Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.


In his book Ken-Iti Sato ("Lévy Processes and Infinitely Divisible Distributions") provides the theory for measure change for Lévy processes in Theorems 33.1 and 33.2. It can be summarised as:

PROPOSITION (Sato Theorems 33.1 and 33.2)

Let $X_t$ be a Lévy process under the probability measure $\mathbb{P}$ with the characteristic triplet $(A, \nu, \gamma)$, and a Levy process under the probability measure $\mathbb{Q}$ with the triplet $(A^Q, \nu^Q, \gamma^Q)$.

Then the measures $\mathbb{P}|_{\mathcal{F}_t}$ and $\mathbb{Q}|_{\mathcal{F}_t}$ are equivalent for all $t$ if and only if the three conditions are satisfied:

  1. $A=A^{Q}$
  2. The \levy{} measures are equivalent with \begin{equation} \int_{-\infty}^{\infty} (e^{\phi(x)/2}-1)^2 \nu(dx) < \infty \end{equation} where $\phi(x) = \ln(\frac{d \nu^Q}{d \nu})$

  3. If $A=0$ then in addition we must have: \begin{equation} \gamma^Q = \gamma+\int_{|x|\leq 1} x(\nu^Q-\nu)(dx) \end{equation}

When $\mathbb{P}$ and $\mathbb{Q}$ are equivalent, then the Radon-Nikodym derivative is given by the following exponential martingale: \begin{equation} \left. \frac{d \mathbb{Q}}{d \mathbb{P}} \right|_{\mathcal{F}_t} = e^{U_t}, \end{equation} where $U_t$ is a Lévy process with characteristic triplet $(A_U, \nu_U, \gamma_U)$ given by: \begin{equation} A_U = <\eta, A\eta> \end{equation} \begin{equation} \nu_U = \nu\phi^{-1} \end{equation} \begin{equation} \gamma_U = -0.5<\eta, A\eta> - \int_{-\infty}^{\infty} (e^y -1 -y \mathbf{1}(0<|y|<1))\nu_U(dx) \end{equation} where $\eta=0$ if $A=0$, otherwise it solves the equation: \begin{equation} \gamma^Q - \gamma+\int_{|x|\leq 1} x(\nu^Q-\nu)(dx) = A\eta \end{equation}

Note, that $\gamma_U$ is determined in the equation above by the condition that $e^{U_t}$ is a martingale.


By a gamma process $\{\gamma_t\}$ on a given probability space with shape $m$ and scale $\kappa$ we mean a process with independent increments, such that $\gamma_0 = 0$ and the random variable $\gamma_t$ has a gamma distribution with mean $\kappa mt$ and variance $\kappa^2 mt$. It has the density function: \begin{equation}\label{eq:gamma_density} g(x) = \frac{\kappa^{-mt}x^{mt-1}e^{-x/\kappa}}{\Gamma[mt]}\quad \textit{for }x\geq 0, \end{equation} where $\Gamma[a]$ denotes the standard gamma function. The characteristic function of the gamma process is given by: \begin{equation} \mathbb{E}\left[e^{i\lambda\gamma_t}\right]=\frac{1}{(1-i\kappa\lambda)^{mt}} = \exp(t m \log (1 + \kappa u)) \end{equation} By doing some algebraic transformation (cf. Protter, 2003, p. 33) one can show that the associated Lévy measure is given by: \begin{equation} \nu(x) = mx^{-1}e^{-x/\kappa} \end{equation} Let $\gamma_t$ be a standard gamma process with shape parameter $m$ and scale parameter $\kappa=1$ under the probability measure $\mathbb{P}$. For a given $\kappa>0$ we can define a measure change to an equivalent probability measure $\mathbb{Q}$ as: \begin{equation} \left. \frac{d \mathbb{Q}}{d \mathbb{P}} \right|_{\mathcal{F}_t} = \kappa^{-mt}e^{\frac{\kappa-1}{\kappa}\gamma_t}. \end{equation} It is straight forward to check that the Radon-Nikodym process is a martingale. Under the $\mathbb{Q}$ measure the process $\gamma_t$ is a gamma process with the same shape $m$, but with scale given by $\kappa$. To show this, lets calculate the characteristic function: \begin{equation} \mathbb{E}^Q[e^{ia\gamma_t} = \mathbb{E}[e^{ia\gamma_t}\kappa^{-mt}e^{\frac{\kappa-1}{\kappa}\gamma_t}] \end{equation} \begin{equation} =\kappa^{-mt}\mathbb{E}[\exp\left(\frac{ia\kappa + \kappa -1}{\kappa}\gamma_t\right)] \end{equation} \begin{equation} =\kappa^{-mt}\frac{1}{\left[\frac{1-ia\kappa}{\kappa}\right]^{mt}} \end{equation} \begin{equation} =\frac{1}{\left[1-ia\kappa\right]^{mt}} \end{equation} which indeed shows that $\gamma_t$ has the scale parameter $\kappa$ under $\mathbb{Q}$.

We can calculate the measure change above using the proposition above. Because $\frac{d \nu^Q}{d \nu} = e^{x/\kappa - x}$, we have: \begin{equation} \phi^{-1}(x) = \frac{\kappa y}{1-\kappa}. \end{equation} The Lévy density of $U_t$ is thus given by: \begin{equation} \nu_U = \nu(\phi^{-1}(x)) = m(\frac{1-\kappa}{\kappa})\exp(\frac{\kappa x}{\kappa -1}) \end{equation} which is also a gamma process with the same scale as in the previous method given by $\frac{1-\kappa}{\kappa}$. However, the shape parameter is different! To my best knowledge $U_t$ should be unique up to null sets, so what am I doing wrong here?

share|improve this question
@Louigi, how did you managed to change the latex at the top of the question? I tried, but couldn't get it to work.... –  Grzenio Oct 3 '11 at 14:46

1 Answer 1

Figured it out finally. For the gamma process the quantity: \begin{equation} \nu(x) = mx^{-1}e^{-x/\kappa} \end{equation} is not really the Levy measure, but the density of the Levy measure wrt. the Lebesgue measure - so the correct equation should be: \begin{equation} \nu(A) = \int_A mx^{-1}e^{-x/\kappa} dx \end{equation} and \begin{equation} \nu(\phi^{-1}(A)) = \int_A mx^{-1}e^{-x/\kappa} d\phi^{-1}(x) = \int_A \kappa^{-mt}e^{\frac{\kappa-1}{\kappa}\gamma_t} dx \end{equation} an we get the correct result.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.