Solution
= Solution
For an isotropic three-dimensional ideal walk, each Cartesian component contributes $Nb^2/3$ to the mean square. Keeping two components gives $\langle R_{\rm proj}^2\rangle=2Nb^2/3$. The <planar projection of an isotropic random vector> therefore has $R_{{\rm proj},{\rm rms}}=b\sqrt{2N/3}$, smaller than the unprojected planar ideal walk by $\sqrt{2/3}$. The corresponding two-dimensional Gaussian mean distance has the same ratio.