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Two independent random vectors are almost surely non-orthogonal

  1. Jan 26, 2012 #1
    Hi all,

    I got stuck with the following problem:

    Let X, Y and Z be three random vectors of the same length drawn from a continuous random distribution.

    Z is independent of X and Y but [itex]Y=f(X)[/itex] with a non-linear function [itex]f[/itex].

    Can I claim that:

    1. [itex]Z^{T}X\neq 0[/itex] almost surely (i.e., vector X wouldn't almost surely lay on the null space of vector Z),


    2. [itex]Y^{T}X= 0[/itex] almost surely (i.e., vector X would almost surely lay on the null space of vector Y).

    If so, could you give me a hint to the proof or a citation.

    Thank you for your help in advance.

  2. jcsd
  3. Jan 26, 2012 #2
    What do you mean by a random distribution? Are you randomly selecting from a set of distributions? Individual probability distributions are described by functions which are not themselves random. That is, their parameters are specified. Secondly, if three vectors are randomly selected from a population with a known or unknown probability distribution, how do two of these vectors (X and Y) come to be related by a non linear function? I assume you're talking about a population of three component vectors, all of the same length and sharing the same vector space.

    I do agree that if you randomly select three vectors from a population where the orientations are randomly distributed on a continuous distribution, the probability that they would be orthogonal is zero. The probability of an exact value on a continuous distribution is zero by the definition of a continuum, in this case: all real numbers on the interval [0,1]. The same would be true for any pair from {X,Y,Z}.

    EDIT: If you are talking about three random vectors in a 2-space, then only 2 vectors can be mutually orthogonal, the third being a linear combination of the other two. Was that part of your question?
    Last edited: Jan 26, 2012
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