Lévy Models - Computational Methods for Quantitative Finance - page 125

Proof See [3, Chap. 1.2.4].

Note that in ( 10.2 ) the integral with respect to the Lévy measure exists since the
integrand is bounded outside of any neighborhood of 0 and

e iξz (z 2 ) 1 − + iξz 1

} = O

as | z |→ 0 .

{| z |≤ 1

But there are many other ways to obtain an integrable integrand. We could, for
example, replace 1 by any bounded measurable function f : R → R satisfying

{| z |≤ 1 }

= + O | | | |→ = O | | | |→∞

f(z) . Different choices

of f do not affect σ 2 and ν .But γ depends on the choice of the truncation function.

1 ( z ) as z 0 and f(z) ( 1 / z ) as z

If the Lévy measure satisfies

1 |

z | ν( d z) < ∞ , we can use the zero function as f

| z |≤

and get

1

e iξz ν( d z),

1
2 σ 2 ξ 2

=− + + −

ψ(ξ) iγ 0 ξ (10.3)

R

. We denote this representation by the triplet (σ 2 ,ν,γ 0 ) 0 . Furthermore,

with γ 0 ∈ R

if

ν( d z) < ∞ ,i.e. X is a compound Poisson process, we can rewrite ( 10.3 )as

R

+ λ

1
2 σ 2 ξ 2

− e iξz )ν 0 ( d z), ψ(ξ) =− iγ 0 ξ + ( 1 (10.4)

R

=

with λ ν/λ . We say that X is of finite activity with

jump intensity λ and jump size distribution ν 0 . If the Lévy measure satisfies

ν( d z) and ν 0 =

R

> 1 | z | ν( d z) < ∞

, then, letting f be a constant function 1, we obtain

| z |

1

iξz ν( d z),

1
2 σ 2 ξ 2

e iξz ψ(ξ) =− iγ c ξ + + − + (10.5)

R

with triplet (σ 2 ,ν,γ c ) c where γ c is called the center of X since E[ X t ]= γ c t .We

use the representation ( 10.5 ) instead of ( 10.2 ) throughout this work but omit the sub-
script c for simplicity. No arbitrage considerations require Lévy processes employed
in mathematical finance to be martingales. The following result gives sufficient con-
ditions of the characteristic triplet to ensure this.

Lemma 10.1.5 Let X be a Lévy process with characteristic triplet (σ 2 ,ν,γ) . As-

sume , and

> 1 e z ν( d z) <

. Then , e X

> 1 |

| ∞ ∞

z ν( d z) < is a martingale with

| z | | z |

respect to the filtration F of X if and only if

σ 2
2 +

(e z γ + − 1 − z)ν( d z) = 0 .

R

Proof We obtain for 0 ≤ t<s using the independent and stationary increments

property

e X s e X t + X s − X t e X t e X s − X t E[ | F t ]=E[ | F t ]= E[ ]

e X t e X s − t e X t e (t − s)ψ( − i) . = E[ ]=