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Let $W_t$ be a 1-dimensional Brownian Motion. For $x>0$, we define: $$\tau_{x} = inf \{ t \geq 0; |W_t| = x\}$$ Compute $E[e^{-s\tau_x}]$ and prove that $\tau_x$ is equal in distribution to $x^2\tau_{1}$.

I would like some hint to get started. I am unsure how to prove equality in distribution.

To compute the expectation, I believe I need to first determine the distribution of $\tau_x$. I believe this is very complicated.

Attempt: To avoid cumbersome notation, I will let $\tau = \tau_{x}$ for some $x>0$.

Consider exponential martingale $Z_t = e^{\lambda W_t - \lambda ^2t/2 } + e^{-\lambda W_t - \lambda ^2t/2}$.

By optional sampling theorem, $E[Z_0] = E[Z_{\tau}]$.

Hence,

$$2 = E[e^{- \lambda ^2 \tau /2}(e^{\lambda W_{\tau}} + e^{-\lambda W_{\tau} })] = E[E[e^{- \lambda ^2 \tau /2}(e^{\lambda W_{\tau}} + e^{-\lambda W_{\tau} })|\tau]] = E[e^{- \lambda ^2 \tau /2}E[(e^{\lambda W_{\tau}} + e^{-\lambda W_{\tau} })|\tau]]$$

By a symmetrical argument for fixed $\tau$, we have that: $$E[e^{\lambda W_{\tau}} + e^{-\lambda W_{\tau} }|\tau ] = 0.5 (e^{\lambda x} + e^{-\lambda x}) + 0.5 (e^{\lambda (-x)} + e^{-\lambda (-x)}) = e^{\lambda x} + e^{-\lambda x}$$ Combining, $$2= E[e^{- \lambda ^2 \tau /2} (e^{\lambda x} + e^{-\lambda x})]= (e^{\lambda x} + e^{-\lambda x})E[e^{- \lambda ^2 \tau /2}]$$

So, $$E[e^{- \lambda ^2 \tau /2}] = \dfrac{2}{(e^{\lambda x} + e^{-\lambda x})}$$

And let $s=\lambda^2/2$ to find that: $$E[e^{- s \tau}] = \dfrac{2}{(e^{\sqrt{2s}x} + e^{-\sqrt{2s}x})}$$

Reverting to original notation, we have proved $\forall x>0$ that $$E[e^{- s \tau_{x}}] = \dfrac{2}{(e^{\sqrt{2s}x} + e^{-\sqrt{2s}x})}$$

Kerry
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    For finding the Laplace transform of $\tau_x$, you can consider the exponential martingale $Z_t = \exp(\lambda W_t - \frac{\lambda^2}{2}t) + \exp(-\lambda W_t - \frac{\lambda^2}{2}t) $. – Nocturne Nov 01 '15 at 17:03
  • @Nocturne What is the connection between $Z_t$ and $\tau_x$? – Kerry Nov 01 '15 at 17:18
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    Do you know the Optional Stopping Theorem? – John Dawkins Nov 01 '15 at 17:40
  • @JohnDawkins I know the statement, and can understand (but not reproduce) the proof. If 1 of 3 conditions hold, then $E[X_{\tau}] = E[X_0]$. I don't see how this theorem applies here. – Kerry Nov 01 '15 at 17:41
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    Hint: Apply the Optional Stopping Theorem to the martingsale $Z_t$ at the stopping time $\tau_x$ and see what develops. – John Dawkins Nov 01 '15 at 17:46
  • @JohnDawkins I believe I solved it. How does my solution look? – Kerry Nov 01 '15 at 19:00
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    Looks okay, but there's no need to condition on $\tau$. As you observe, by symmetry $e^{-\lambda W_\tau}+e^{\lambda W_\tau}=e^{-\lambda x}+e^{\lambda x}$. – John Dawkins Nov 01 '15 at 19:06
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    @Ryan Also, you have to justify the use of the Optional Stopping Theorem. $E[X_0] \neq E[X_{\tau}]$ in general. – Nocturne Nov 01 '15 at 19:08
  • @JohnDawkins I thought that symmetrical argument was only true for a fixed $\tau$. – Kerry Nov 01 '15 at 19:09
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    @Ryan Your conditioning on $\tau$ looks a bit confusing. The best would be to say that $W_{\tau}=x$ almost surely by (right) continuity of the trajectories of $\mathbf{W}=(W_t)_{t \geq 0}$. – Nocturne Nov 01 '15 at 19:16
  • @Nocturne Looking at the theorem statement, https://en.wikipedia.org/wiki/Optional_stopping_theorem, I observe that condition 1 does not hold since $x$ is not necessarily finite implies that $\tau$ is not a.s. bounded. I would like to prove condition 2, i.e. showing $E[\tau] < \infty$, but wouldn't I need the distribution of $\tau$. – Kerry Nov 01 '15 at 19:19
  • @Ryan $E[\tau]=\infty$ actually. You can use condition 3. The theorem is stated for a discrete time martingale, but it remains true in your case since the exponential martingale is (right) continuous. – Nocturne Nov 01 '15 at 19:25
  • @Nocturne - Thank you! I was also wondering if you knew how to prove the other part of this question, i.e. showing that $\tau_x$ is equal in distribution to $x^2 \tau_1$. Does this follow from computing the Laplace Transform? I have never shown equality in distribution. How does one do this in general? – Kerry Nov 01 '15 at 19:29
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    @Ryan This fact is a consequence of the Brownian scaling property. You can see it on the Laplace transforms. The Laplace transform of $x^2\tau_1$ for a fixed $x$ is just the Laplace transform of $\tau_1$ evaluated at $\lambda x^2$ which is the same as the Laplace transform of $\tau_x$ since $x>0$ (we are using the fact the the Laplace transform characterizes the distribution, i.e. two random variables having the same Laplace transform are equal in distribution). – Nocturne Nov 01 '15 at 19:41
  • @Nocturne - THANK YOU SO MUCH! – Kerry Nov 01 '15 at 20:22
  • @Nocturne - One last question - What is the connection between the characteristic function and the Laplace transform? One can state that two distributions are the same if their characteristic functions agree. One could also say the same if their Laplace transforms agree. Are both these facts equivalent? – Kerry Nov 01 '15 at 20:26

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