How to show the root mean square deviation

[arcTomato]

Homework Statement

prove that the mean position after a given number of steps is the starting position

Relevant Equations

random walk

Hi
I tried like this.
$σ^2=<(λ_1+λ_2+,,,+λ_i)^2>=<λ_1^2>+<λ_2^2>+,,,<λ_i^2>$
And I know $σ^2=Σ_in_iλ_i^2$from equation (4-12) (so this is cheat ).

So I know also $<λ_i^2>=n_iλ_i^2$, But why??

I know if I take $λ=1 ,σ^2=n$,But I don't understand $λ≠1$ version.

Sorry my bad english.
Please help me!

 

[RPinPA]

I tried like this.
$σ^2=<(λ_1+λ_2+...+λ_i)^2>=<λ_1^2>+<λ_2^2>+...+<λ_i^2>$

Why do you think that is true? If I understand the problem correctly, the $\lambda_i$ are fixed parameters and so are the $n_i$. You need a random variable here.

Are you studying random walks in one dimension? Then let's define a random variable $\Lambda_{ij}$, $j = 1,n_i$ which is either $\lambda_i$ or $-\lambda_i$ with equal probability. That is, corresponding to each $\lambda_i$ (fixed) there are $n_i$ (fixed number) of steps, each of which is either $\lambda_i$ to the right or $\lambda_i$ to the left (randomly).

Let's start with a simpler example. Let's suppose all the steps have the same size $\lambda_1$ and there are $n_1$ of them. So $N = n_1$. So we have $\Lambda_{11}, \Lambda_{12}, ..., \Lambda_{1,n_1} = \pm\lambda_1$. Obviously each has zero mean.

The total square deviation we are interested in is then $\sigma^2 = <\left ( \sum_{j=1}^{n_1} \Lambda_{1j} \right )^2 > = \sum_{j=1}^{n_1} < \Lambda_{1j}^2 >$ since the $\Lambda_{1j}$ are independent and the expectation of the cross terms is 0. Thus $\sigma^2$ in this case reduces to $n_1 <\Lambda_{11}^2>$.

This is the familiar result that $\sigma^2$ is proportional to the number of steps. The final thing here (I leave it to you) is to replace the expectation value of $\Lambda_{11}^2$ by something in terms of the fixed parameters $\lambda$.

You may think that this is what you were doing, but it's not. You didn't have a random variable in your expression, and you didn't have the right number of terms. You need to have a random variable for each of the $n_i$ steps of each size, as I did even when there is only one step size. So your steps will be $\Lambda_{11}, \Lambda_{12}, ..., \Lambda_{1n_1} = \pm\lambda_1$ and $\Lambda_{21}, \Lambda_{22}, ..., \Lambda_{2n_2} = \pm \lambda_2$, etc., and you will need double sums.

 

[arcTomato]

At first, Thank you for reply. Your explanation is so easy to understand.
I think I didn't understand the problem correctly.

I have to take the length like$Λ_{i,1},Λ_{i,2}...,Λ_{i,n_i}=±λ_i$.
so
$\sigma^2 = <\left ( \sum_{j=1}^{n_i}\sum_{i} \Lambda_{ij} \right )^2 > = \sum_{i} \sum_{j=1}^{n_i} < \Lambda_{ij}^2 >=\sum_{i} n_i(±\lambda_{i})^2$
I think this is almost the answer. Right??