See [here](https://mathoverflow.net/questions/221823/poisson-kernel-ex-y-textexp-i-theta-x-t-theta-y-t-ei-theta-x). 



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


Does it follow that for $x$, $\theta \in \mathbb{R}$ and $y > 0$,$$E^{(x, y)} e^{i\theta X\_\tau} = e^{i\theta x - |\theta|y}?$$



Poisson kernel, expectation, an absolute value comes in
Yes, if $\tau$ is defined as $\min \{t\ge 0: Y\_t=0 \}$. Condition on $\tau =t$ and use that $Ee^{i\theta X\_t}=e^{i\theta x-\theta^2 t/2}$, as explained in the answer to the linked question. Thus
$$
Ee^{i\theta X\_{\tau}} = e^{i\theta x} E e^{-\theta^2\tau/2} .
$$
The distribution of $\tau$ and $Ee^{-b\tau}$ are discussed for example [in this MSE question.](https://math.stackexchange.com/questions/133628/distribution-of-hitting-time-of-line-by-brownian-motion) In particular, we have that $Ee^{-b\tau}=e^{-y\sqrt{2b}}$, and this gives the desired result.


