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Theorem 12.16 (Berry-Esseen) Let \( \\left( {X}_{n}\\right) \) be a sequence of i.i.d. random variables such that \( E\\left\\lbrack {X}_{1}\\right\\rbrack = 0 \) and \( {\\sigma }^{2} = \\operatorname{var}\\left( {X}_{1}\\right) ,\\varrho = E\\left\\lbrack {\\left| {X}_{1}\\right| }^{3}\\right\\rbrack \) are finite. I...
For the proof we refer to, for example, Durrett [109].
No
Lemma 13.1 For any \( n \in \mathbb{N} \), the random variable \( {T}_{n} \) has probability density\n\n\[ \n{f}_{{T}_{n}}\left( t\right) = \lambda {e}^{-{\lambda t}}\frac{{\left( \lambda t\right) }^{n - 1}}{\left( {n - 1}\right) !}{\mathbb{1}}_{{\mathbb{R}}_{ \geq 0}}\left( t\right) ,\;t \in \mathbb{R}. \n\]
Proof. We prove the thesis by induction. For \( n = 1 \) it is obvious. Next we assume that \( {T}_{n} \) has probability density given by (13.2): by the independence of the variables \( \left\{ {\tau }_{n}\right\} \) and Corollary A.54, we have\n\n\[ \n{f}_{{T}_{n + 1}}\left( t\right) = {f}_{{T}_{n} + {\tau }_{n + 1}}...
Yes
Proposition 13.3 Let \( {\left( {N}_{t}\right) }_{t \geq 0} \) be a Poisson process. Then:\ni) the trajectories \( t \mapsto {N}_{t}\left( \omega \right) \) are right continuous with finite left limits, that is \( N \) is a càdlà \( {g}^{1} \) process;\n\nii) for any positive \( t \), almost all trajectories are contin...
Proof. Property \( i \) ) follows from the definition of Poisson process. Secondly, the discontinuities of \( N \) are at the jump times \( {T}_{n}, n \in \mathbb{N} \) : however, by Lemma 13.1 for any \( t > 0 \) we have\n\n\[ \nP\left( {{T}_{n} = t}\right) = 0\n\]\n\nand therefore, with probability one, \( t \) is no...
Yes
Proposition 13.4 Let \( {\left( {N}_{t}\right) }_{t \geq 0} \) be a Poisson process with intensity \( \lambda \) . Then: i) for any \( t \geq 0,{N}_{t} \) has the distribution\n\n\[ P\left( {{N}_{t} = n}\right) = {e}^{-{\lambda t}}\frac{{\left( \lambda t\right) }^{n}}{n!},\;n \in \mathbb{N}, \]
Proof. We only prove \( i \) ): for the other properties, that are consequence of the absence of memory of the exponential distribution (cf. Example A.29), we refer for instante to [76]. We first observe that by (13.2) we have\n\n\[ P\left( {t \geq {T}_{n + 1}}\right) = {\int }_{0}^{t}\lambda {e}^{-{\lambda s}}\frac{{\...
No
Lemma 13.12 Let \( f \) be a càdlàg function defined on a compact interval \( \left\lbrack {0, T}\right\rbrack \) . Then, for any \( n \in \mathbb{N} \), the number of jumps of \( f \) of size greater than \( \frac{1}{n} \) is finite:\n\n\[ \left. {\left. {\# \{ t \in \rbrack 0, T}\right\rbrack \left| \right| {\Delta f...
Proof. By contradiction, assume that for some \( n \in \mathbb{N} \) the number of jumps of size greater than \( \frac{1}{n} \) is infinite: then, since the domain is compact, there exists a sequence \( \left( {t}_{k}\right) \) in \( \left\lbrack {0, T}\right\rbrack \), strictly increasing or decreasing, which converge...
Yes
Lemma 13.17 If \( X \) is Lévy process, then \( {X}_{t} \) is infinitely divisible for each \( t \geq 0 \) and we have\n\n\[{\varphi }_{{X}_{t}}\left( \xi \right) = {\left( {\varphi }_{{X}_{\frac{t}{n}}}\left( \xi \right) \right) }^{n},\;t \geq 0, n \in \mathbb{N}.\]
Proof. The thesis follows by the properties \( L - i \) ) and \( L - {ii} \) ) of Definition 13.10: indeed, for any \( n \geq 2 \) we set\n\n\[{Y}_{i}^{\left( n\right) } \mathrel{\text{:=}} {X}_{\frac{it}{n}} - {X}_{\frac{\left( {i - 1}\right) t}{n}}\overset{d}{ = }{X}_{\frac{t}{n}},\;i = 1,\ldots, n,\]\n\nand we remar...
Yes
Lemma 13.18 If \( {\left( {X}_{t}\right) }_{t \geq 0} \) is stochastically continuous, then the map \( t \mapsto \) \( {\varphi }_{{X}_{t}}\left( \xi \right) \) is continuous for each \( \xi \in {\mathbb{R}}^{d} \) .
Proof. Let \( \xi \in {\mathbb{R}}^{d} \) be fixed: for any \( \varepsilon > 0 \) we consider \( {\delta }_{\varepsilon } > 0 \) such that\n\n\[ \mathop{\sup }\limits_{{\left| y\right| \leq {\delta }_{\varepsilon }}}\left| {{e}^{{i\xi } \cdot y} - 1}\right| < \frac{\varepsilon }{2} \]\n\nIf \( X \) is stochastically co...
Yes
Lemma 13.19 If \( {\left( {X}_{t}\right) }_{t \geq 0} \) is a Lévy process, then \( {\varphi }_{{X}_{t}}\left( \xi \right) \neq 0 \) for any \( \xi \in {\mathbb{R}}^{d} \) and \( t \geq 0 \) .
Proof. See Sato [297], Lemma 7.5.
No
Lemma 13.20 Let \( \varphi \in C\left( {{\mathbb{R}}^{d},\mathbb{C}}\right) \) such that \( \varphi \left( 0\right) = 1 \) and \( \varphi \left( \xi \right) \neq 0 \) for any \( \xi \in {\mathbb{R}}^{d} \) . Then there exists a unique continuous function \( g \in C\left( {{\mathbb{R}}^{d},\mathbb{C}}\right) \) such tha...
Proof. See Sato [297], Lemma 7.6.
No
Example 13.21 (Brownian motion with drift) Let \( {X}_{t} = {\mu t} + \sigma {W}_{t} \) where \( W \) is a standard real Brownian motion: as a consequence of Example A.60 and Lemma A.70, we have
\[ E\left\lbrack {e}^{{i\xi }{X}_{t}}\right\rbrack = {e}^{i\mu t\xi }E\left\lbrack {e}^{{i\xi \sigma }{W}_{t}}\right\rbrack = {e}^{{i\mu t\xi } + \frac{1}{2}{\left( i\xi \sigma \right) }^{2}t}, \] that is \[ \psi \left( \xi \right) = {i\mu \xi } - \frac{{\sigma }^{2}{\xi }^{2}}{2} \] In the \( d \) -dimensional case wh...
Yes
Example 13.22 (Poisson process) Let \( {N}_{t} \) denote a Poisson process with intensity \( \lambda \) (cf. Definition 13.2). We have\n\n\[ \n{\varphi }_{{N}_{t}}\left( \xi \right) = E\left\lbrack {e}^{{i\xi }{N}_{t}}\right\rbrack = \mathop{\sum }\limits_{{n \geq 0}}E\left\lbrack {{e}^{i\xi n}{\mathbb{1}}_{\left\{ {N}...
\[ \n{\varphi }_{{N}_{t}}\left( \xi \right) = E\left\lbrack {e}^{{i\xi }{N}_{t}}\right\rbrack = \mathop{\sum }\limits_{{n \geq 0}}E\left\lbrack {{e}^{i\xi n}{\mathbb{1}}_{\left\{ {N}_{t} = n\right\} }}\right\rbrack = \n\]\n\n(by (13.3))\n\n\[ \n= {e}^{-{\lambda t}}\mathop{\sum }\limits_{{n \geq 0}}\frac{{\left( {e}^{i\...
Yes
Example 13.23 (Compound Poisson process) Let\n\n\[ \n{X}_{t} = \mathop{\sum }\limits_{{n = 1}}^{{N}_{t}}{Z}_{n},\;t \geq 0 \n\] \n\nbe a \( d \) -dimensional compound Poisson process (cf. Definition 13.8) with intensity \( \lambda \) and distribution of jumps \( \eta \) . We denote by \n\n\[ \n\widehat{\eta }\left( \xi...
\[ \n= \mathop{\sum }\limits_{{n \geq 0}}{\left( E\left\lbrack {e}^{{i\xi } \cdot {Z}_{1}}\right\rbrack \right) }^{n}P\left( {{N}_{t} = n}\right) = \n\] \n\n(by (13.3)) \n\n\[ \n= {e}^{-{\lambda t}}\mathop{\sum }\limits_{{n \geq 0}}\frac{{\left( \lambda t\widehat{\eta }\left( \xi \right) \right) }^{n}}{n!} = {e}^{-{\la...
