Distribution of $\tau/x^2$ where $\tau$ is first hitting time

Question

I am considering the hitting time $\tau=\inf\{t:|B_t|\geq x\}$. I am told to consider the martingale $\cosh(\lambda B_t)\textrm{e}^{-\frac12\lambda^2t}$. I know that by letting $\lambda=\sqrt{2\theta}$, we can prove that the moment generating function of $\tau$ is $\mathbb{E}(\textrm{e}^{-\theta\tau})=\frac1{\cosh\sqrt{2\theta}x}$. The follow-up question from this is to show that $\tau/x^2$ therefore has a distribution independent of $x$, but I am not sure how to go about this. Am I supposed to be differentiating the MGF? All help is appreciated, thanks!

Answer 1

By your equation, $$E(e^{-\theta(\tau/x^2)}) = E(e^{-(\theta/x^2)\tau})= \frac{1}{\cosh(\sqrt{2\theta/x^2}x)}=\frac{1}{\cosh(\sqrt{2\theta})},$$ which is independent of $x.$