Let $d = 2$, and consider the domain $D = \mathbb{H}$, the upper half-plane. Let $W\_t = (X\_t, Y\_t)$. How do I see that for any $\theta \in \mathbb{R}$ and any $t \ge 0$, we have$$E^{(x, y)}\text{exp}\{i\theta X\_t - \theta Y\_t\} = e^{i\theta x - \theta y}?$$



Poisson kernel, $E^{(x, y)}\text{exp}\{i\theta X_t - \theta Y_t\} = e^{i\theta x - \theta y}$
Ignoring first the issue of boundary conditions, we note that by adding and subtracting an artificial $\frac{\theta^2}{2}t$ term in the exponential,


$\mathbb{E}\bigl[e^{i \theta X\_t - \theta Y\_t} \bigr] = \mathbb{E}\bigl[e^{i \theta X\_t + \frac{\theta^2}{2} t} \bigr]\mathbb{E}\bigl[e^{- \theta Y\_t - \frac{\theta^2}{2}t}\bigr] = e^{i \theta x}e^{- \theta y} $


as the two exponentials are Doleans-Dade exponentials. If we impose absorbing boundary conditions, the result stays the same as stopped martingales are martingales, too.


