# Lévy's triplet of $Y_{t}=aB_{t}+Z_{t}+rt$

I'm trying to find the triplet of the next Lévy process: $$Y_{t}=aB_{t}+Z_{t}+rt,$$ where $\{B_{t}\}_{t\geq 0}$ is a standard Brownian Motion, $\{Z_{t}\}_{t\geq 0}$ is a Compound Poisson Process and $a,r\in\mathbb{R}.$

I know such triplet is determined by the formula of characterist exponent and is unique. So, such triplet exists, because for each Lévy process there is such triplet. I'd like to make explicit each term of the triplet but I feel confused about it.

To obtain this, I was thinking the triplet is given by $(r,t,\pi),$ where $\pi$ is the measure of jumps of the Compound Poisson but I'm not sure about this. How can we make these terms explicit?

Any kind of help is thanked in advanced.

## 1 Answer

Every Lévy process $Y$ can be decomposed as the sum of three independent processes $Y=B+Z+D$, respectively a Brownian motion, a Lévy jump process $Z$ and a drift $D$. Hence one can choose three objects that characterize each of $(B,\,Z,\,D)$ to characterise the Lévy process $Y$. Your Brownian motion is determined by $a\ge0$, your compound Poisson process is determined by its Lévy measure $\pi$, and your drift by the coefficient $r\in\mathbb R$. You can make $\pi$ more explicit, where $\pi(A)=\lambda \int_A \pi^X(dy)$, where $\lambda$ is the intensity of the exponential jumps, and $\pi^X$ is the law of each jump.

• Thanks to answer @Ton. So is enough to compute characteristic function of Y, using independence of $B,Z,D$ right? – Squird37 May 15 '18 at 23:03
• Yes, or equivalently you can compute their Markovian generators. – Ton May 15 '18 at 23:30
• Thanks again @Ton. – Squird37 May 15 '18 at 23:36