 Mean displacement magnitude of three-dimensional Brownian motion

For isotropic Brownian motion started at zero with independent coordinate variances $2Dt$, the Gaussian distribution has density $(4\pi Dt{)}^{-3/2}{e}^{-|x{|}^2/(4Dt)}$. Integrating the radius with spherical volume element $4\pi r^{2}dr$ gives $\mathbb{E}|X_t|=4\sqrt{Dt/\pi }$. The root mean square displacement is instead $\sqrt{6Dt}$, and the mean vector is zero. None of these is the microscopic path length.