[Math] Measure changes for gamma process

pr.probabilitystochastic-calculusstochastic-processes

GENERAL THEORY

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.

GAMMA PROCESS

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?

Best Answer

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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