Dispersion of rightward steps in 1-D random walks

[f3sicA_A]

Homework Statement

What is the relation between the dispersion of the rightward steps ($\overline{(\Delta n_1)^2}$) and the displacement of the random walker in a simple 1-D random walk?

Relevant Equations

$$\overline{(\Delta n_1)^2}=Npq$$

$$m=n_1-n_2$$

I am currently going through the first chapter of Fundamentals of Statistical and Thermal Physics by F. Reif which is on random walks, various mean values, etc. and specifically unit 1.4 mentions "The quantity $\overline{(\Delta n_1)^2}$ is quadratic in the displacement." This is with reference to a simple 1-D random walk where the probability of going to the right is $p$ and the probability of going to the left is $q\equiv1-p$, where $n_1$ is the number of steps taken to the right and $n_2$ is the number of steps taken to the left, such that $n_1+n_2\equiv N$ is the total number of steps taken.

I am not able to verify as to why $\overline{(\Delta n_1)^2}$ must be quadratic in displacement, $m=n_1-n_2$ (the rightward displacement in units of step length $l$), and I am not sure how to approach this problem. One thing we have shown just before this statement is:

$$\overline{(\Delta n_1)^2}\equiv\overline{(n_1-\overline{n_1})^2}$$
$$\implies\overline{(\Delta n_1)^2}=\overline{n_1^2-2n_1\overline{n_1}+(\overline{n_1})^2}$$
$$=\overline{n_1^2}-\overline{n_1}^2$$

where we have:

$$\overline{n_1}=\sum_{n_1=1}^{N}P(n_1)n_1$$
$$=\sum_{n_1=1}^{N}\frac{N!}{n_1!(N-n_1)!}p^{n_1}q^{N-n_1}n_1$$
here we have $n_1p^{n_1}=p\partial {\left(p^{n_1}\right)}/\partial p$, so:

$$\overline{n_1}=\sum_{n_1=1}^{N}\frac{N!}{n_1!(N-n_1)!}\left[p\frac{\partial\left(p^{n_1}\right)}{\partial p}\right]q^{N-n_1}$$
$$=p\frac{\partial}{\partial p}\sum_{n_1=1}^{N}\frac{N!}{n_1!(N-n_1)!}p^{n_1}q^{N-n_1}$$
$$=p\frac{\partial(p+q)^{N}}{\partial p}$$
$$=pN(p+q)^{N-1}$$

and since we have $p+q=1$, in our case, $\overline{n_1}=Np$. Similarly, we can follow a similar procedure to show that $\overline{n_1^2}=\overline{n_1}^2+Npq$, and hence showing that:

$$\overline{(\Delta n_1)^2}=Npq$$

However, this does not convey to me the idea that $\overline{(\Delta n_1)^2}$ must be quadratic in $m$, how should I start attempting this problem?

 

[haruspex]

"The quantity $\overline{(\Delta n_1)^2}$ is quadratic in the displacement."

I don’t even know what that statement means.
$\overline{(\Delta n_1)^2}$ is a characteristic of the distribution, whereas the displacement, m, is a single value of a random variable.
Is there any ensuing text that clarifies it?

 

[f3sicA_A]

Is there any ensuing text that clarifies it?

This is the exact paragraph from unit 1.4 of the book:

The quantity $\overline{(\Delta n_1)^2}$ is quadratic in the displacement. Its square root, i.e., the rms (root-mean-square) deviation $\Delta^*n_1\equiv[\overline{(\Delta n_1)^2}]^{\frac{1}{2}}$, is a linear measure of the width of the range over which $n_1$ is distributed.

Before this paragraph is the proof for $\overline{(\Delta n_1)^2}=Npq$ and after this it talks about $\Delta^*n_1/\overline{n_1}$ being a good measure of relative width of the distribution.

 

[haruspex]

This is the exact paragraph from unit 1.4 of the book:

Before this paragraph is the proof for $\overline{(\Delta n_1)^2}=Npq$ and after this it talks about $\Delta^*n_1/\overline{n_1}$ being a good measure of relative width of the distribution.

What if you simply ignore the statement that is bothering you/us? Does the rest make sense without it?