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 = \langle\eta, A\eta\rangle \end{equation} \begin{equation} \nu_U = \nu\phi^{-1} \end{equation} \begin{equation} \gamma_U = -0.5\langle\eta, A\eta\rangle - \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 \text{ 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} \left[e^{ia\gamma_t} \kappa^{-mt} e^{\frac{\kappa-1}{\kappa}\gamma_t} \right] \end{equation} \begin{equation} =\kappa^{-mt}\mathbb{E} \left[ \exp\left(\frac{ia\kappa + \kappa -1}{\kappa}\gamma_t\right) \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\left( \frac{1-\kappa}{\kappa}\right) \exp\left(\frac{\kappa x}{\kappa -1} \right) \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?

  • $\begingroup$ $$ \begin{align} & A_U = <\eta,A\eta> \\ {} \\ & A_U = \langle\eta, A\eta\rangle \end{align} $$ The second item above is the right way to do it, and I edited accordingly. $\endgroup$ Apr 7, 2022 at 1:48

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.


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