Where does the n-1 come from in 1-dimensional random walks?

[SansaStark]

Hello!
I'm struggling with a probably easy physics question concerning random walks. Here I have the slide regarding this:
Delta is the distance that a particle moves.

Can someone explain where the n-1 initially comes from? Does it have to do wtih the concept of the degrees of freedom?
Than you already! Regards, Vera

 

[Dr.AbeNikIanEdL]

Not sure if I understand the question. So your first equation states that after $n$ steps the particle is at the position it was after the previous step (which is the $(n-1)$th step) plus some $\delta$. If you put in $n=1$ you get that the position after the first step is the initial position plus some $\delta$: $x_1(1) - x_i(0) \pm \delta$. Every next step a new $\delta$ is added, so of course the position after $n$ steps depends on the position after $(n-1)$ steps. Why do you think this has to do anything with degrees of freedom?

 

[Samy_A]

After $n-1$ steps, particle i is at position $x_i(n-1)$.
One step after that, at step $n$, particle i has moved by $\pm \delta$: therefore $x_i(n)=x_i(n-1)\pm \delta$.

 

[fresh_42]

The notations $x_n^{(i)}$ and $<x_n^{(i)} | 1≤i≤N >$ might have been less confusing. Or more.

 

[SansaStark]

Oh okay, thank you. After some thinking ad not giving that much emphasis of on this n-1 (which so muchly looks like the n-1 from the df concept) I finally realized what both of you mean ;) I was really confused about why suddenly this n-1 would appear. But its probably quite reasonable. oO Thanks a lot!

 

[SansaStark]

@fresh42: More (for I have no clue what the vertical bar means AND SO ON) ;)

 

[SansaStark]

Another question:
Equation 1.) is followed by equation 2.). In equation 2.) the delta is treated separately as encircled in red colour. Is this just a process of pulling the expression after the sigma sign in equ. 1 apart?
But then, why is the second expression (in the circle) SUBTRACTED from the first part of equ. 2.)? I know it doesn't really matter as delta is 0... but just assume that the encircled expression without the minus in the beginning would equal a positive number. Then the outcome would be different compared to when the expression was negative. I know there's a mistake in thinking but I I just can figure out what's behind this. ^^

1.)

2.)

Just how I'd approach this: The first part of equ. 2.) is the average of all preceding steps (n-1) and the second part is the average of all distances after one step. Then, if I subtract all deltas and assume their average was negative (which just means this whole particle is moving in the opposite direction compared to a positive delta) this would result in a position (x) actually in the opposite direction of the negative delta average for it is subtracted from the first average...

Is that right or complete bullsh**? ;) Thanks!