Yes
Example 13.24 (Compensated compound Poisson process) Let\n\n\[ \n{\\widetilde{X}}_{t} = {X}_{t} - {m\\lambda t},\;m = {\\int }_{{\\mathbb{R}}^{d}}{x\\eta }\\left( {dx}\\right) ,\n\]\n\nbe a compensated compound Poisson process with intensity \( \\lambda \) and distribution of jumps \( \\eta \) (cf. Definition 13.9). Th...
\[ \n\\psi \\left( \\xi \\right) = \\lambda \\left( {\\widehat{\\eta }\\left( \\xi \\right) - 1 - {i\\xi } \\cdot m}\\right) = {\\int }_{{\\mathbb{R}}^{d}}\\left( {{e}^{{i\\xi } \\cdot x} - 1 - {i\\xi } \\cdot x}\\right) {\\lambda \\eta }\\left( {dx}\\right) .\n\]
Yes
Theorem 13.30 Let \( X \) be the jump-diffusion process in (13.17) with jump measure \( J \) and Lévy measure \( \nu \) . For any function \( f = f\left( {t, x}\right) \) we have\n\n\[ \mathop{\sum }\limits_{\substack{{0 < s \leq t} \\ {\Delta {X}_{s} \neq 0} }}f\left( {s,\Delta {X}_{s}}\right) = {\int }_{0}^{t}{\int }...
Proof. For simplicity, we only consider the case \( f = f\left( x\right) \) . We first remark that, for a jump-diffusion process \( X \), we have\n\n\[ \mathop{\sum }\limits_{\substack{{0 < s \leq t} \\ {\Delta {X}_{s} \neq 0} }}f\left( {\Delta {X}_{s}}\right) = \mathop{\sum }\limits_{{n = 1}}^{{N}_{t}}f\left( {Z}_{n}\...
Yes
Lemma 13.33 Let \( X \) be a Lévy process with jump measure \( J \) and Lévy measure \( \nu \) . Then\ni) if \( H \in \mathcal{B}\left( {\mathbb{R}}^{d}\right) \) is such that \( 0 \notin \bar{H} \) then the process\n\n\[t \mapsto {J}_{t}\left( H\right) \mathrel{\text{:=}} J\left( {\left\lbrack {0, t}\right\rbrack \tim...
Proof. Part i) can be verified directly using the definition of Poisson process: see, for instance, Protter [287], p. 26.
No
Theorem 13.34 Let \( {\left( {X}_{t}\right) }_{t > 0} \) be a d-dimensional Levy process with Levy measure \( \nu \) and jump measure \( J \) . For any measurable function \( f \) such that\n\n\[ \n{\int }_{0}^{t}{\int }_{\left| x\right| \leq \varepsilon }\left| {f\left( {s, x}\right) }\right| \nu \left( {dx}\right) {d...
Proof. Formulas (13.43) and (13.44) follows from (13.23)-(13.24) by limit arguments (for further details, see Section 2.4 in Applebaum [11] and Section I-4 in Protter [287]).\n\nNow assume that \( f \) satisfies condition (13.40). The idea is that \
No
Example 13.37 Let\n\n\[ \n{X}_{t} = {\mu t} + {B}_{t} + \mathop{\sum }\limits_{{n = 1}}^{{N}_{t}}{Z}_{n} \]\n\nbe a jump-diffusion process: \( \mu \in {\mathbb{R}}^{d}, B \) is \( d \) -dimensional correlated Brownian motion with correlation matrix \( \mathcal{C}, N \) is a Poisson process with intensity \( \lambda \) ...
We remark that the second and fourth terms in (13.47) (i.e. Brownian motion and compensated small jumps) form the martingale part of \( X \), while the first and third terms (i.e. drift term and large jumps) govern the drift of the process. More precisely, it is always possible to split a Lévy process into the sum of a...
No
Corollary 13.38 Let \( X \) be a Lévy process. Then \( X = M + Z \) where \( M \) and \( Z \) are Lévy processes, \( M \) is a martingale such that \( {M}_{t} \in {L}^{p}\left( \Omega \right) \) for any \( p \geq 1 \) and \( Z \) has (locally in time) bounded variation.
Proof. By the Lévy-Ito decomposition (13.47), it is suffices to set\n\n\[ \n{Z}_{t} = {\mu }_{R}t + {X}_{t}^{R} = {\mu }_{R}t + {\int }_{0}^{t}{\int }_{\left| x\right| \geq R}{xJ}\left( {{ds},{dx}}\right) , \n\] \n\n\[ \n{M}_{t} = {B}_{t} + {M}_{t}^{R} = {B}_{t} + {\int }_{0}^{t}{\int }_{\left| x\right| < R}x\widetilde...
Yes
Theorem 13.40 (Lévy-Khintchine representation) Let \( X \) be a Lévy process in \( {\mathbb{R}}^{d} \) with characteristic triplet \( \left( {{\mu }_{1},\mathcal{C},\nu }\right) \). Then we have\n\n\[ \n{\varphi }_{{X}_{t}}\left( \xi \right) = E\left\lbrack {e}^{{i\xi } \cdot {X}_{t}}\right\rbrack = {e}^{t{\psi }_{X}\l...
Proof. By the Lévy-Itô decomposition \( {X}_{t} \) can be represented as the independent sum of is the a.s. limit of the sum of \( {\mu }_{1}t + {B}_{t},{X}_{t}^{1} \) and \( {\widetilde{X}}_{t}^{\varepsilon ,1} \) in (13.51), as \( \varepsilon \rightarrow {0}^{ + } \). Since these terms are independent, by Remark 13.2...
Yes
Corollary 13.42 Let \( X \) be a Lévy process with characteristic triplet \( \left( {{\mu }_{1},\mathcal{C},\nu }\right) \) and Lévy measure \( \nu \) such that\n\n\[ \nu \left( {\mathbb{R}}^{d}\right) < \infty \]\n\nThen \( X \) is a jump-diffusion process with intensity \( \lambda = \nu \left( {\mathbb{R}}^{d}\right)...
Proof. Under condition (13.66), we can let \( R \) go to zero in the Lévy-Khintchine representation (13.62): we get\n\n\[ {\psi }_{X}\left( \xi \right) = i{\mu }_{0} \cdot \xi - \frac{1}{2}\langle \mathcal{C}\xi ,\xi \rangle + {\int }_{{\mathbb{R}}^{d}}\left( {{e}^{{i\xi } \cdot x} - 1}\right) \nu \left( {dx}\right) \]...
Yes
Proposition 13.43 Let \( X \) be a Lévy process with triplet \( \left( {{\mu }_{1},\mathcal{C},\nu }\right) \) . Then \( X \) has (locally in time) bounded variation if and only if
Proof. We only prove the \
No
Corollary 13.44 Let \( X \) be a Lévy process with (locally in time) bounded variation and characteristic triplet \( \left( {{\mu }_{1},0,\nu }\right) \). Then we have the Lévy-Itô decomposition
\[ {X}_{t} = {\mu }_{0}t + {\int }_{{\mathbb{R}}^{d}}{xJ}\left( {{ds},{dx}}\right) \] where \[ {\mu }_{0} = {\mu }_{1} - {\int }_{\left| x\right| \leq 1}{x\nu }\left( {dx}\right) \] Moreover the characteristic exponent takes the form \[ {\psi }_{X}\left( \xi \right) = i{\mu }_{0} \cdot \xi + {\int }_{{\mathbb{R}}^{d}}\...
Yes
The Lévy measure of a stable distribution is of the form:
\[ \nu \left( {dx}\right) = \left( {\frac{{C}_{1}}{{x}^{1 + \alpha }}{\mathbb{1}}_{\{ x > 0\} } + \frac{{C}_{2}}{{\left( -x\right) }^{1 + \alpha }}{\mathbb{1}}_{\{ x < 0\} }}\right) {dx} \] where \( {C}_{1},{C}_{2} > 0 \) . By conditions (13.45) and (13.46), we necessarily have \( \alpha \in \rbrack 0,2\lbrack \) . Mor...
Yes
Proposition 13.49 Let \( X \) be a Lévy process on \( \mathbb{R} \) with characteristic triplet \( \left( {{\mu }_{1},{\sigma }^{2},\nu }\right) \). The exponential moment \( E\left\lbrack {e}^{\xi {X}_{t}}\right\rbrack ,\xi \in \mathbb{R} \), is finite if and only if \[ {\int }_{\left| x\right| \geq 1}{e}^{\xi x}\nu \...
In this case \[ E\left\lbrack {e}^{\xi {X}_{t}}\right\rbrack = {e}^{{t\psi }\left( {-{i\xi }}\right) } \] where \( \psi \) is the characteristic exponent of \( X \) .
No
Theorem 13.50 Let \( X \) be a real valued Lévy process with characteristic triplet \( \left( {{\mu }_{1},{\sigma }^{2},\nu }\right) \) . We have:\n\ni) if \( E\left\lbrack \left| {X}_{1}\right| \right\rbrack < \infty \) then \( {\left( {X}_{t} - E\left\lbrack {X}_{t}\right\rbrack \right) }_{t > 0} \) is a martingale;
Proof. i) Since \( E\left\lbrack {X}_{t}\right\rbrack = {tE}\left\lbrack {X}_{1}\right\rbrack \) (cf. (13.72)), if \( E\left\lbrack \left| {X}_{1}\right| \right\rbrack < \infty \) then \( {X}_{t} - E\left\lbrack {X}_{t}\right\rbrack \) is integrable. Moreover, by the independence of increments we have\n\n\[ E\left\lbra...
Yes
Example 13.51 (Merton model) In the Merton model [251], the log-price is modeled by a process of the form (13.76) where \( \eta = {\mathcal{N}}_{m,{\delta }^{2}} \) . Thus the 0 -triplet is \( \left( {\mu ,{\sigma }^{2},\nu }\right) \) with Lévy measure\n\n\[ \nu \left( {dx}\right) = \frac{\lambda }{\sqrt{{2\pi }{\delt...
