 Solution

There is a source error to resolve first. For a drag coefficient defined by $F_{\mathrm{drag}}=-\xi v$, the velocity relaxation time is $\tau =m/\xi$. The mass and drag coefficient in the PDF's exponential are interchanged: $\xi t/m$ is dimensionless, whereas $mt/\xi$ is not. The consistent mean relaxation law is therefore $⟨v(t)⟩=v_0{e}^{-t/\tau }$.

A mean relaxation law alone does not determine a variance. Use the usual thermal Langevin equation, with a bath at temperature $T$ and the fluctuation-dissipation relation for a Langevin particle:Here the calculation is for one component in Cartesian coordinates. With fixed initial position and velocity, integration gives the mean displacementIntegrating the stochastic part of the velocity once more gives the centered displacementThe delta correlation reduces its variance to a single integral:This is the integrated Ornstein-Uhlenbeck displacement for a prescribed initial velocity. Its short-time variance is $2D{t}^3/(3{\tau }^2)+O(t^4)$; at long times it is $2Dt+O(1)$. Thus the diffusion coefficient isIf the requested mean-squared fluctuation instead refers to the conventional equilibrium mean squared displacement, the initial velocity is also averaged over its thermal distribution, with $⟨v_0⟩=0$ and $⟨v_0^2⟩=k_{B}T/m$. Its contribution adds $D\tau (1-e^{-t/\tau }{)}^2$, givingThis has the ballistic limit $(k_{B}T/m)t^2$ and the same diffusive limit $2Dt$. For fixed nonzero $v_0$, the uncentered mean square instead equals the first boxed variance plus $v_0^2{\tau }^2(1-e^{-t/\tau }{)}^2$. In three independent spatial dimensions, the centered variances add, and the long-time result is $⟨|\mathbf{r}-⟨\mathbf{r}⟩{|}^2⟩\sim 6Dt$. Stating the initial-velocity convention is necessary to distinguish these finite-time formulas.