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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.

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    $\begingroup$ For $T=\inf\{t; B_t = a\}$ it holds that $B_T = a$ (i.e. $B_T$ is a deterministic constant), and therefore the independence of $T$ and $B_T$ is somewhat trivial, right? $\endgroup$ – saz Apr 17 at 6:38
  • $\begingroup$ @saz Okay yeah that makes sense, I still am not sure how to deal with the incorrect factor I want, perhaps a substitution would work, like $\sqrt{2h} = k, h >0$ ? $\endgroup$ – all.over Apr 17 at 14:08
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    $\begingroup$ well, yes, perhaps? Why not give it a try? $\endgroup$ – saz Apr 17 at 14:09
  • $\begingroup$ @saz one thing I realized I missed though insuring I can use optional sampling. $T_0$ doesn't seem to be bounded need to show $E|B_{T_0}| < \infty, P(T_0 < \infty ) = 1$ and $lim_n E[|B_{T_0}|1_{T_0 > n}] = 0$. But from the fact 1-dimensional BM is recurrent, it has continuous sample paths and using intermediate value property and $B_{T_0}$ is deterministic it seems to follow. $\endgroup$ – all.over Apr 17 at 14:35
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    $\begingroup$ Mind that $T_0$ is no integrable. Apply the optional stopping theorem to the bounded stopping time $T_0 \wedge k$ and then let $k \to \infty$ using dominated convergence. $\endgroup$ – saz Apr 17 at 14:38
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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} $

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