As already proved in Example 13.26, the characteristic exponent is\n\n\[ \psi \left( \xi \right) = {i\mu \xi } - \frac{1}{2}{\sigma }^{2}{\xi }^{2} + \lambda \left( {{e}^{{im\xi } - \frac{1}{2}{\delta }^{2}{\xi }^{2}} - 1}\right) . \]
Yes
In the Kou model [217], the distribution of jumps is defined in terms of an asymmetric double exponential density: more precisely, we have\n\n\[ \eta \left( {dx}\right) = \left( {p{\lambda }_{1}{e}^{-{\lambda }_{1}x}{\mathbb{1}}_{\{ x > 0\} } + \left( {1 - p}\right) {\lambda }_{2}{e}^{{\lambda }_{2}x}{\mathbb{1}}_{\{ x...
\[ {\psi }_{X}\left( \xi \right) = {i\mu \xi } - \frac{{\sigma }^{2}{\xi }^{2}}{2} + {\int }_{\mathbb{R}}\left( {{e}^{i\xi x} - 1}\right) \nu \left( {dx}\right) \] \n\n\[ = {i\mu \xi } - \frac{{\sigma }^{2}{\xi }^{2}}{2} + {i\lambda \xi }\left( {\frac{p}{{\lambda }_{1} - {i\xi }} - \frac{1 - p}{{\lambda }_{2} + {i\xi }...
Yes
A Poisson processes is a subordinator. A compound Poisson process is a subordinator if and only if all the \( {Z}_{n} \) (cf. Definition 13.8) take only non-negative values.
Since a subordinator \( S \) takes only non-negative values, it is convenient to characterize it by the Laplace transform instead of the Fourier transform: if \( {\psi }_{S} \) denotes as usual the characteristic exponent of \( S \), by Proposition 13.49 the Laplace exponent\n\n\[ \n{\ell }_{S}\left( \xi \right) \mathr...
No
Example 13.60 (Gamma subordinator) We use slightly different notations and consider a Lévy process \( S \) with 0 -triplet \( \left( {0,0,\varrho }\right) \) and Lévy measure\n\n\[ \varrho \left( {dx}\right) = \frac{a{e}^{-{bx}}}{x}{\mathbb{1}}_{\{ x > 0\} }{dx} \]\n\nwhere \( a, b \) are positive parameters: we call \...
The density of \( {S}_{t} \) can be recovered from the characteristic function by Fourier inversion:\n\n\[ {f}_{{S}_{t}}\left( x\right) = \frac{1}{2\pi }{\int }_{\mathbb{R}}{e}^{-{ix\xi }}{\varphi }_{{S}_{t}}\left( \xi \right) {d\xi } = \frac{{e}^{-{bx}}{\left( bx\right) }^{at}}{{x\Gamma }\left( {at}\right) },\;x > 0, ...
Yes
Example 13.61 (Variance-Gamma process) By subordinating a Brownian motion with drift \( \mu \) and volatility \( \sigma \) by a Gamma process \( S \) with variance \( v \) (and unitary mean), we obtain the so-called Variance-Gamma (VG) process
\[ {X}_{t} = \mu {S}_{t} + \sigma {W}_{{S}_{t}} \] This is a three-parameter process: the variance \( v \) of the subordinator, the drift \( \mu \) and the volatility \( \sigma \) of the Brownian motion. By Theorem 13.58, the characteristic exponent of \( X \) is \[ {\psi }_{X}\left( \xi \right) = - \frac{1}{v}\log \le...
Yes
Example 13.62 (Inverse Gaussian subordinator) The Inverse Gaussian (IG) subordinator \( S \) is a tempered stable subordinator with \( \alpha = \frac{1}{2} \) : thus \( S \) has 0-triplet \( \left( {0,0,\varrho }\right) \) and Lévy measure\n\n\[ \varrho \left( {dx}\right) = \frac{a{e}^{-{bx}}}{{x}^{\frac{3}{2}}}{\mathb...
The density of \( {S}_{t} \) can be recovered from the characteristic function by Fourier inversion:\n\n\[ {f}_{{S}_{t}}\left( x\right) = \frac{1}{2\pi }{\int }_{\mathbb{R}}{e}^{-{ix\xi }}{\varphi }_{{S}_{t}}\left( \xi \right) {d\xi } = \frac{at}{{x}^{3/2}}\exp \left( {-\frac{{\left( at\sqrt{\pi } - x\sqrt{b}\right) }^...
Yes
Proposition 13.64 Assume that \( X \) is a Lévy process with characteristic exponent \( {\psi }_{Q} \) under \( Q \) . The discounted price process \( {\widetilde{S}}_{t} = {S}_{0}{e}^{{X}_{t} - {rt}} \) is a \( Q \) - martingale if and only if\n\n\[ \n{E}^{Q}\left\lbrack {S}_{t}\right\rbrack = {E}^{Q}\left\lbrack {{S}...
Proof. It suffices to recall that, by Theorem 13.50-iii), \( {\widetilde{S}}_{t} = {S}_{0}{e}^{{X}_{t} - {rt}} \) is a martingale if and only if \( 1 = {e}^{-{rt}}{E}^{Q}\left\lbrack {e}^{{X}_{t}}\right\rbrack = {e}^{t\left( {{\psi }_{Q}\left( {-i}\right) - r}\right) } \) .
Yes
Theorem 13.66 Let \( X = {\left( {X}_{t}\right) }_{t \in \left\lbrack {0, T}\right\rbrack } \) be a Lévy process. Then \( X \) is a Lévy process also with respect to \( {P}_{T}^{\theta } \) in (13.100) and its Laplace exponent \( {\ell }_{\theta } \) under \( {P}_{T}^{\theta } \) is given by \[ {\ell }_{\theta }\left( ...
Proof. In view of Bayes’s formula (Theorem A.113), for all \( 0 \leq s \leq t \leq T \) we have \[ {E}^{{P}_{T}^{\theta }}\left\lbrack {{e}^{z\left( {{X}_{t} - {X}_{s}}\right) } \mid {\mathcal{F}}_{s}}\right\rbrack = \frac{{E}^{P}\left\lbrack {{e}^{z\left( {{X}_{t} - {X}_{s}}\right) }{Z}_{T}^{\theta } \mid {\mathcal{F}...
Yes
Theorem 13.67 Let \( \left( {\mu ,{\sigma }^{2},\nu }\right) \) be the triplet of a Lévy process \( X \) with respect to the measure \( P \) . Then the triplet \( \left( {{\mu }_{\theta },{\sigma }_{\theta }^{2},{\nu }_{\theta }}\right) \) of \( X \) with respect to the measure \( {P}_{T}^{\theta } \) in (13.100) is de...
Proof. In view of (13.101) we have:\n\n\[ \n{\ell }_{\theta }\left( z\right) = {\mu z} + \frac{{\sigma }^{2}}{2}\left( {{\left( z + \theta \right) }^{2} - {\theta }^{2}}\right) + {\int }_{\mathbb{R}}\left( {\left( {{e}^{zx} - 1}\right) {e}^{\theta x} - {zx}{\mathbb{1}}_{\{ \left| x\right| < 1\} }}\right) \nu \left( {dx...
Yes
Theorem 13.68 Suppose that \( {\theta }^{ * } \) is a solution to\n\n\[ \ell \left( {1 + {\theta }^{ * }}\right) - \ell \left( {\theta }^{ * }\right) = r \]\n\n(13.103)\n\nIf \( {E}^{P}\left\lbrack {e}^{{\theta }^{ * }{X}_{T}}\right\rbrack < \infty \) (to assure that \( {P}_{T}^{{\theta }^{ * }} \) exists) and \( {E}^{...
Under the assumptions of Theorem 13.68, \( {P}_{T}^{{\theta }^{ * }} \) is called the Esscher martingale transform of the objective measure \( P \) . Let us denote \( {P}_{T}^{{\theta }^{ * }} \) by \( Q \) : in view of (13.101), with obvious notation we have\n\n\[ {\ell }_{Q}\left( z\right) = {\ell }_{P}\left( {z + {\...
Yes
Example 13.69 (Brownian motion with drift) Let \( {\ell }_{P}\left( z\right) = {\mu z} + \frac{{\sigma }^{2}{z}^{2}}{2} \) be the Laplace exponent under the historic measure \( P \) . The solution to (13.103)
is\n\[ \n{\theta }^{ * } = \frac{1}{{\sigma }^{2}}\left( {r - \mu - \frac{{\sigma }^{2}}{2}}\right) \]\n\nand by (13.102) the characteristic exponent is\n\n\[ \n{\psi }_{Q}\left( \xi \right) = i\left( {r - \frac{{\sigma }^{2}}{2}}\right) \xi - \frac{{\sigma }^{2}{\xi }^{2}}{2}. \]\n\nIt is clear that in this case the E...
