# Variance of 2-D random walker

I've made a 2D walker to compare different RNG's. I'm measuring the succes of each walk as the distance from the origin to the endpoint, using the regular 2-norm. The thing I can't seem to work out is the variance.

$$D_n=\sqrt(x_n^2+y_n^2)$$

$$Var(D_n)=E[D_n^2]=E[Z_1^2+...+Z_n^2]$$

Since $Var(Z_i)=\sqrt{2}$ does this mean that the variance is $2n$? Seems too easy...

Ps. I'm not sure how to make the formatting prettier, if someone can tell me, I'll edit it naturally!
Ps2. Thanks Stephen Tashi!

Last edited:

Stephen Tashi
Ps. I'm not sure how to make the formatting prettier, if someone can tell me, I'll edit it naturally!

On this forum, surround the LaTex with "tags" rather than the dollar sign.

Code:
$$D_n=\sqrt{x_n^2+y_n^2}$$

$$Var(D_n)=E[D_n^2]=E[Z_1^2+...+Z_n^2]$$

Since $Var(Z_i)=\sqrt{2}$ does this mean that the variance is $2n$ ?

$$D_n=\sqrt{x_n^2+y_n^2}$$

$$Var(D_n)=E[D_n^2]=E[Z_1^2+...+Z_n^2]$$

Since $Var(Z_i)=\sqrt{2}$ does this mean that the variance is $2n$ ?

Stephen Tashi
Another thing about the forum: When you edit a post, sometimes "Save" doesn't display the LaTex. You must refresh the page to accomplish that.

$Var(z_i) = \sqrt{2}$ for the random variable $z_i$ that uses the square of the distance between the current position and the previous position. But this is not the same as using the distance between the current position and $(x_0,y_0)$.

For example, there is the distinction between
$$z_2 = \sqrt{(x_1-x_0)^2 + (y_1-y_0)^2} + \sqrt{( x_2-x_1)^2 + (y_2-y_1)^2}$$

and

$$Z_2 = \sqrt{ (x_2-x_0)^2 + (y_2-y_0)^2}$$

I've made a 2D walker to compare different RNG's. I'm measuring the succes of each walk as the distance from the origin to the endpoint, using the regular 2-norm. The thing I can't seem to work out is the variance.

$$D_n=\sqrt(x_n^2+y_n^2)$$

$$Var(D_n)=E[D_n^2]=E[Z_1^2+...+Z_n^2]$$

Since $Var(Z_i)=\sqrt{2}$ does this mean that the variance is $2n$? Seems too easy...

Ps. I'm not sure how to make the formatting prettier, if someone can tell me, I'll edit it naturally!
Ps2. Thanks Stephen Tashi!

The sum of the variances of independent random variables is the variance of the sum. That should make it easy, unless I'm missing something.