# Decomposition of semimartingales

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.

Q1: When is it true that

$X_t^d = \sum_{0 \leq u \leq t} \Delta X_u$

(where $\Delta X_t \equiv X_t - X_{t-}$ and $X_{t-} \equiv \lim_{u \nearrow t} X_u$)?

Is it always true?

Pro05, pg.221 seems to require:

Hypothesis A. $\sum_{0 \leq u \leq t} | \Delta X_u | \lt \infty$ a.s., each $t \gt 0$

But I thought this was trivially satisfied by semimartingales since their quadratic variation exists and is finite.

Q2: When is it true that $X^d$ is a finite variation process?

I suppose if someone answers my first question the second is necessarily answered?

(Thx for bearing with me. This material is way over my head. Just formulating a good question has been a challenge.)

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Really, nothing on this? Anyone have suggestions for a better way to pose my question or different forum? Is this more appropriate for mathoverflow? –  lowndrul Apr 22 '12 at 22:26
This is fine for math.stackexchange. I'll answer this properly later this week if nobody else does. Btw, your Hypothesis A is missing some absolute-value signs. If you put them in then the hypothesis is equivalent to $X^d$ being a finite variation process. This is not satisfied by all semimartingales (e.g., Cauchy process). Q1 does not have a good answer though, it's just true when it is true. In general there is no reason to expect this to hold. –  George Lowther Apr 25 '12 at 23:53
Good catch! I added the absolute value signs in the question. For Q1, I'd be content with an example for when it holds and when it doesn't. I'm really just trying to wrap my head around what these different processes are supposed to look like. –  lowndrul Apr 26 '12 at 0:12
@GeorgeLowther: If you add a short answer below, I'll almost surely accept it. Still have something? :) –  lowndrul May 1 '12 at 14:17

Q1: In general it is not true that $X^d_t$ is equal to $\sum_{s\le t}\Delta X_s$. In order for this equality to make sense, you need assume something like that the sum of the jumps of $X$ is absolutely convergent (Hypothesis A), although that still does not imply that the identity holds. The term $X^d$ is the purely discontinuous part of the semimartingale $X$, but this terminology can be confusing. A purely discontinuous semimartingale is not in general equal to the sum of its jumps, and need not be discontinuous at all! All (adapted and right-continuous) finite variation processes are "purely discontinuous" semimartingales, regardless of whether they are continuous or not. So, for example, take any continuous finite variation process $X$, such as $X_t=t$. Then, you simply have $X^d_t=X_t-X_0$ whereas $\sum_{s\le t}\Delta X_s=0$. The terminology "purely discontinuous" does make a bit more sense for local martingales, where you can show that purely discontinuous local martingales are, in a sense, orthogonal to the continuous local martingales and are themselves (almost surely) continuous only if they are constant. I've also seen the term "quadratic pure jump semimartingale" to describe purely discontinuous semimartingales (in, e.g., Protter, Stochastic Integration and Differential Equations). This does make some sense, as a semimartingale is purely discontinuous if and only if its quadratic variation is a pure jump process, $[X]_t=\sum_{s\le t}(\Delta X_s)^2$.

Anyway, the short answer to Q1 is that $X^d_t=\sum_{s\le t}\Delta X_s$ if and only if $X$ is the sum of a continuous local martingale and an (absolutely convergent) sum of jump terms. This is really just tautological and saying that the identity holds only when it holds, but I doubt you are going to get any better general statement than that.

You also mention Hypothesis A: $\sum_{s\le t}\vert \Delta X_s\vert < \infty$ (a.s.). This property is not true for all semimartingales. In general, the existance of the quadratic variation ensures that $$\sum_{s\le t}(\Delta X_s)^2\le[X]_t < \infty.$$ This does not imply Hypothesis A. You do have the inequality $\sum_{s\le t}\vert\Delta X_s\vert\ge\sqrt{\sum_{s\le t}(\Delta X_s)^2}$, but that is the wrong way round to tell us anything useful here. Examples of semimartingales which fail this hypothesis include the Cauchy process (I can link you to my blog post on properties of Levy processes for this, although any book on Levy processes should give more details), which are purely discontinuous semimartingales with finite variation.

Q2: This question has a very nice answer, $X^d$ is a finite variation process if and only if Hypothesis A holds! This can be proven without too much difficulty (assuming some facts about continuous semimartingales).

Proof: Suppose that $X^d$ is a finite variation process. As $\Delta X_t=\Delta X^d_t$, the sum $\sum_{s\le t}\vert\Delta X_s\vert$ is bounded by the variation of $X^d$ on the interval $[0,t]$, so is finite (a.s.)

In the other direction, suppose that Hypothesis A holds, so we can define the finite variation process $Y_t=\sum_{s\le t}\Delta X_s$. Then $X-Y$ is a continuous semimartingale. By a standard result on decomposition of continuous semimartingales, we can decompose $X-Y=M+V$ for a continuous local martingale $M$ and continuous finite variation process $V$. Then, $Y+V$ is a finite variation process, so is a purely discontinuous semimartingale. Also, $X=M+(Y+V)$ implies that $X^d=Y+V$ is a finite variation process.

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Awesome answer. This was exactly what I was looking for. Thanks George. By the way, your blog has been a huge help for me. –  lowndrul May 2 '12 at 14:00
1. Protter defines purely discontinuous to mean orthogonal to all continuous path martingales. As a result he would have (problem 6.9) $N_t - \lambda t, N_t \sim \;poiss(\lambda)$, purely discontinuous.

2. quadratic variation would control sum of square of the jumps.

3. for Levy processes q2 has a nice answer in terms of the Levy measure, Cont and Tankov, chapter 3 is a readable source.
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