Yes
Let \( {N}_{t} \) be a Poisson process with intensity parameter \( \lambda \) and let \( {X}_{t} = \alpha {N}_{t} - {\beta t} \) with \( \alpha ,\beta > 0 \) . Since\n\n\[ \n{\ell }_{P}\left( z\right) = - {\beta z} + \lambda \left( {{e}^{\alpha z} - 1}\right) \n\]
the solution to (13.103) is \( {\theta }^{ * } = \frac{1}{\alpha }\log \frac{r + \beta }{\lambda \left( {{e}^{\alpha } - 1}\right) } \) . Thus\n\n\[ \n{\ell }_{Q}\left( z\right) = - {\beta z} + {\lambda }^{ * }\left( {{e}^{\alpha z} - 1}\right) \n\]\n\nwith \( {\lambda }^{ * } = \lambda {e}^{{\theta }^{ * }y} = \frac{r...
Yes
Example 13.71 (NIG process) Let\n\n\\[ \n{\\ell }_{P}\\left( z\\right) = {\\mu z} + \\delta \\left( {{\\left( {\\alpha }^{2} - {\\beta }^{2}\\right) }^{\\frac{1}{2}} - {\\left( {\\alpha }^{2} - {\\left( \\beta + z\\right) }^{2}\\right) }^{\\frac{1}{2}}}\\right) \n\\]\n\nwith \\( - \\alpha - \\beta \\leq \\operatorname{...
Then (13.104) yields\n\n\\[ \n{\\ell }_{Q}\\left( z\\right) = {\\mu z} + \\delta \\left( {{\\left( {\\alpha }^{2} - {\\beta }^{*2}\\right) }^{1/2} - {\\left( {\\alpha }^{2} - {\\left( {\\beta }^{ * } + z\\right) }^{2}\\right) }^{1/2}}\\right) \n\\]\n\nwith \\( {\\beta }^{ * } = - \\frac{1}{2} - \\frac{\\mu - r}{2\\delt...
Yes
Theorem 13.74 Let \( X \) be a Lévy process with triplet \( \left( {\mu ,{\sigma }^{2},\nu }\right) \) under some probability measure \( P \) . Then the following two conditions are equivalent:\n\ni) there is a probability measure \( Q \), equivalent to \( P \), such that \( X \) is a Lévy process with triplet \( \left...
Proof. See Sato [297], Theorem 33.1.
Yes
Example 13.75 (Brownian motion) In this case \( \nu \equiv 0 \) and \( \sigma > 0 \) . The unique solution is \( \eta = \frac{r - \mu }{\sigma } - \frac{\sigma }{2} \) .
This implies that the new drift and volatility parameters under \( Q \) are: \( \widetilde{\mu } = r - \frac{{\sigma }^{2}}{2} \) and \( \widetilde{\sigma } = \sigma \) .
Yes
Let \( {X}_{t} = \alpha {N}_{t} + {\mu t} \) with \( \mu < r \) and \( \alpha > 0 \) . For simplicity, we take \( \alpha > 1 \) . In this case \( \sigma = 0 \) and \( \nu = \lambda {\delta }_{\alpha } \) where \( \lambda \) is the intensity parameter. Then \( H \) is constant
\[ H = \frac{r - \mu }{\lambda \left( {{e}^{\alpha } - 1}\right) } \] and the new Lévy measure under \( Q \) is \( \widetilde{\nu } = {\lambda }^{ * }{\delta }_{\alpha } \) with \( {\lambda }^{ * } = \frac{r - \mu }{{e}^{\alpha } - 1} \) ; moreover \( \widetilde{\mu } = \mu \) .
Yes
Let \( {T}_{1} > 0 \) and consider the deterministic function\n\n\[ t \mapsto {S}_{t} = {\mathbb{1}}_{\left\lbrack {T}_{1},\infty \lbrack \left( t\right) .\right. } \]\n\nThen, as in Example 3.68, for any continuous function \( u \), the Riemann-Stieltjes integral is well-defined and
\[ {\int }_{0}^{t}{u}_{s}d{S}_{s} = \left\{ \begin{array}{ll} 0 & \text{ if }t < {T}_{1} \\ {u}_{{T}_{1}} & \text{ if }t \geq {T}_{1} \end{array}\right. \]
Yes
We now examine the convergence of Riemann-Stieltjes sums in case \( u \) is discontinuous and \( S \) as in Example 14.1. We consider \( t \geq {T}_{1} \) and a sequence of partitions \( {\varsigma }_{n} = \left( {{t}_{0}^{n},\ldots ,{t}_{{N}_{n}}^{n}}\right) \) of \( \left\lbrack {0, t}\right\rbrack \) such that \( {T...
\[ \mathop{\sum }\limits_{{k = 1}}^{N}{u}_{{\tau }_{k}^{n}}\left( {{S}_{{t}_{k}^{n}} - {S}_{{t}_{k - 1}^{n}}}\right) = {u}_{{\tau }_{{k}_{n}}^{n}} \] be the Riemann-Stieltjes sum where as usual \( {\tau }_{k}^{n} \in \left\lbrack {{t}_{k - 1}^{n},{t}_{k}^{n}}\right\rbrack \) for \( k = 1,\ldots ,{N}_{n} \) and in parti...
Yes
Let \( {S}_{t} = {\lambda t} - {N}_{t} \) where \( N \) is a Poisson process with intensity \( \lambda \) . Then \( S \) is a martingale because \( - S \) is a compensated Poisson process: intuitively, \( S \) is a fair investment giving zero gain in average, because the deterministic increase \( {\lambda t} \) is comp...
We denote by \( {T}_{n} \) the jump times of \( S \) and consider the strategy\n\n\[ \n{u}_{t} = {\mathbb{1}}_{\left\lbrack 0,{T}_{1}\right\rbrack }\left( t\right) \]\n\n(14.2)\n\nwhich consists in buying (at zero price) the asset at \( t = 0 \) and selling it at the time of the first jump. Note that \( u \) is left-co...
Yes
Example 14.8 Let \( N \) be a Poisson process with jump times \( {\left( {T}_{n}\right) }_{n \geq 1} \) (we also set \( {T}_{0} = 0 \) ). Then\n\n\[ \n{N}_{t} = n\;\text{ for }\;t \in \left\lbrack {{T}_{n},{T}_{n + 1}\lbrack, n \geq 0,}\right. \]\n\nand \( N \in \mathbb{D} \) . We already noted that\n\n\[ \n\left. {{N}...
\[ \n{\int }_{0}^{t}{N}_{s - }d{N}_{s} = \frac{{N}_{t}\left( {{N}_{t} - 1}\right) }{2},\;t \geq 0. \]\n\n(14.9)
Yes
If a process \( S \in \mathbb{D} \) has bounded variation a.s. then it is a semimartingale.
Indeed, for any simple predictable process \( u \), we have\n\n\[ \left| {{\int }_{0}^{T}{u}_{t}d{S}_{t}}\right| \leq {V}_{\left\lbrack 0, T\right\rbrack }\left( S\right) \mathop{\sup }\limits_{{\left\lbrack {0, T}\right\rbrack \times \Omega }}\left| u\right| \]\n\nwhere \( {V}_{\left\lbrack 0, T\right\rbrack }\left( S...
Yes
Every square integrable martingale \( S \in \mathbb{D} \) is a semimartingale.
Indeed, for any simple predictable process \( u \), we have\n\n\[ E\left\lbrack {\left( {\int }_{0}^{T}{u}_{t}d{S}_{t}\right) }^{2}\right\rbrack = E\left\lbrack {\left( \mathop{\sum }\limits_{{k = 1}}^{N}{e}_{k}\left( {S}_{{T}_{k}} - {S}_{{T}_{k - 1}}\right) \right) }^{2}\right\rbrack = \]\n\n(by the orthogonality of t...
Yes
A Lévy process is a semimartingale.
Indeed, by the decomposition in Corollary 13.38, every Lévy process is the sum of a càdlàg \( {L}^{2} \) - martingale with a BV process. Since the set of semimartingales forms a vector space, the thesis is a consequence of the two examples above.
Yes
The space of simple predictable processes is dense in \( \mathbb{L} \) under the uniform convergence in probability: for any \( u \in \mathbb{L} \) there exists a sequence \( \left( {u}^{n}\right) \) of simple predictable processes such that (14.10) holds. The stochastic integral with respect to a semimartingale \( S \...
Proof. See Protter [287], Theorems II-10 and II-11.
No
Theorem 14.15 Let \( S \) be a semimartingale and \( u \) be a process in \( \mathbb{D} \) or in \( \mathbb{L} \) . Then the Riemann-Stieltjes sum \( {}^{5} \n\n\[ \n\mathop{\sum }\limits_{{k = 1}}^{{N}_{n}}{u}_{{T}_{k - 1}^{n}}\left( {{S}_{{T}_{k}^{n}}^{t} - {S}_{{T}_{k - 1}^{n}}^{t}}\right) \n\] \n\nconverges uniform...
Proof. See Theorem II-21 in Protter [287].
No
Proposition 14.19 For any \( \varphi \in {\mathbb{L}}_{\nu }^{2} \), the process \( M \) in (14.13) is a square-integrable martingale such that\n\n\[ E\left\lbrack {M}_{t}\right\rbrack = 0,\;\operatorname{var}\left( {M}_{t}\right) = E\left\lbrack {{\int }_{0}^{t}{\int }_{\mathbb{R}}{\varphi }^{2}\left( {t, x}\right) \n...
Proof. The thesis follows from Proposition 14.17 by a limit argument: for full details see, for instance, Theorem 4.2.3 in Applebaum [11].
No
Proposition 14.20 For any \( \varphi \in {\mathbb{L}}_{\nu ,\text{ loc }}^{2} \), the integral process \( M \) in (14.13) is a local martingale.
Proof. See, for instance, Theorem 4.2.12 in Applebaum [11].
No
Proposition 14.24 Let \( S \) be a one dimensional Lévy process with Lévy-Itô decomposition\n\n\[ \n{S}_{t} = {\mu }_{R}t + \sigma {W}_{t} + {S}_{t}^{R} + {M}_{t}^{R} \]\n\n(14.18)\n\nwhere \( W \) is a standard Brownian motion and\n\n\[ \n{S}_{t}^{R} = {\int }_{0}^{t}{\int }_{\left| y\right| \geq R}{yJ}\left( {{ds},{d...
Proof. Any \( u \in \mathbb{L} \) is progressively measurable and such that\n\n\[ \n{\int }_{0}^{T}{u}_{t}^{2}{dt} < \infty \;\text{ a.s. } \]\n\nor in other terms \( u \in {\mathbb{L}}_{\text{loc }}^{2} \) (cf. Definition 4.1) so that the Brownian integral in (14.21) is well-defined and it is a local martingale. Moreo...
Yes
Theorem 14.26 (Deterministic Itô formula) Let \( f \in {C}^{1}\left( \mathbb{R}\right) \) and \( X \) be a càdlàg (deterministic) function with bounded variation. Then we have\n\n\[ f\left( {X}_{t}\right) - f\left( {X}_{0}\right) = {\int }_{0}^{t}{f}^{\prime }\left( {X}_{s - }\right) d{X}_{s} \]\n\n\[ + \mathop{\sum }\...
Proof. We show that the series in (14.27) is convergent: first of all, it has a countable number of terms because \( X \in \mathrm{{BV}} \) . For the same reason, \( \parallel X{\parallel }_{\infty } = \) sup \( \left| {X}_{t}\right| < \infty \) and therefore we have \( t \in \left\lbrack {0, T}\right\rbrack \)\n\n\[ \...
Yes
Theorem 14.28 If \( S \) is a semimartingale, then \( \langle S\rangle \) is a càdlàg, adapted and increasing process such that \( \langle S{\rangle }_{0} = {S}_{0}^{2} \) and \( \Delta \langle S{\rangle }_{t} = {\left( \Delta {S}_{t}\right) }^{2},\;t \in \left\lbrack {0, T}\right\rbrack \)
Proof. We use repeatedly the elementary equality \[ {\left( b - a\right) }^{2} = {b}^{2} - {a}^{2} - {2a}\left( {b - a}\right) ,\;a, b \in \mathbb{R}. \] It is clear that, by definition of stochastic integral, \( \langle S\rangle \in \mathbb{D} \). Moreover, by (14.32) we have \[ {\left( \Delta {S}_{t}\right) }^{2} = {...
Yes
Example 14.30 By (14.31) and Theorem 3.74, for a real Brownian motion \( W \) we have\n\n\[ \langle W{\rangle }_{t} = \langle W{\rangle }_{t}^{c} = t \]
By (14.31) and Proposition 4.24, the quadratic variation of a Brownian integral\n\n\[ {X}_{t} = {\int }_{0}^{t}{u}_{s}d{W}_{s} \]\n\nwith \( u \in {\mathbb{L}}_{\text{loc }}^{2} \), is given by\n\n\[ \langle X{\rangle }_{t} = \langle X{\rangle }_{t}^{c} = {\int }_{0}^{t}{u}_{s}^{2}{ds} \]
No
If \( N \) is a Poisson process, then by definition of quadratic variation and Example 14.8, we have\n\n\[\n\langle N{\rangle }_{t} = {N}_{t}\n\]
Moreover\n\n\[\n\langle N{\rangle }_{t}^{c} = {N}_{t} - \mathop{\sum }\limits_{{0 < s \leq t}}{\left( \Delta {N}_{s}\right) }^{2} = 0.\n\]
Yes
If \( S \) is a continuous semimartingale with bounded variation then \( \langle S{\rangle }_{t} = {S}_{0}^{2}, t \in \left\lbrack {0, T}\right\rbrack \) .
The proof is the same as in the deterministic case (cf. Proposition 3.73) and uses the characterization (14.31) of the quadratic variation process.
No
If \( S \) is a one-dimensional Lévy process with characteristic triplet \( \left( {{\mu }_{1},\sigma ,\nu }\right) \) then
\[ \langle S{\rangle }_{t} = {\sigma }^{2}t + \mathop{\sum }\limits_{{0 < s \leq t}}{\left( \Delta {S}_{s}\right) }^{2} = {\sigma }^{2}t + {\int }_{\mathbb{R}}{x}^{2}J\left( {{ds},{dx}}\right) ,\] where \( {J}_{t} \) denotes the jump measure of \( S \) (cf. (13.59)). Moreover \[ \langle S{\rangle }_{t}^{c} = {\sigma }^...
Yes
Example 14.35 We apply the Itô formula with \( f\left( x\right) = {x}^{2} \) and \( {X}_{t} = {N}_{t} \) , Poisson process. We get
\[ {N}_{t}^{2} = 2{\int }_{0}^{t}{N}_{s - }d{N}_{s} + \mathop{\sum }\limits_{{0 < s \leq t}}\left( {{N}_{s}^{2} - {N}_{s - }^{2} - 2{N}_{s - }\Delta {N}_{s}}\right) \] \[ = 2{\int }_{0}^{t}{N}_{s - }d{N}_{s} + \mathop{\sum }\limits_{{k = 1}}^{{N}_{t}}\left( {{k}^{2} - {\left( k - 1\right) }^{2} - 2\left( {k - 1}\right)...
Yes
Lemma 14.36 Let \( X \) be a one dimensional Lévy process with R-triplet \( \left( {{\mu }_{R},{\sigma }^{2},\nu }\right) \) and \( f = f\left( {t, x}\right) \in {C}^{1,2}\left( {\left\lbrack {0, T}\right\rbrack \times \mathbb{R}}\right) \) . Then we have\n\n\[ \n{df}\left( {t,{X}_{t}}\right) = \left( {{\mathcal{A}}_{R...
Proof. First of all we show that each term in (14.39) and (14.40) is well defined. Let us recall that (cf. (13.46))\n\n\[ \n{\int }_{\left| y\right| < 1}{y}^{2}\nu \left( {dy}\right) < \infty \n\]\n\n(14.41)\n\nand let us denote by\n\n\[ \nD\left( {z, R}\right) = \rbrack z - R, z + R\lbrack \n\]\n\nthe interval centere...
Yes
Theorem 14.37 (Itô formula) Let \( X \) be a one dimensional Lévy process with \( R \) -triplet \( \left( {{\mu }_{R},{\sigma }^{2},\nu }\right) \) and \( f = f\left( {t, x}\right) \in {C}^{1,2}\left( {\left\lbrack {0, T}\right\rbrack \times \mathbb{R}}\right) \) . If \( f \) is bounded then we have\n\n\[ \n{df}\left( ...
Proof. Formulas (14.44)-(14.45) follow directly by adding to (14.40), and subtracting from (14.39), the integral term\n\n\[ \n{\int }_{\left| y\right| \geq R}\left( {f\left( {t, y + {X}_{t - }}\right) - f\left( {t,{X}_{t - }}\right) }\right) \nu \left( {dy}\right) {dt} \]\n\nthat is well defined because \( f \) is boun...
Yes
Let \( X \) be a one dimensional Lévy process with \( R \) -triplet \( \left( {{\mu }_{R},{\sigma }^{2},\nu }\right) \) and characteristic exponent \( {\psi }_{X} \). We assume that \[ {\int }_{\left| y\right| \geq 1}{e}^{2y}\nu \left( {dy}\right) < \infty \]
By Proposition 13.49, condition (14.47) is equivalent to the existence of the second moment of \( X \): \[ E\left\lbrack {e}^{2{X}_{t}}\right\rbrack = {e}^{t{\psi }_{X}\left( {-{2i}}\right) }.\]
Yes
By the Lévy-Itô decomposition, a Lévy process \( X \) with triplet \( \left( {{\mu }_{1},{\sigma }^{2},\nu }\right) \) can be written as\n\n\[ d{X}_{t} = {\mu }_{1}{dt} + {\sigma d}{W}_{t} + {\int }_{\left| y\right| < 1}y\widetilde{J}\left( {{dt},{dy}}\right) + {\int }_{\left| y\right| \geq 1}{yJ}\left( {{dt},{dy}}\rig...
and therefore it can be considered as a solution of a SDE of type (14.55) where\n\n\[ \bar{b}\left( {t, x}\right) = {\mu }_{1},\;\sigma \left( {t, x}\right) = \sigma ,\;\widetilde{a}\left( {t, x, y}\right) = \bar{a}\left( {t, x, y}\right) = y. \]
Yes
Under the hypotheses of Theorem 14.43, let \( X \) be a strong solution of the SDE (14.55) and \( f = f\left( {t, x}\right) \in {C}^{1,2}\left( {\left\lbrack {0, T}\right\rbrack \times \mathbb{R}}\right) \) . Then we have\n\n\[ {df}\left( {t,{X}_{t}}\right) = \left( {\mathcal{A} + {\partial }_{t}}\right) f\left( {t,{X}...
where \( \mathcal{A} \) is the integro-differential operator with variable coefficients\n\n\[ \mathcal{A}f\left( {t, x}\right) = \bar{b}\left( {t, x}\right) {\partial }_{x}f\left( {t, x}\right) + \frac{{\sigma }^{2}\left( {t, x}\right) }{2}{\partial }_{xx}f\left( {t, x}\right) \]\n\n\[ + {\int }_{\left| y\right| < 1}\l...
