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I know that the Ito integral is defined in general for continuous semimartingales. But it can also be defined only for Ito processes. My question is if every process $X_t$ satisying a SDE of the form $dX_t=f(X_t)dt+g(X_t)dB_t$ where $f,g$ are "well-behaved" functions is a semimartingale. If so, why is that true?

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Let $dX_t = \sigma_t \, dB_t+b_t \, dt$ the solution of the SDE, i.e. $$X_t= X_0 + \int_0^t \sigma_s dB_s + \int_0^t b_s \, ds$$ Then

  1. $M(t,w) :=X_0(w) + \left(\int_0^t \sigma_s \, dB_s\right)(w)$ is a local martingale (by the definition of the stochastic integral!)
  2. $A(t,w) := \int_0^t b_s(w) \, ds$ is a finite variation process since $$\sum_{k=1}^n |A_{t_k}-A_{t_{k-1}}| \leq \sum_{k=1}^n \int_{t_{k-1}}^{t_k}|b(s)| \, ds = \int_0^t |b(s)| \, ds < \infty \quad \text{a.s.}$$ for all partitions $0=t_0<\ldots=t_n=t$.

Hence you obtain the decomposition $X_t = M_t+A_t$ which implies that $X_t$ is a semimartingale.

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Because the integration with respect to Brownian motion preserves the local martingale property and the first term gives you a finite variation process. This is precisely the canonical decomposition of a semimartingale, for which the finite variation property must hold only locally.

You can also define stochastic integrals for processes involving jumps!

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However, we must a priori know that $X_t$ is a semimartingale, don't we? Or is this true in general for a general process satisfying a SDE? –  Nick Papadopoulos Dec 9 '12 at 13:07
    
No, also the converse is true. Check Thoerem II.3.9 in Protter's book. A cadlag, locally square integrable local martingale is a semimartingale, and a cadlag process with finite variation on compacts also is a semimartingale. Moreover, semimartingales form a vector space. –  user13655 Dec 9 '12 at 20:49
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