Root Mean Square Displacement of Diffusion and Radial Diffusion Function

I read, that for normal diffusion the root mean square displacement $sqrt{langle x^2(t)rangle}$ (for particles at the origin) can be interpreted as the mean distance the particles have with respect to the origin at time $t$. For normal diffusion, the MSD is $$langle x^2rangle=2Dt$$ and therefore the RMSD equals $$sqrt{langle x^2rangle}=sqrt{2Dt}.$$ But if I say, this is more or less the expectation value of the origin-to-endpoint distance of a particle diffusing, couldn’t I just compute it with a "radialized" diffusion function. That is, with the original diffusion function $$p(vec{X})=frac{1}{(4pi Dt)^{frac{d}{2}}}mathrm{e}^{,-lVert vec{X}rVert^2/4Dt}$$ and transforming this: $$iiint_{mathbb{R}^d}d^dX,p(vec{X})=int dOmega_dint_0^infty dR,R^{d-1}p(R)$$ $$Rightarrow p(R)=S_dfrac{R^{d-1}}{(4pi Dt)^{frac{d}{2}}}mathrm{e}^{,-R^2/4Dt}$$ where $S_d$ is the surface of the $d$ dimensional unit sphere. Computing $langle Rrangle$ has the same meaning then as RMSD: $boldsymbol{langle Rrangle}$ and RMSD are both the expected distance from the origin of a particle diffusion at time $t$. But if I compute $langle Rrangle$ e.g. in one dimension I get $$langle Rrangle=frac{2}{sqrt{pi}}sqrt{Dt},$$ both are proportional to $sqrt{t}$ but the constants are different. Is my understanding of the RMSD not right or is there a deeper meaning, e.g. both are under the respective conditions the mean distance to the origin?