1
$\begingroup$

I have to prove that a Lévy process is a semimartingale.

In general we say that $X$ is a semimartingale if it is an adapted process such that, for each $t ≥ 0$,

$$X (t) = X (0) + M(t) + C(t)$$ where $M = (M(t), t ≥ 0)$ is a local martingale and $C = (C(t), t ≥ 0)$ is an adapted process of finite variation.

By the Lévy–Itô decomposition we have, for each $t ≥ 0$,

$X (t) = M(t) + C(t)$,

where $M(t) = B_A(t) + \int_{|x|\leq1} x \tilde{N} (t, dx)$ and $C(t) = bt +\int_{|x|>1} xN(t, dx)$, with $$\int_{A} xN(t,dx)= \sum_{0 \leq u \leq t} \Delta X_u 1_A(\Delta X_u)$$ and $$\int_A x\tilde{N}(t,dx):= \int_A xN(t,dx)-t\int_A x\nu(dx).$$

Now, I think I can prove that $M$ is a martingale and that $C$ is of finite variation, but with respect to which filtration?

Because the book "Lévy Processes and Stochastic Calculus" of Applebaum doesn't say anything about the filtration but in another book I read that $M$ and $C$ are to be adapted to the same filtration to which X is adapted.

But, for example I can prove that $(B_A(t),t\geq 0)$ is a martingale wrt his natural filtration, not wrt the filtration to which the Lévy processes $X$ is adapted.

Similarly, $\int_{|x|\leq1} x \tilde{N} (t, dx)$ is a martingale wrt his natural filtration?

Thanks.

$\endgroup$

1 Answer 1

1
$\begingroup$

The statement that cadlag Levy is semimartingale is proved here in theorem 4. Namely they prove that a cadlag Lévy process $X$ decomposes as $X_t=bt+W+Y$ where $Y$ is a semimartingale and $W$ is a continuous centered Gaussian process with independent increments, hence a martingale.

Filtration question

When one starts with a semimartingale $X$ with filtration $F_{t}$, then the Doobs-Meyer decomposition theorem gives decomposition $X=M+A$, where these are adapted to $F_{t}$ since we use projections of $X$ in order to construct them.

Conversely, if one starts with the decomposition, one can use Stricker's theorem

if you take a semimartingale X according to some filtration F and if G is a subfiltration of F for which X is adapted, then X is also a G-semimartingale

Decomposition

Just as an aside, for a semimartingales to have the decomposition you mentioned one needs some additional assumptions eg. see MSE answer here

From Kal97, pg. 446:

Theorem 23.14 Any semimartingale $X$ has an a.s. unique decomposition $X=X_0 + X^c + X^d$ where $X^c$ is a continuous local martingale with $X_0^c=0$ and $X^d$ is a purely discontinuous semimartingale.

And if we want to have $X^{d}$ further be finite variation one needs

$$\sum_{s\le t}\vert \Delta X_s\vert < \infty\text{ (a.s.) }.$$

(eg. Cauchy process is a semimartingale that fails to have it here).

Filtration shrinkage

Also, as another aside there can be issues of losing semimartingale if we project into a smaller filtration. For some discussion and references see here Semimartingale decomposition and filtrations and also online article "Local Martingales and Filtration Shrinkage".

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .