# Independence between Brownian motion and hitting time.

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.

• 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? – saz Apr 17 at 6:38
• @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 $\sqrt{2h} = k, h >0$ ? – all.over Apr 17 at 14:08
• well, yes, perhaps? Why not give it a try? – saz Apr 17 at 14:09
• @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. – all.over Apr 17 at 14:35
• 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. – saz Apr 17 at 14:38

## 1 Answer

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

$$\displaystyle E[\exp(-hT_0)] = \frac{e^{\sqrt{2h}x}}{E[e^{\sqrt{2h}B_{T_0}}]} = e^{\sqrt{2h}x}$$