= Solution
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 $S_n=\sum_{j=1}^nV_j$, expand the <squared Euclidean norm>:
$$
\mathbb E|S_n|^2=\sum_{j=1}^n\mathbb E|V_j|^2+2\sum_{i<j}\mathbb E[V_i\cdot V_j].
$$
By <independence>, each cross term equals $\mathbb E[V_i]\cdot\mathbb E[V_j]=0$. Hence
$$
\boxed{\mathbb E|S_n|^2=n.}
$$
This <mean-square displacement of an isotropic planar random walk> grows linearly even though the <expected value> of the displacement vector remains zero.
Back to article page
