# Math Help - random walk

1. ## random walk

I have a question similar to #10 here Probability: An Introduction - Google Book Search

I have a 2D symmetric random walk starting from zero. Again, D_n is the distance from the origin after n steps. I need to show that E((D_n)^2) like #10 is equal to n.

So I know that euclidean distance = sqrt((x-0)^2 + (y-0)^2), so let X_n and Y_n be the coordinates of the particle.
It seems that E(X_n)=E(Y_n)=n/2, but how do I show this?
For n = 1, E(X_1)=(.25)(0^2 + 0^2 + 1^2 + -1^2) = .5 which holds
I can do this similarly for E(X_2), E(X_3) and I get the right answers (i.e 1 and 1.5 respectively), but I'm missing the pattern. Can someone show me how to do this? Does induction work?

2. Originally Posted by PvtBillPilgrim
I have a question similar to #10 here Probability: An Introduction - Google Book Search

I have a 2D symmetric random walk starting from zero. Again, D_n is the distance from the origin after n steps. I need to show that E((D_n)^2) like #10 is equal to n.

So I know that euclidean distance = sqrt((x-0)^2 + (y-0)^2), so let X_n and Y_n be the coordinates of the particle.
It seems that E(X_n)=E(Y_n)=n/2, but how do I show this?
For n = 1, E(X_1)=(.25)(0^2 + 0^2 + 1^2 + -1^2) = .5 which holds
I can do this similarly for E(X_2), E(X_3) and I get the right answers (i.e 1 and 1.5 respectively), but I'm missing the pattern. Can someone show me how to do this? Does induction work?
Notice that $E[D_n^2]=E[X_n^2]+E[Y_n^2]={\rm Var}(X_n)+{\rm Var}(Y_n)$ and $X_n$ is the sum of $n$ independent steps, so that ${\rm Var}(X_n)$ equals $n$ times the variance of the x-component of one step. So you should compute this variance. With probability 1/2, a step goes $\pm 1$ in the x-direction, and with probability 1/2, it goes in the y-direction so its x-component is 0...

3. Is there anyway to do it in terms of just the expected value, the way I was trying? I don't think I'm far off, and I don't really understand how computing the variance will help me. Can you show me, or tell me what I'm missing with the expected value?
Thanks for the response.

4. First, you are rightly computing the E[Xn^2] and E[Yn^2], yet you say you're computing E[Xn] and E[Yn].

He has explained throughly how the variance appears, since E[Xn^2] = Var[Xn] (actually Var[Xn] = E[(Xn-u)^2] but because the average is 0 it gives this), since each step is independent, you'll have Var[Xn] = n*Var[X] (the variance of each step).

So now all you need is Variance - Wikipedia, the free encyclopedia