Let $D_N$ be the expected average of the displacement of a random walk on $\mathbb Z$ from the origin, where $N$ is the number of steps, each of which is either $-1$ or $1$.
We take the definition of the expected average of a variable $A$ to be $\sum_{i=1}^N a_ip_i$ where $A$ can take any value $a_1,a_2,\ldots,a_N$ with probability $p_1,p_2,\ldots,p_N$.
In one step, ${D_1}^2 = \frac121^2+\frac12(-1)^2=1$.
If we know $D_{N-1}$, then ${D_N}^2 = \frac12(D_{N-1}+1)^2 + \frac12(D_{N-1}-1)^2 = {D_{N-1}}^2 + 1$.
This means that ${D_N}^2 = N$ or $D_N = \sqrt N$.
However, if we look at the distribution of end paths after $N$ steps, we can represent the expected average $D_N = \frac{\binom{0}{N}d_1+\binom{1}{N}d_2 + \cdots + \binom{N}{N}d_N}{2^N}$ where $d_1,d_2,\ldots,d_{N}$ are the actual displacements from the origin.
ie:
$$D_1 = \frac{1}{2}\left[\binom011+\binom111\right] = 0.5 \\ D_2 = \frac{1}{4}\left[\binom022+\binom120+\binom222\right] = 1.0 \\ D_3 = \frac{1}{8}\left[\binom033+\binom131+\binom231+\binom333\right] = 1.5 \\ \vdots$$
All of which are different from $\sqrt N$.
I know that $\sqrt N$ is really only an approximation of $D_N$ and not an exact value, so where does the proof that $D_N$ is equal to $\sqrt N$ go wrong?