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Start at $x=(x_1,\dots ,x_n)$ with all $x_i>0$, and let ${\tau }_i$ be the first zero of the $i$th coordinate. The orthant exit time is $T={\min }_i{\tau }_i$. The coordinate processes are independent, soHere the Brownian reflection principle gives ${\mathbb{P}}_a(\tau \le t)=2\mathbb{P}(B_t\le -a)$; $\Phi$ is the standard normal distribution function. Since $\Phi (z)-1/2=z/\sqrt{2\pi }+o(z)$ as $z\to 0$,The tail integral formula for expectation, ${\mathbb{E}}_{x}T={\int }_0^{\infty }{\mathbb{P}}_x(T>t)dt$, has a bounded integrand on $(0,1)$ and converges at infinity exactly when $n/2>1$. Therefore the Brownian exit-time expectation in an orthant givesThe divergence for $n=2$ is logarithmic; almost-sure finiteness of the exit time does not imply a finite expectation.