For any given random walker, we cannot say much definite about its path. We can however look at an ensemble of walkers and study their average behavior.

Random walks obey the diffusion equation

\[ \langle x^2 \rangle = 2Dt \]

The expectation value of the position squared has a linear relationship with the time that the walkers have been walking, times a proportionality constant.

For the coin flip scenario, the walkers can only move two ways, \(\pm 1\). Let \(x_n\) be the position of a single walker after \(n\) steps at a given \(\Delta t\), which will here be considered 1 (one "time step", without physical units). The final position of the walker will be

\[ x_n = \sum_{i=1}^N s_i \qquad s_i = \pm 1 \]

and the final value for \(x^2_n\) is
\[ x^2_n = \sum_{j=1}^N\sum_{i=1}^N (s_i s_j) \qquad s_i, s_j = \pm 1 \]

What can we say about this? For any one walk, not much. But if we consider the expectation value of this operator, we can get much more mileage.

\[ \langle x^2_n \rangle = \bigg \langle \begin{matrix} s_1 s_1 & + \quad s_1 s_2 & + \quad ... & + \quad s_1 s_n \\ s_2 s_1 & + \quad s_2 s_2 & + \quad ... & + \quad s_2 s_n \\ ... & + \quad ... & + \quad ... & + \quad ... \\ s_n s_1 & + \quad s_n s_2 & + \quad ... & + \quad s_n s_n \end{matrix} \bigg \rangle \]

The quantity \(s_i s_j\) has four outcomes:

\[ s_i s_j = \left\{ \begin{array}{cl} 1 & : s_i = s_j \qquad (+1)(+1) \& (-1)(-1) \\ -1 & : \ s_i \neq s_j \qquad (+1)(-1) \& (-1)(+1) \end{array} \right. \]

Each state has multiplicity two; if all possible arrangements are equally probable, then the average over them will be \((2 (+1) + 2(-1)) / 4 = 0\). Since \((+1)^2\) and \((-1)^2\) are both 1, we can succinctly summarize the expectation values of the above "matrix".

\[ \langle s_i s_j \rangle = \left\{ \begin{array}{cl} i \neq j & : 0 \\ i = j & : 1 \end{array} \right. \]

The only terms that will survive the expectation value are the diagonal terms of the "matrix" above; all off-diagonal terms will average to zero.

Because the surviving terms are all unity, and because they are on the diagonal, we know there are \(N\) of these terms and we have a clean result for our diffusion equation.

\[ \langle x^2_n \rangle = \sum_{j=1}^N\sum_{i=1}^N \langle s_i s_j \rangle = N = \frac{t}{\Delta t} = a t \qquad a \equiv 2D \\ \boxed{\langle x^2_n \rangle = 2 D t} \]