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Independent planar unit steps with uniformly distributed direction have zero vector expected value. The squared Euclidean norm of their sum has expectation $n$, because each diagonal term is one and all cross terms vanish by independence. More generally, zero-mean independent steps with common second moment $s^2$ give mean-square displacement $n{s}^2$; nonzero drift adds a quadratic contribution.

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.