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Math Help - Guassian distribution and polar co-ordinates

  1. #1
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    Post Guassian distribution and polar co-ordinates

    Hi
    Not sure about this problem

    Let X and Y be independent random variable each with a standard gaussian distribution. Let 0<M calculate P ((X^2)+(Y^2)<=(M^2)) using polar co-ordinates??????

    my solution is as follows

    let
    x=rcos(p)
    y=rsin(p)
    then want to find P ((X^2)+(Y^2)<=(M^2)) = P(r<=m)

    Joint distribution of X,Y is

    FXY(x,y) = (1/(2*Pi))Exp[-0.5*((x^2)+(y^2))]

    and so

    FXY(r,p) = (1/(2*Pi))Exp[-0.5*(r^2)]

    then P(r<=m) = to the double intergral of FXY(r,p) w.r.t r and p between the limits [0,M] and [0,2*pi]

    this however isn't very nice

    is this right or have i gone wrong some where . . . .

    thankssss
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  2. #2
    MHF Contributor matheagle's Avatar
    Joined
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    Quote Originally Posted by silk23 View Post
    Hi
    Not sure about this problem

    Let X and Y be independent random variable each with a standard gaussian distribution. Let 0<M calculate P ((X^2)+(Y^2)<=(M^2)) using polar co-ordinates??????

    my solution is as follows

    let
    x=rcos(p)
    y=rsin(p)
    then want to find P ((X^2)+(Y^2)<=(M^2)) = P(r<=m)

    Joint distribution of X,Y is

    FXY(x,y) = (1/(2*Pi))Exp[-0.5*((x^2)+(y^2))]

    and so

    FXY(r,p) = (1/(2*Pi))Exp[-0.5*(r^2)]

    then P(r<=m) = to the double intergral of FXY(r,p) w.r.t r and p between the limits [0,M] and [0,2*pi]

    this however isn't very nice

    is this right or have i gone wrong some where . . . .

    thankssss

    YOU left out the Jacobian, r, which makes this a very simple integral.
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