Yes
Theorem 14.45 Assume the following Lipschitz and growth conditions:\n\ni) for every \( n \in \mathbb{N} \) there exists a constant \( {K}_{n} \) such that\n\n\[ \n{\left| b\left( t,{x}_{1}\right) - b\left( t,{x}_{2}\right) \right| }^{2} + {\left| \sigma \left( t,{x}_{1}\right) - \sigma \left( t,{x}_{2}\right) \right| }...
Moreover, there exists a positive constant \( C \), depending on \( K,{K}_{n} \) and \( T \) only, such that\n\n\[ \nE\left\lbrack {X}_{t}^{2}\right\rbrack \leq C\left( {1 + {X}_{0}^{2}}\right) ,\;t \in \left\lbrack {0, T}\right\rbrack . \n\]\n\n(14.61)
Yes
The SDE (14.49) for an exponential Lévy process \( {S}_{t} = {e}^{{X}_{t}} \) is of the form (14.58), that is\n\n\[ d{S}_{t} = b\left( {t,{S}_{t - }}\right) {dt} + \sigma \left( {t,{S}_{t - }}\right) d{W}_{t} + {\int }_{\mathbb{R}}a\left( {t,{S}_{t - }, y}\right) \widetilde{J}\left( {{dt},{dy}}\right) \]\n\nwhere\n\n\[...
Under condition (14.47) on the Lévy measure \( \nu \), the Lipschitz and growth conditions (14.59)-(14.60) are satisfied.
No
Example 14.51 (Cauchy distribution) We consider a \( \alpha \) -stable process with \( \alpha = 1 \) : this is a pure jump process with no diffusion component, see Section 13.4.2. By (13.80), the characteristic exponent takes the form\n\n\[ \n\psi \left( \xi \right) = {i\mu \xi } - \sigma \left| \xi \right| \left( {1 +...
Even if the expression of \( {\psi }_{X} \) is known, in most cases it is not possible to compute explicitly \( \Gamma \) by Fourier inversion. However, formula (14.66) is used \( {}^{8} \) in the practical applications because it can be inverted numerically in various efficient ways: numerical methods in option pricin...
No
Proposition 14.52 Let \( Z \) be a Lévy process with characteristic exponent \( \psi \) . Then the characteristic function of \( X \) in (14.69) is equal to\n\n\[ \n{\varphi }_{{X}_{t}}\left( \xi \right) = E\left\lbrack {e}^{{i\xi }{X}_{t}}\right\rbrack = \exp \left( {{i\xi }{X}_{0}{e}^{-{Bt}} + {\int }_{0}^{t}\psi \le...
Proof. The thesis is a consequence of the identity\n\n\[ \nE\left\lbrack {e}^{i{\int }_{0}^{t}f\left( s\right) d{Z}_{s}}\right\rbrack = {e}^{{\int }_{0}^{t}\psi \left( {f\left( s\right) }\right) {ds}} \]\n\n(14.71)\n\nwhich holds for any continuous function \( f : \left\lbrack {0, T}\right\rbrack \rightarrow \mathbb{R}...
Yes
If \( {Z}_{t} = {\mu t} + \sigma {W}_{t} \) is a Brownian motion with drift and characteristic exponent\n\n\[ \n{\psi }_{Z}\left( \xi \right) = {i\mu \xi } - \frac{{\sigma }^{2}{\xi }^{2}}{2} \n\]\n\nthen, by (14.70), the characteristic function of the solution \( X \) in (14.68) is equal to\n\n\[ \n{\varphi }_{{X}_{t}...
In this case, as we already showed in Section 9.5, \( {X}_{t} \) is a Gaussian random variable.
No
In an exponential Lévy model, the underlying asset is of the form (14.72) where \( X \) is a Lévy process satisfying the SDE\n\n\[ d{X}_{t} = {\mu }_{\infty }{dt} + {\sigma d}{W}_{t} + {\int }_{\mathbb{R}}y\widetilde{J}\left( {{dt},{dy}}\right) \]\n\nunder condition\n\n\[ {\int }_{\left| y\right| \geq 1}{e}^{2y}\nu \le...
More precisely, as in Example 14.40, we have that \( \widetilde{S} \) is a martingale if and only if (14.78) is satisfied. Thus, under an EMM the drift parameter \( {\mu }_{\infty } \) is determined by (14.78):\n\n\[ {\mu }_{\infty } = r - \frac{{\sigma }^{2}}{2} - {\int }_{\mathbb{R}}\left( {{e}^{y} - 1 - y}\right) \n...
Yes
In the Black-Scholes model, we have\n\n\[ \n{S}_{T} = {S}_{0}{e}^{{X}_{T}},\;{X}_{T} = \left( {r - \frac{{\sigma }^{2}}{2}}\right) T + \sigma {W}_{T}, \n\]
and\n\n\[ \n{\varphi }_{{X}_{T}}\left( \xi \right) = {e}^{i\left( {r - \frac{{\sigma }^{2}}{2}}\right) {T\xi } - \frac{{\sigma }^{2}{\xi }^{2}}{2}T}. \n\]
Yes
Example 15.3 We have\n\n\[ \n{f}_{\alpha }^{\mathrm{{Call}}}\left( x\right) = {e}^{-{\alpha x}}{\left( {e}^{x} - K\right) }^{ + },\;{f}_{\alpha }^{\mathrm{{Put}}}\left( x\right) = {e}^{-{\alpha x}}{\left( K - {e}^{x}\right) }^{ + }, \n\]\n\nand therefore \( {f}_{\alpha }^{\text{Call }} \in {L}^{1}\left( \mathbb{R}\righ...
No
Lemma 15.4 If \( g \in {W}^{1,2}\left( \mathbb{R}\right) \), that is \( g \in {L}^{2}\left( \mathbb{R}\right) \) and the weak derivative \( {Dg} \in {L}^{2}\left( \mathbb{R}\right) \), then \( \widehat{g} \in {L}^{1}\left( \mathbb{R}\right) \) .
Proof. It is well known that if \( g \in {W}^{1,2}\left( \mathbb{R}\right) \) then \( g,\widehat{g} \in {L}^{2}\left( \mathbb{R}\right) \) and\n\n\[ \n\widehat{Dg}\left( \xi \right) = - {i\xi }\widehat{g}\left( \xi \right) \n\]\n\nThen we have\n\n\[ \n{\int }_{\mathbb{R}}\left( {{\left| \widehat{g}\left( \xi \right) \r...
Yes
Example 15.5 By Lemma 15.4, \( {f}_{\alpha }^{\text{Call }},\widehat{{f}_{\alpha }^{\text{Call }}} \in {L}^{1}\left( \mathbb{R}\right) \) for any \( \alpha > 1 \) . Indeed \( {f}_{\alpha }^{\text{Call }} \in {W}^{1,2}\left( \mathbb{R}\right) \) because \( {f}_{\alpha }^{\text{Call }} \in {L}^{2} \) and
\[ D{f}_{\alpha }^{\text{Call }}\left( x\right) = \left\{ \begin{array}{ll} 0 & \text{ if }x < \log K, \\ \left( {1 - \alpha }\right) {e}^{\left( {1 - \alpha }\right) x} + {\alpha K}{e}^{-{\alpha x}} & \text{ if }x > \log K, \end{array}\right. \] is a square integrable function for \( \alpha > 1 \) .
Yes
Proposition 15.6 Assume that there exists \( \alpha \in \mathbb{R} \) such that\ni) \( {f}_{\alpha },{\widehat{f}}_{\alpha } \in {L}^{1}\left( \mathbb{R}\right) \) ;\nii) \( {E}^{Q}\left\lbrack {S}_{T}^{\alpha }\right\rbrack \) is finite.\n\nThen the following pricing formula holds:\n\n\[ H\left( {{S}_{0}, T}\right) = ...
Proof. First of all we show that conditions \( i \) ) and \( {ii} \) ) guarantee that the integral in (15.8) converges. Indeed, by \( i \) ) we have that \( \xi \mapsto \widehat{f}\left( {\xi + {i\alpha }}\right) \) is integrable because\n\n\[ \widehat{f}\left( {\xi + {i\alpha }}\right) = {\int }_{\mathbb{R}}{e}^{i\lef...
Yes
Corollary 15.8 (Delta) Under the assumptions of Proposition 15.6, if in addition one of the functions \( \xi \mapsto \left( {1 + \left| \xi \right| }\right) {\varphi }_{{X}_{T}}\left( {-\left( {\xi + {i\alpha }}\right) }\right) \) or \( \widehat{D{f}_{\alpha }} \) is integrable, then we have\n\n\[ \operatorname{Delta}\...
Proof. The thesis follows by differentiating formula (15.8) with respect to \( {S}_{0} \) : the additional assumptions guarantee that we can exchange the integral and differential signs.
No
Theorem 15.9 (Call option) For any \( \alpha > 1 \) such that \( {E}^{Q}\left\lbrack {S}_{T}^{\alpha }\right\rbrack \) is finite, we have the following pricing formula for a Call option with strike \( K \) and maturity \( T \) :\n\n\[ \operatorname{Call}\left( {{S}_{0}, K, T}\right) = \frac{{e}^{-{rT}}{S}_{0}^{\alpha }...
