Consider a point on a coordinate line. The starting coordinate of it is $0$, and it performs $n$ walks, each one shifts the point either by $1$ to the left or by $1$ to the right. Each direction is chosen with probability $p=\frac{1}{2}$. Prove that the mean coordinate modulo is $O(\sqrt n)$.
My approach is as follows: let $\xi$ be the coordinate modulo of the point after $n$ walks. $$P(\xi=a)=\begin{cases}0, & n \text{ and } a \text{ are of different parity or }a>n \\ 2\frac{1}{2^n}\binom{n}{a+\frac{n-a}{2}}, & \text{ otherwise}\end{cases}$$ Explanation for $2\frac{1}{2^n}\binom{n}{a+\frac{n-a}{2}}$: $\frac{1}{2^n}$ is the probability of a fixed sequence of $n$ walks happening, for $a$ to be the final coordinate there needs to be $a+\frac{n-a}{2}$ walks left (or right), therefore the binomial is the amount of walk sequences that result in the final coordinate being $a$.
$$\mathbb E\xi=\sum\limits_{a=0}^n aP(\xi=a)$$
With both even and uneven $n$ Wolfram shows that $\mathbb E\xi=O(\sqrt n)$, but I'm looking for a more elegeant proof.