Is the Variance of a 2D Random Walk Simply 2n?

[ragnabob]

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!

 

[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.

[tex]D_n=\sqrt{x_n^2+y_n^2}[/tex]

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

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

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}

 

[PatrickPowers]

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.