Proof. We recall Example 15.5 and use the pricing formula (15.8) of Proposition 15.6. To this end, we compute \( \widehat{{f}^{\text{Call }}}\left( {\xi + {i\alpha }}\right) \) : by (15.9), we have\n\n\[ \widehat{{f}^{\text{Call }}}\left( {\xi + {i\alpha }}\right) = \widehat{{f}_{\alpha }^{\text{Call }}}\left( \xi \rig...
Yes
Example 15.15 (Heston model) We consider another example where the option prices are sensitive to the choice of the damping parameter \( \alpha \) . In the Heston stochastic volatility model (cf. Example 10.33) the risk-neutral dynamics of the asset and its variance is given by (15.4), that is\n\n\[ d{S}_{t} = r{S}_{t}...
To avoid complex discontinuities (cf. Paragraph 15.1), we use the risk-neutral characteristic function given by Bakshi, Cao and Chen [17], which takes the form\n\n\[ {\varphi }_{{X}_{T}}\left( \xi \right) = \exp \left( {{i\xi rT} + \frac{{\nu }_{0}}{{\eta }^{2}}\left( \frac{1 - {e}^{-D\left( \xi \right) T}}{1 - G\left(...
Yes
Example 15.17 (Call price and Delta) For a Call option we have\n\n\[ \n{B}_{0}^{\mathrm{{Call}}}\left( {S}_{0}\right) = {\int }_{a}^{b}{\left( {e}^{x + \log {S}_{0}} - K\right) }^{ + }{dx} = \n\]\n\n(assuming \( a < - \log \frac{{S}_{0}}{K} \) )\n\n\[ \n= {\int }_{-\log \frac{{S}_{0}}{K}}^{b}\left( {{e}^{x + \log {S}_{...
The coefficients of the Delta of a Call option can be obtained by differentiating with respect to \( {S}_{0} \) :\n\n\[ \n{B}_{0}^{\text{Delta }}\left( {S}_{0}\right) = \frac{d}{d{S}_{0}}{B}_{0}^{\text{Call }}\left( {S}_{0}\right) = {e}^{b} - \frac{K}{{S}_{0}}, \n\]\n\nand, for \( k \geq 1 \) ,\n\n\[ \n{B}_{k}^{\text{D...
Yes
Example 15.20 (Heston) We consider the Heston model (cf. Example 15.15) with\n\n\[ r = 0, k = {1.5768},{\nu }_{0} = {0.0175},{\nu }_{\infty } = {0.0398},\eta = {0.5751},\varrho = - {0.5711}\text{.} \]\n\nWe analyze the Fourier-cosine approximation with \( a, b \) as in (15.36) and different choices of \( L \) and \( N ...
In particular, we compute the price of a Call with \( {S}_{0} = K = {100} \) and maturity \( T = 1 \) : for \( L = {10},{20},{30},{40} \) and \( N = {1000} \), we obtain the same price that is our reference value \( \mathrm{{RV}} = {5.785155434} \) . Figure 15.17 shows the percentage difference (15.37) between RV and t...
Yes
Example 15.22 (CGMY) The CGMY model is based on a particular tempered stable process (cf. Section 13.4.3) and encompasses the VG and Black-Scholes models: as usual, the underlying asset is in the form \( {S}_{T} = {S}_{0}{e}^{{X}_{T}} \) and by (13.98)-(13.99) the risk-neutral characteristic exponent given by
\[ \psi \left( \xi \right) = i{\mu }_{\infty }^{Q}\xi + C\left( {{\left( M - i\xi \right) }^{Y} - {M}^{Y} + {\left( G + i\xi \right) }^{Y} - {G}^{Y}}\right. \]\n\[ \left. {+{i\xi Y}\left( {{M}^{Y - 1} - {G}^{Y - 1}}\right) }\right) \Gamma \left( {-Y}\right) \]\n\[ {\mu }_{\infty }^{Q} = r + {C\Gamma }\left( {-Y}\right)...
Yes
Example 16.10 Let \( u \in {L}^{2}\left( {0, T}\right) \) be a (deterministic) function and\n\n\[ \nX = {\int }_{0}^{t}u\left( r\right) d{W}_{r} \n\]\n\nThen \( X \in {\mathbb{D}}^{1,2} \) and\n\n\[ \n{D}_{s}X = \left\{ \begin{array}{ll} u\left( s\right) & \text{ for }s \leq t \\ 0 & \text{ for }s > t \end{array}\right...
Indeed the sequence defined by\n\n\[ \n{X}_{n} = \mathop{\sum }\limits_{{k = 1}}^{{{k}_{n}\left( t\right) }}u\left( {t}_{n}^{k - 1}\right) {\Delta }_{n}^{k} \n\]\n\n is such that\n\n\[ \n{D}_{s}{X}_{n} = \varphi \left( {t}_{n}^{{k}_{n}\left( s\right) }\right) \n\]\n\n if \( s \leq {t}_{n}^{{k}_{n}\left( t\right) } \) a...
Yes
Proposition 16.12 (Chain rule) Let \( {}^{5}\varphi \in {C}_{\text{pol }}^{\infty }\left( \mathbb{R}\right) \). Then:\ni) if \( X \in {\mathbb{D}}^{1,\infty } \), then \( \varphi \left( X\right) \in {\mathbb{D}}^{1,\infty } \) and\n\n\[{D\varphi }\left( X\right) = {\varphi }^{\prime }\left( X\right) {DX}\]\n\n(16.3)\n\...
Proof. We prove only \( {ii} \) ) since the other parts can be proved essentially in an analogous way. If \( X \in \mathcal{S},\varphi \in {C}^{1} \) and both \( \varphi \) and its first-order derivative are bounded, then \( \varphi \left( X\right) \in \mathcal{S} \) and the claim is obvious.\n\nIf \( X \in {\mathbb{D}...
Yes
Example 16.13 By the chain rule, \( {\left( {W}_{t}\right) }^{2} \in {\mathbb{D}}^{1,\infty } \) and
\[ {D}_{s}{W}_{t}^{2} = 2{W}_{t}{\mathbb{1}}_{\left\lbrack 0, t\right\rbrack }\left( s\right) \]
Yes
Let \( u \in {\mathbb{L}}^{2} \) such that \( {u}_{t} \in {\mathbb{D}}^{1,2} \) for every \( t \) . Then \[ X \mathrel{\text{:=}} {\int }_{0}^{t}{u}_{r}d{W}_{r} \in {\mathbb{D}}^{1,2} \] and for \( s \leq t \) \[ {D}_{s}{\int }_{0}^{t}{u}_{r}d{W}_{r} = {u}_{s} + {\int }_{s}^{t}{D}_{s}{u}_{r}d{W}_{r} \]
Indeed, for fixed \( t \), we consider the sequence defined by \[ {X}_{n} \mathrel{\text{:=}} \mathop{\sum }\limits_{{k = 1}}^{{{k}_{n}\left( t\right) }}{u}_{{t}_{n}^{k - 1}}{\Delta }_{n}^{k},\;n \in \mathbb{N}, \] approximating \( X \) in \( {L}^{2}\left( \Omega \right) \) . Then \( {X}_{n} \in {\mathbb{D}}^{1,2} \) a...
Yes
Let us consider the solution \( \left( {X}_{t}\right) \) of the SDE\n\n\[ \n{X}_{t} = x + {\int }_{0}^{t}b\left( {r,{X}_{r}}\right) {dr} + {\int }_{0}^{t}\sigma \left( {r,{X}_{r}}\right) d{W}_{r},\n\]\n\nwith \( x \in \mathbb{R} \) and the coefficients \( b,\sigma \in {C}_{b}^{1} \) . Then \( {X}_{t} \in {\mathbb{D}}^{...
We do not go into the details of the proof of the first claim. The idea is to use an approximation argument based on the Euler scheme (cf. Paragraph 12.2): more precisely, the claim follows from the fact that \( \left( {X}_{t}\right) \) is the limit of the sequence of piecewise constant processes defined by\n\n\[ \n{X}...
No
Lemma 16.17 Let \( Y \) be as in (16.7) and \( Z \) be solution of the SDE\n\n\[ \n{Z}_{t} = 1 + {\int }_{0}^{t}\left( {{\left( {\partial }_{x}\sigma \right) }^{2} - {\partial }_{x}b}\right) \left( {r,{X}_{r}}\right) {Z}_{r}{dr} - {\int }_{0}^{t}{\partial }_{x}\sigma \left( {r,{X}_{r}}\right) {Z}_{r}d{W}_{r}.\n\]\n\n(1...
Proof. We have \( {Y}_{0}{Z}_{0} = 1 \) and, omitting the arguments, by the Itô formula we have\n\n\[ \nd\left( {{Y}_{t}{Z}_{t}}\right) = {Y}_{t}d{Z}_{t} + {Z}_{t}d{Y}_{t} + d\langle Y, Z{\rangle }_{t}\n\]\n\n\[ \n= {Y}_{t}{Z}_{t}\left( {\left( {{\left( {\partial }_{x}\sigma \right) }^{2} - \left( {{\partial }_{x}b}\ri...
Yes
Proposition 16.18 Let \( X, Y, Z \) be the solutions of the SDEs (16.4),(16.7) and (16.8), respectively. Then\n\n\[ \n{D}_{s}{X}_{t} = {Y}_{t}{Z}_{s}\sigma \left( {s,{X}_{s}}\right) \n\]\n\n(16.9)
Proof. We recall that, for fixed \( s \), the process \( {D}_{s}{X}_{t} \) verifies the SDE (16.5) over \( \left\lbrack {s, T}\right\rbrack \) and we prove that \( {A}_{t} \mathrel{\text{:=}} {Y}_{t}{Z}_{s}\sigma \left( {s,{X}_{s}}\right) \) verifies the same equation: the claim will then follow from the uniqueness res...
