Expected value in Probability (heads or tails)

Question

I was reading an article about probability. I saw a case in which you move a step forward if you get heads and a step backwards if you get tails. Since both these cases have equal probabilities, the average distance travelled $D=0$. But the expected value of the absolute value of the distance travelled $\langle |D|\rangle=N$ where $N$ is the number of trials. It was asserted that it was easier to use $D^2$ instead of$|D|$. But the expected value of $D^2$ remains the same (i.e.) $\langle D^2 \rangle $ =N. My question is shouldn't it be $N^2$. This question might be absurd but please educate me. Thank You.

Answer 1

Denote $+1$ as a step forward and $-1$ as a step backward. Then if $D$ is a variable which denotes the location, we can write as $D=X_1+\cdots+X_N$, where $X_i$s are independent variables which denotes the result of $i$th move, that is;

$P(X_i=1)=P(X_i=-1)=\frac{1}{2}$ (so that the cases respectively denote a step forward and a step backward on $i$th move)

Since $D^2=\sum_{i=1}^{N}X_i^2+2\sum_{i<j}X_iX_j$, by linearity of expectation we have $E(D^2)=E(\sum_{i=1}^{N}X_i^2+2\sum_{i<j}X_iX_j)=\sum_{i=1}^{N}E(X_i^2)+2\sum_{i<j}E(X_iX_j)$. Since $P(X_i^2=1)=1$ we have $E(X_i^2)=1$ and $E(X_iX_j)=E(X_i)E(X_j)=0$ as $X_i, X_j$ are independent for all $i<j$. Thus we have $E(D^2)=N$, as desired.

Also for me it seems that the $E(|D|)$ is not $N$; of course since there are $N$ moves the total distance of movement is $N$, but it does not fit with the definition of $|D|$.