The mean-squared dislacement of a particle is defined simply as, \[\rho(t) = \langle r^2(t) \rangle = \int c(r) p(r,t) dr,\] where \(c(r) = 1\), \(2 \pi r\), and \(4 \pi r^2\) in 1, 2, and 3 dimensions, respectively. This evaluates to \(\langle r^2(t) \rangle = 2dDt\), where \(d\) is the dimension (1, 2, or 3).
One can also compute the variance of the MSD, as \[\text{var}(\rho) = \langle \rho^2(t) \rangle - \left(\langle \rho(t) \rangle\right)^2.\] This can be evaluated as,
\begin{align}
1D:& 2\,(2Dt)^2 = 8D^2t^2\\
2D:& \dfrac{2}{2} (4Dt)^2 = 16 D^2 t^2\\
3D:& \dfrac{2}{3} (6Dt)^2 = 24 D^2 t^2
\end{align}
This can be simplified into a common expression as, \[\text{var}(\rho) = \dfrac{2}{d} \rho^2\]
One can also compute the variance of the MSD, as \[\text{var}(\rho) = \langle \rho^2(t) \rangle - \left(\langle \rho(t) \rangle\right)^2.\] This can be evaluated as,
\begin{align}
1D:& 2\,(2Dt)^2 = 8D^2t^2\\
2D:& \dfrac{2}{2} (4Dt)^2 = 16 D^2 t^2\\
3D:& \dfrac{2}{3} (6Dt)^2 = 24 D^2 t^2
\end{align}
This can be simplified into a common expression as, \[\text{var}(\rho) = \dfrac{2}{d} \rho^2\]
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