= Solution
Let $\xi_j$ be the displacement in step $j$, so the <random walk> position is $x(n)=\sum_{j=1}^n\xi_j$. The single-step <expected value> and second <moment> are
$$
\mathbb E[\xi_j]=\tfrac13\delta-\tfrac13\delta+\tfrac13(0)=0,\qquad
\mathbb E[\xi_j^2]=\tfrac13\delta^2+\tfrac13\delta^2=\tfrac23\delta^2.
$$
By linearity of <expected value>, \b[$\boxed{\langle x(n)\rangle=0}$.] The steps are <independent random variables>, so for $i\ne j$, $\mathbb E[\xi_i\xi_j]=\mathbb E[\xi_i]\mathbb E[\xi_j]=0$. Expanding the square gives the <mean squared displacement>
$$
\boxed{\langle x^2(n)\rangle=\sum_{j=1}^n\mathbb E[\xi_j^2]
+2\sum_{i<j}\mathbb E[\xi_i\xi_j]=\frac23n\delta^2}.
$$
These are ensemble <expected values>; the empirical average over a finite number $N$ of independent cells fluctuates about them. The expected <mean squared displacement> itself does not depend on $N$.
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