The variance is not necessarily 0. Consider first $$\frac{\partial \langle x^2\rangle}{\partial t} = Dx^2\frac{\partial w(x,t)}{\partial x}|_{x=\pm\infty}- 2D\int_{-\infty}^{\infty}x\frac{\partial w(x,t)}{\partial x}dx,$$ where we have performed an integration by parts as we did for $\frac{\partial \langle x\rangle}{\partial t}$.