I have been doing a computation but I am making a mistake somewhere and cannot figure out where. The question I have is: where am I making my mistake?
I will run through my reasoning:
Consider standard Brownian motion $((W_x)_t,(W_y)_t)$ in $\mathbb{R}^2$ and consider the barrier that is the square centered at $0$ of side length $d$. We want to compute the expected first hitting time. My idea was to approximate by the $L^{2p}$ balls of radius $\frac{d}{2}$.
Since $\mathbb{E}[(W_x)_t^{2p}] = \frac{1}{\sqrt{\pi}} t^{p} \cdot 2^p \cdot \Gamma(p+\frac{1}{2})$, $$(W_x)_t^{2p} + (W_y)_t^{2p} - \frac{2}{\sqrt{\pi}} t^{p} \cdot 2^p \cdot \Gamma(p + \frac{1}{2})$$ is a martingale, and thus, optional stopping theorem gives us that
$$\mathbb{E}[(\tau_p)^p] = \frac{\sqrt{\pi}}{2} \frac{(d/2)^{2p}}{2^p \Gamma(p+\frac{1}{2})}$$
where $\tau_p$ is the stopping time to hit the $L^p$-ball of radius $d/2$.
By Jensen's inequality, $\mathbb{E}[\tau_p]^p \le \mathbb{E}[(\tau_p)^p]$ and therefore, we have that:
$$\mathbb{E}[(\tau_p)] \le (\frac{\sqrt{\pi}}{2})^{\frac{1}{p}} \frac{(d/2)^{2}}{2 (\Gamma(p+\frac{1}{2}))^\frac{1}{p}}$$
We are interested in taking $p$ to infinity to approximate the square by the $L^p$ balls, and here is the trouble: $\Gamma(p+\frac{1}{2})^\frac{1}{p}$ diverges to infinity, so I get that $\mathbb{E}[\tau] \le 0$ which is clearly false.
What went wrong here? I am sure I am doing something silly, but i cannot figure out what!