Random Walk: Calculating Standard Deviation from Mean Position

[Someone1987]

Homework Statement

I need to determine the standard deviation from the mean position value <x> in a random walk scenario. We are given that the probability of taking a step to the right is p and the probability of taking a step to left is q. Please note that p and q need not be equal. The starting point is x = 0 and we are only moving in one dimension.

Homework Equations

The Attempt at a Solution

I would just like to see if this solution seems valid. In a random walk, after taking N steps one's position is x = n_R - n_L where n_R is the number of steps taken to the right and n_L is the number of steps taken to the left. Both n_R and n_L of course follow binomial probability distributions. Using the fact that the variance = <x²> - <x>², I went ahead and calculated these two expectation values for x² and x.

It would seem to me that the expectation value of x is
<x> = <n_R> - <n_L>, thus using the fact that n_R and n_L follow binomial probability distributions, I get <x> = Np - Nq = N(2p-1). This formula seems reasonable to me as if p = 0.5, then <x> = 0 as one would expect.

Now for <x²>, I simplifed n_R - n_L to 2n_R - N. Therefore x² = 4n_R² - 4Nn_R + N². Thus,
<x²> = <4N_R²> - <4Nn_R> + <N²>

Using the formulas for the binomial distribution, I get
<x²> = 4(Np)² - 4Np² + 4Np - 4pN² +N².

Putting all this together and taking the square root to get the standard deviation, I get
\sigma = 2\sqrt{Npq}.

This all seems reasonable to me but I am not really an expert on probability and statistics so I was hoping to get some thoughts on this solution. Any help is very much appreciated.

 

[turin]

It would seem to me that the expectation value of x is
<x> = <n_R> - <n_L>, ...
...
Now for <x²>, I simplifed n_R - n_L to 2n_R - N.
...
... I was hoping to get some thoughts on this solution. Any help is very much appreciated.

I'm not an expert either, but I'm wondering about your separation of <x>. I believe that n_R and n_L are strongly correlated (r=-1), as demonstated by your treatment of <x²>. Doesn't this prevent you from separating <x> as you have done? I think that you should use x=2n_R-N for <x> as well.

 

[Someone1987]

I understand your concern but I checked and taking the expected value of <2n_R -N> gives you the exact same answer. I actually didn't separate them when I did my homework and in the process of computing <x> I could see that <x> = <n_R> - <n_L> is the same as splitting them apart and I just used this simplifying fact when I made this post. Thank you very much for your comment though. I appreciate your taking the time to post your thoughts.