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 $$E[e^{- s \tau_{x}}] = \dfrac{2}{(e^{\sqrt{2s}x} + e^{-\sqrt{2s}x})}$$