 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 areBy linearity of expected value, $\boxed{⟨x(n)⟩=0}$. 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 displacementThese 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 $N$.