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Let the independent unit-step vectors be $V_j=(\cos {\Theta }_j,\sin {\Theta }_j)$. The uniform distribution of each angle gives $\mathbb{E}[V_j]=0$, while $|V_j{|}^2=1$. With $S_n={\sum }_{j=1}^n{V}_j$, expand the squared Euclidean norm:By independence, each cross term equals $\mathbb{E}[V_i]\cdot \mathbb{E}[V_j]=0$. HenceThis mean-square displacement of an isotropic planar random walk grows linearly even though the expected value of the displacement vector remains zero.

For a real vector $v$, the square of its Euclidean norm is $\Vert v{\Vert }_2^2=v\cdot v={\sum }_j{v}_j^2$. Unlike the norm itself it is a quadratic function, so expanding sums directly exposes diagonal and cross terms. This is useful for the mean-square displacement of an isotropic planar random walk.