)}{\varepsilon \log \delta }\int _{0}^{u} z^2 \nu (dz) + |x-x'| \int _{u}^{\infty } z \nu (dz) \right\} . \end{aligned}$$

(5.4)

In particular, if $x'=0$, then

$$\begin{aligned}&\int _{0}^{\infty } \{\phi _{\delta ,\varepsilon }(y+xz)-\phi _{\delta ,\varepsilon }(y)-xz\phi _{\delta ,\varepsilon }'(y)\} \nu (dz)\nonumber \\&\quad \leqslant 2 \left\{ \frac{\delta |x|^2}{\varepsilon \log \delta }\int _{0}^{u} z^2 \nu (dz) + |x| \int _{u}^{\infty } z \nu (dz) \right\} . \end{aligned}$$

(5.5)

1.3 Estimates of the probability that D is negative

In the case where Z is an $\alpha$-stable compensated Lévy process with $\alpha \in (1,2)$, where for $q\geqslant 0$, the moment generating function is given by

$$\begin{aligned} \mathbb {E}\left[ e^{-qZ_t}\right] = \exp \left( \frac{tq^{\alpha }}{\sin (\pi (\alpha -1)/2)}\right) , \end{aligned}$$

see Jiao et al. [17]. The support of $Z_t$ is not bounded below and it is not possible to find conditions on the parameters which guarantee that the process D is non-negative.

Proof of Lemma 2.1

One can estimate the conditional probability using the conditional Laplace transform, and for any $m>0$,

$$\begin{aligned}&\mathbb {E}\big [e^{-mD_{t_{i+1}}}\big |\, \mathscr {F}_{t_i}\big ] \\&\quad = \exp \left( -m\left( a - \sigma _1^2/2\right) \varDelta t_i\right) \mathbb {E}\big [\exp \big (-m\big (x + \sigma _2 (|x|^\frac{1}{\alpha }\wedge H) \varDelta Z_{t_i} \big ) \big ) \big ] \Big |_{x= X^{H,n}_{t_i}}\\&\quad = \exp \left( -m\left( a - \sigma _1^2/2\right) \varDelta t_i\right) \exp \left( -m X^{H,n}_{t_i} \right) \exp \left( \frac{m^\alpha \varDelta t_i \sigma _2^\alpha (|X^{H,n}_{t_i}|\wedge H^\alpha )}{\sin (\pi (\alpha -1)/2)}\right) \\&\quad \leqslant \exp \left( -m\left( a - \sigma _1^2/2\right) \varDelta t_i\right) \exp \left( \left( \frac{m^{\alpha -1} \varDelta t_i \sigma _2^\alpha }{\sin (\pi (\alpha -1)/2)} - 1 \right) m X^{H,n}_{t_i} \right) . \end{aligned}$$

By setting m as the solution to

$$\begin{aligned} \frac{m^{\alpha -1} \varDelta t_i \sigma _2^\alpha }{\sin (\pi (\alpha -1)/2)} - 1 = 0 \end{aligned}$$

we eliminate $X^{H,n}_{t_i}$ from the above expression, giving the upper bound

$$\begin{aligned} \exp \left( -\left( a-\sigma _1^2/2\right) \left( \sin \left( \pi (\alpha -1)/2\right) \right) ^{\frac{1}{\alpha -1}}\sigma _2^{-\frac{\alpha }{\alpha -1}}\left( 1/\varDelta t_i\right) ^{\frac{1}{\alpha -1}-1} \right) . \end{aligned}$$

$\square$