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Newbie question: Algebra of Mahalanobis distance

  1. Nov 11, 2013 #1

    The Mahalanobis distance or rather its square is defined as :

    [itex](X-\mu)^2/\Sigma[/itex] which is then written as

    [itex](X-\mu)^{T} Ʃ^{-1}(X-\mu)[/itex]

    Ʃ is the covariance matrix. My silly question is why is the sigma placed in the middle of the dot product of the (X-μ) vector with itself. I am sure this makes sense mathematically (this reduces the output to a scalar) but I would like to know the intuitive reason behind it.

    Thanks a lot!
  2. jcsd
  3. Nov 11, 2013 #2


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    The idea behind the Mahalanobis distance is that you are measuring how many standard deviations from the mean X is in the one dimensional case. In multidimensional cases, [itex] \Sigma[/itex] is going to be a positive (semi)definite matrix, which will have a unique positive (semi)definite square root which I will call S. S serves the same role as the standard deviation. Then the expression above is the same as

    [tex] \left( S^{-1}(X-\mu) \right)^T \left(S^{-1}(X-\mu) \right) [/tex]

    basically, you scale the random vector [itex] X-\mu[/itex] by the standard deviation, the same as you would in the one dimensional case.
  4. Nov 11, 2013 #3

    D H

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    The expression ##(X-\mu)^T \Sigma^{-1}(X-\mu) = \sigma^2## defines a family of hyperellipsoids in the N-dimensional space in which X and μ live, characterized by the scalar parameter σ. I used σ intentionally. Think of σ as representing "standard deviations". For example, ##(X-\mu)^T \Sigma^{-1}(X-\mu) = 1## is the one sigma hyperellipsoid.

    The Mahalanobis distance is essentially a measure of how many standard deviations a point X is from the mean μ.
  5. Nov 11, 2013 #4
    Thank you guys!
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