A question about random walk in 1 dimension

Question

For a simple random walk problem in 1 D, the expected position of the particle in $n$ step is $E(X_n)=n(p-q)$ so the distance from origin should be $=E(X_n)$ but according to Mean distance from origin after $N$ equal steps of Random-Walk in a $d$-dimensional space. the expected distance from the origin is different. So why is it?

Answer 1

Position can be negative, distance is the absolute value of position.
So the expected distance is the expected value of the absolute value of the position, not the absolute value of the expected position.

Answer 2

Once the confusion between $\mathbb E(|X_n|)$ (the mean distance from the origin) and $|\mathbb E(X_n)|$ (the absolute value of the mean position) is cleared, it seems a second kind of confusion might explain your trouble. There are two very different cases.

Either $p\ne q$, then $X_n\sim n(p-q)$ almost surely and both $|\mathbb E(X_n)|$ and $\mathbb E(|X_n|)$ are asymptotically $n|p-q|$ (then one says the walk is ballistic). Or $p=q=\frac12$, then $X_n$ behaves like $\sqrt{n}$ times a standard normal random variable and $|\mathbb E(X_n)|=0$ for every $n$ while $\mathbb E(|X_n|)\sim\sqrt{n}$ (then one says the walk is diffusive).

A more usual formulation of the diffusive property uses the root mean square displacement $\sqrt{\mathbb E(X_n^2)}$ and states that $\mathbb E(X_n^2)\sim n$. The nearly universal use of the root mean square displacement rather than the mean displacement is mainly due to computational feasibility, the computation of the mean square displacement being often easy while the computation of the mean displacement is often impossible.