$\chi^{2}$ probability for 6 degrees of freedom is calculated as:

$$\chi^{2} =A+B$$

$$A=\frac{(U_{o} - U_{g})^{2}}{\sigma^{2}_{U}} + \frac{(V_{o} - V_{g})^{2}}{\sigma^{2}_{V}} + \frac{(W_{o} - W_{g})^{2}}{\sigma^{2}_{W}}$$

$$\sigma_{i}=\sqrt{\sigma_{i,o}^{2}+\sigma_{i,g}^{2}+\sigma_{i,d}^{2}}$$

where $i$ is indexed as $U,V$ or $W$, $o$ is the component for 2M0355, $g$ is the component of the group, and $d$ is the intrinsic 1-D $i$-velocity dispersion of the group.

$$B=\frac{(X_{o} - X_{g})^2}{\Delta_{X}^{2}} + \frac{(Y_{o} - Y_{g})^{2}}{\Delta_{Y}^{2}} + \frac{(Z_{o} - Z_{g})^2}{\Delta_{Z}^{2}}$$

$$\Delta_{j} = \sqrt{\Delta_{j,o}^{2} + \Delta_{j,g}^{2}}$$

where $j$ is indexed as $X,Y$, or $Z$, $\Delta_{j}$ is defined as the 1$\sigma$ dispersion in the Galactic cartesian coordinates; $o$ is the component for 2M0355, $g$ is the component for the group (we ignore the uncertainties in the group centroids which are negligible compared to the 1$\sigma$ dispersions).

Using this method, we estimate that 73% of AB Doradus members would have velocities and positions more discrepant than that for 2M0355, while only 0.06% of $\beta$ Pictoris members would have more discrepant values. Approximately 99.9% of field stars would have velocities and positions more discrepant than that of 2M0355, although this is likely skewed by the fact that the field star centroid (as well as our source) is so close to the Sun.

The $\chi^{2}$ probabilities for the 