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Math Help - random walk

  1. #1
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    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?
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  2. #2
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    Quote Originally Posted by PvtBillPilgrim View Post
    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...
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  3. #3
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    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.
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  4. #4
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    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
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