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Suppose I have a Levy process $Z_t\triangleq A*X_t$, where $A$ is a $d\times d$-matrix and $X_t$ is Levy also. Then under what time-change $T(t)$ is $$ Z_{T(t)}=X_t? $$

For example, if $X_t$ is (d=1)-dimensional I know that $$ T(t)=\frac{t}{A}, $$ works but how can I translate this to $d\geq 1$ dimensions?

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As the following example shows such a time change does, in general, not exist - even in dimension $d=1$.

Let $(X_t)_{t \geq 0}$ be a Poisson process with intensity $\lambda>0$ and set $Z_t := 2X_t$. By the very definition, $Z_t$ takes values only in $2 \mathbb{N}_0 =\{0,2,\ldots\}$, and therefore $$\mathbb{P}(\forall t \geq 0: Z_{T(t)}\in 2 \mathbb{N}_0)=1$$ for any time change $(T(t))_{t \geq 0}$. Since the Poisson process $X_t$ takes values in $\mathbb{R} \backslash (2 \mathbb{N}_0)$ with probability $>0$, this means that there can't exist a time change such that

$$Z_{T(t)} \stackrel{d}{=} X_t.$$

In particular, we can't find $(T(t))_{t \geq 0}$ such that $Z_{T(t)} = X_t$ almost surely.

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