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Thread: Uncorrelated, orthogonal and independent???

  1. #1
    Jun 2010

    Uncorrelated, orthogonal and independent???

    Hi, i have a couple of questions...

    I know that when two random vectors are independent, they are uncorrelated. The opposite it isn't true necessarily.

    What about orthogonal random vectors??
    Orthongonality implies independence??
    Independence implies orthogonality?


    i know for example, that orthogonality exists when the correlation matrix is diagonal, and uncorrelation exists when covariance matrix is diagonal. But i don't understand clearly the relation with independence...

    It's a little confusing to me... Thanks....
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  2. #2
    Jun 2010
    Orthogonality imposes independence. The opposite may not hold.
    To prove it:
    Take two vectors X and Y such that they are orthogonal.
    i.e. x^Ty=0
    let us assume that if possible they are linearly dependent.
    i.e. there exists scalar c(not equal to 0)
    such that,
    this means, c Y^TY=0
    implies, Y^TY=0
    implies, each element of Y vector is 0.which is absurd.
    hence x and y are independent.
    The opposite can be proved taking any counter example
    x=(1 0 2)
    y=(0 1 1)
    the above two vectors are independent but not orthogonal.
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  3. #3
    Senior Member
    Oct 2009
    Dooti, I think the OP was referring to independence in the sense of random variables, not in the sense of "linear dependence."

    I would think that what he means by orthogonal is either something along the lines of

    $\displaystyle Ex^\top y = 0$


    $\displaystyle E(x - Ex)^\top (y - Ey) = 0$.

    In neither case does orthogonality imply independence (this is trivial to show if you take x, y to be real numbers). In the first case, independence does not imply orthogonality, while in the second case it does.
    Last edited by theodds; Jun 10th 2010 at 11:40 AM.
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