Given a Brownian motion $X_t$, we know (by definition) that the density is $f(x,t)=\frac{1}{\sqrt{2 \pi t}} e^{-\frac{x^2}{2 t}}$
Now, if our Brownian motion starts from a certain point $y$ instead of $0$, we consider the conditional density
$f(x,t|y)=\frac{1}{\sqrt{2 \pi t}} e^{\frac{-(x-y)^2}{2 t}}$
Then, we can observe that (with a little abuse of notation)
$f(x,t|y)=f(x-y,t)$ or, omitting the time,
$f(x|y)=f(x-y)$.
My question is: Is this true for more general processes, such as Lévy processes? If so, why?
I have the impression that this should be true for any process with independent increments, but I actually don't know how to prove it. Somehow related to this, I also read the following: if we have a process with independent increments, such as a Lévy process, then the (conditional) characteristic function can be written as
$\psi(u|x)=e^{i u x} \phi(u)$.
The two statements are connected, aren't they ?
Could you provide me with some insights into this topic ?
Thank you very much for your help!! :)