Yes
Theorem 16.26 (Clark-Ocone formula) If \( X \in {\mathbb{D}}^{1,2} \), then\n\n\[ X = E\left\lbrack X\right\rbrack + {\int }_{0}^{T}E\left\lbrack {{D}_{t}X \mid {\mathcal{F}}_{t}^{W}}\right\rbrack d{W}_{t} \]
Proof. It is not restrictive to suppose \( E\left\lbrack X\right\rbrack = 0 \) . For every simple adapted process \( U \in \mathcal{P} \) we have, by the duality relation of Theorem 16.23,\n\n\[ E\left\lbrack {X{D}^{ * }U}\right\rbrack = E\left\lbrack {{\int }_{0}^{T}\left( {{D}_{t}X}\right) {U}_{t}{dt}}\right\rbrack =...
Yes
Theorem 16.28 (Stochastic integration by parts) Let \( F \in {C}_{b}^{1} \) and let \( X \in {\mathbb{D}}^{1,2} \) . Then the following integration by parts holds:\n\n\[ E\left\lbrack {{F}^{\prime }\left( X\right) Y}\right\rbrack = E\left\lbrack {F\left( X\right) {\int }_{0}^{T}\frac{{u}_{t}Y}{{\int }_{0}^{T}{u}_{s}{D}...
Sketch of the proof. By the chain rule we have\n\n\[ {D}_{t}F\left( X\right) = {F}^{\prime }\left( X\right) {D}_{t}X \]\n\nmultiplying by \( {u}_{t}Y \) and integrating from 0 to \( T \) we get\n\n\[ {\int }_{0}^{T}{u}_{t}Y{D}_{t}F\left( X\right) {dt} = {F}^{\prime }\left( X\right) Y{\int }_{0}^{T}{u}_{t}{D}_{t}{Xdt} \...
No
Example 16.30 (Delta) We observe that \( {D}_{s}{S}_{T} = \sigma {S}_{T} \) and \( {\partial }_{x}{S}_{T} = \frac{{S}_{T}}{x} \) . Then, by (16.17) we have the following expression for the Black-Scholes Delta
\[ \Delta = {e}^{-{rT}}{\partial }_{x}E\left\lbrack {F\left( {S}_{T}\right) }\right\rbrack \] \[ = {e}^{-{rT}}E\left\lbrack {F\left( {S}_{T}\right) {\int }_{0}^{T}\frac{{\partial }_{x}{S}_{T}}{{\int }_{0}^{T}{D}_{s}{S}_{T}{ds}}\diamond d{W}_{t}}\right\rbrack \] \[ = {e}^{-{rT}}E\left\lbrack {F\left( {S}_{T}\right) {\in...
Yes
Proposition 16.31 Let \( X \in {\mathbb{D}}^{1,2} \) and let \( U \) be a second-order Skorohod-integrable process. Then\n\n\[ \n{\int }_{0}^{T}X{U}_{t}\diamond d{W}_{t} = X{\int }_{0}^{T}{U}_{t}\diamond d{W}_{t} - {\int }_{0}^{T}\left( {{D}_{t}X}\right) {U}_{t}{dt}.\n\]
Proof. For every \( Y \in \mathcal{S} \), by the duality relation, we have\n\n\[ \nE\left\lbrack {Y{D}^{ * }\left( {XU}\right) }\right\rbrack = E\left\lbrack {{\int }_{0}^{T}\left( {{D}_{t}Y}\right) X{U}_{t}{dt}}\right\rbrack = \n\]\n\n(by the chain rule)\n\n\[ \n= E\left\lbrack {{\int }_{0}^{T}\left( {{D}_{t}\left( {Y...
Yes
[{\int }_{0}^{T}{W}_{T}\diamond d{W}_{t} = {W}_{T}^{2} - T]
By a direct application of (16.20), we have\n\n\[ \n{\int }_{0}^{T}{W}_{T}\diamond d{W}_{t} = {W}_{T}^{2} - T \n\]
Yes
Example 16.33 (Vega) Let us compute the Vega of a European option with payoff function \( F \) in the Black-Scholes model: we first notice that\n\n\[ \n{\partial }_{\sigma }{S}_{T} = \left( {{W}_{T} - {2\sigma T}}\right) {S}_{T},\;{D}_{s}{S}_{T}\sigma {S}_{T}.\n\]\n\nThen\n\n\[ \n\mathcal{V} = {e}^{-{rT}}{\partial }_{\...
\n(by the integration-by-parts formula (16.17))\n\n\[ \n= {e}^{-{rT}}E\left\lbrack {F\left( {S}_{T}\right) {\int }_{0}^{T}\frac{{W}_{T} - {\sigma T}}{\sigma T}\diamond d{W}_{t}}\right\rbrack =\n\]\n\n(by \( \left( {16.20}\right) \) )\n\n\[ \n= {e}^{-{rT}}E\left\lbrack {F\left( {S}_{T}\right) \left( {\frac{{W}_{T} - {\s...
No
Example 16.34 (Gamma) We compute the Gamma of a European option with payoff function \( F \) in the Black-Scholes model:
\[ \Gamma = {e}^{-{rT}}{\partial }_{xx}E\left\lbrack {F\left( {S}_{T}\right) }\right\rbrack = \] \[ = \frac{{e}^{-{rT}}}{\sigma T}E\left\lbrack {{\partial }_{x}\left( \frac{F\left( {S}_{T}\right) }{x}\right) {W}_{T}}\right\rbrack = - \frac{{e}^{-{rT}}}{{\sigma T}{x}^{2}}E\left\lbrack {F\left( {S}_{T}\right) {W}_{T}}\ri...
Yes
We give the expression of the Delta of an arithmetic Asian option with Black-Scholes dynamics (16.18) for the underlying asset. We denote the average by\n\n\\[ \nX = \frac{1}{T}{\\int }_{0}^{T}{S}_{t}{dt} \n\\]\n\nand we observe that \\( {\\partial }_{x}X = \frac{X}{x} \\) and\n\n\\[ \n{\\int }_{0}^{T}{D}_{s}{Xds} = {\...
Then we have\n\n\\[ \n\\Delta = {e}^{-{rT}}{\\partial }_{x}E\\left\\lbrack {F\\left( X\\right) }\\right\\rbrack = \\frac{{e}^{-{rT}}}{x}E\\left\\lbrack {{F}^{\\prime }\\left( X\\right) X}\\right\\rbrack = \n\\]\n\n(by (16.17) and (16.21))\n\n\\[ \n= \\frac{{e}^{-{rT}}}{\\sigma x}E\\left\\lbrack {F\\left( X\\right) {\\i...
Yes
We extend Example 16.30 to the case of a model with local volatility\n\n\[ \n{S}_{t} = x + {\int }_{0}^{t}b\left( {s,{S}_{s}}\right) {ds} + {\int }_{0}^{t}\sigma \left( {s,{S}_{s}}\right) d{W}_{s}.\n\]\n\nUnder suitable assumptions on the coefficients, we prove the following Bismut-Elworthy formula:\n\n\[ \nE\left\lbra...
We recall that, by Proposition 16.18, we have\n\n\[ \n{D}_{s}{S}_{T} = {Y}_{T}{Z}_{s}\sigma \left( {s,{S}_{s}}\right)\n\]\n\n(16.23)\n\nsince\n\n\[ \n{Y}_{t} \mathrel{\text{:=}} {\partial }_{x}{S}_{t} = : {Z}_{t}^{-1}.\n\]\n\nLet us apply (16.16) after choosing\n\n\[ \nX = {S}_{T},\;Y = G{Y}_{T},\;{u}_{t} = \frac{{Y}_{...
Yes
Corollary 6.8 (Sufficiency in Theorem 6.6): If \( N = D \) and \( \sigma \left( t\right) \) is nonsingular for Lebesgue-almost-every \( t \in \left\lbrack {0, T}\right\rbrack \) almost surely, then the financial market is complete.
Proof. We verify the condition of Proposition 6.2. Let \( B \) be an \( \mathcal{F}\left( T\right) \) -measurable random variable satisfying (6.3), and define the Lévy \( {P}_{0} \) -martingale\n\n\[ \n{M}_{0}\left( t\right) = {E}_{0}\left\lbrack {\left. \frac{B}{{S}_{0}\left( T\right) }\right| \;\mathcal{F}\left( t\ri...
Yes
Example 3.1 (Forward contract to purchase a stock that pays no dividends): Suppose the contract is to purchase one share of the first stock, i.e., \( B = {S}_{1}\left( T\right) \) . If the first stock pays no dividends and \( \sigma \left( \cdot \right) \) satisfies the Novikov condition (1.5.17) with \( n = 1 \), then...
\[ {V}^{FC}\left( {t;q}\right) = {S}_{1}\left( t\right) - q{S}_{0}\left( t\right) \cdot {E}_{0}\left\lbrack {1/{S}_{0}\left( T\right) \mid \mathcal{F}\left( t\right) }\right\rbrack ,\;0 \leq t \leq T. \] (3.3) If in addition \( {S}_{0}\left( T\right) \) is nonrandom, the hedging portfolio is particularly simple. The ag...
Yes