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Consider a Levy process $L$ in $\mathbb R^d$ written in the Levy-Kchinchine decomosition as $$L(t)=bt + W(t) + Z(t),$$ where $bt$ is the drift part, $W(t)$ is the Wiener part and $Z(t)$ is the jump part. Take a $d\times d$ matrix $A$ and consider solutions $X$ and $Y$ to the following equations

$$d X(t) = AX(t)\, dt + d L(t)$$

$$ dY(t) = AY(t)\, dt + dZ(t)+dbt$$

Put $U(t) = X(t) - Y(t)$. I was told that the process $U$ is idependent of $Y$ but I find it highly dubious. Where is the catch?

I think they are dependent but not sure how to prove it.

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  • $\begingroup$ Just to be sure: In the SDE for $(X_t)_{t \geq 0}$ it's really supposed to read $L(t) \, dt$ and not $\, dL_t$? $\endgroup$ – saz Jan 21 '18 at 16:31
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A straight-forward computation shows that the process $U_t := X_t-Y_t$ solves the SDE

$$dU_t = A U_t \, dt + dW_t.$$

The SDE has a unique strong solution which is adapted to the canonical filtration $\mathcal{F}_t^W$ of the driving Brownian motion. Consequently, $(U_t)_{t \geq 0}$ is $\mathcal{F}^W$-measurable, and therefore independent from the jump process $(Y_t)_{t \geq 0}$ (see e.g. this question).

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