A Wald's type problem $\mathbb Ee^{\mu\sigma_a}$ about stopping time

Question

Let $B$ be a standard Brownian motion. For any $a>0$ define the stopping time $$\sigma_a=\inf\{t\geqslant0:B_t-t=-a\}.$$ Show that $\mathbb Ee^{\mu\sigma_a}=\infty$ for all $\mu>\frac12$.

I can only handle the case when $\mu\leqslant0$: Take $\alpha=\sqrt{1-2\mu}-1\geqslant0$ and note that $\alpha+\frac{\alpha^2}2=-\mu$. By Ito's formula we know $M_t:=e^{-\alpha B_t-\frac{\alpha^2}2t}$ is a martingale ... By dominated convergence theorem ($M_{\sigma_a\wedge\cdot}\leqslant e^{a\alpha}$ since $\alpha\geqslant0$) we have $\mathbb EM_{\sigma_a}=1$. At last note that $B_{\sigma_a}=\sigma_a-a$, we can conclude that $\mathbb Ee^{\mu\sigma_a}=e^{-(\sqrt{1-2\mu}-1)a}$.

But we can not take such $\alpha$ because $\alpha+\frac{\alpha^2}2=-\mu$ has no real root when $\mu>\frac12$. Even though we take a complex root, we can still not apply dominated convergence theorem since the real part of this root is negative.

Please feel free to discuss.

Answer 1

I think I can prove it now.

Since we have obtained $\mathbb Ee^{-s\sigma_a}=e^{(1-\sqrt{1+2s}~)a}~$ for all $s\geqslant0$, we can use inverse Laplace transform to calculate the density function of $\sigma_a$: \begin{equation} f_{\sigma_a}(t)=\frac{a\exp{\{-\frac{a^2}{2t}+a-\frac{t}{2}}\}}{\sqrt{2\pi t^3}},t>0. \end{equation} Thus we can see that $$\mathbb{E}e^{\mu\sigma_a}=\int_{0}^\infty\frac{a\exp{\{-\frac{a^2}{2t}+a+(\mu-\frac12)t}\}}{\sqrt{2\pi t^3}}dt<\infty$$ if and only if $\mu\leqslant\frac12$.