If given a standard one dimensional Brownian motion $B_t$ and stopping time $T = \inf\{ t : B_t = |a|, a \in \mathbb{R}\}$
We will have independence between $B_T$ and $T$ as $P(B_T = a) = \frac{1}{2} = P(B_T = -a) $
But if we change $T = \inf\{t : B_t = a, a \in \mathbb{R} \}$ Then it seems the independence would change.
I am trying to solve:
If $B_0 = x > 0$ Use the Martingale (By Ito's formula) $\exp(kB_t - \frac{k^2}{2}t)$ to find $Ee^{-kT_0}$ where $T_0 = \inf\{t : B_t = 0\}, k >0$.
If I was working with the first case above then symmetry helps a lot but now I have:
$e^x = E[\exp(kB_{T_0} - \frac{k^2}{2}T_0)] \,\,(?)=(?) \,\, E[\exp(kB_{T_0})]E[\exp(-\frac{k^2}{2}T_0)]$ which helps a bit however still have $-\frac{k^2}{2}$ as a factor not $-k$
NOTE: $(?) = (?)$ denotes whether this is true or not, couldn't get the question mark over the equal sign to work.
If we let $k = \sqrt{2h}, h >0 $ and the martingale given above with optional sampling theorem on $T_0 \wedge k$ which is bounded per the suggestion by Saz, above. we have
$e^{kx} = E[\exp(kB_{T_0} - \frac{k^2}{2}T_0)] = E[\exp(\sqrt{2h}B_{T_0})]E[\exp(-hT_0)]$
Which implies that
$\displaystyle E[\exp(-hT_0)] = \frac{e^{\sqrt{2h}x}}{E[e^{\sqrt{2h}B_{T_0}}]} = e^{\sqrt{2h}x} $