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Bivariate Normal Distribution, contour ellipse containing given % samples?

  1. Jun 23, 2012 #1
    Given a bivariate gaussian distribution,
    I'm attempting to find the probability p for which
    the ellipse of all points (x,y) for which P(X = x, Y= y) = p contains
    a given % of the samples drawn from the distribution.

    I want the 2d equivalent for the 1 dimensional case:
    given a normal distribution N(0,1):
    e.g interval between points with p = 0.24197072 contains 68.2% of all samples
    e.g interval between points with p = 0.05399097 contains 95.4% of all samples
    e.g interval between points with p = 0.00013383 contains 99.6% of all samples
    in two dimensions these interval boundries become an ellipse and I'm interested in finding the p value corresponding to a given % (contained samples in contour ellipse with p) value in the 2 dimensional case.

    Some extra info:
    A matlab, python (using numpy, scipy?) numerical approximation is ok, I don't need an analytic formula.

    Actually I just want to draw the ellipses containing 75%, 95%, 99% of the samples in python (using matlibplot) for a given gaussian distribution (varying mean & covariance). I know how to do this if I obtain p first (contour plots).

    Thank you for reading my question and I hope you can help.
  2. jcsd
  3. Jun 23, 2012 #2
    Hi codiloo,

    The probability hyper-ellipsoid hyper-volume for a multivariate normal follows

    [tex](x-μ)^T \Sigma^{-1}(x-μ) ≤ χ^2_k(p)[/tex]

    Where x is a k-dimensional vector, μ is the k-dimensional mean vector, Ʃ is the variance-covariance matrix and [itex]χ^2_k(p)[/itex] is the p quantile of the chi-square distribution with k degrees of freedom.

    When k = 2 dimensions the expression represents the area of the ellipse you are asking for, and [itex]χ^2_2[/itex] behaves as an exponential distribution.
    Last edited: Jun 23, 2012
  4. Jun 24, 2012 #3


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    After a huge calculation involving rotating co-ordinates I ended up with
    P[(x, y) lies inside the contour pdf(x,y) = k] = 1 - 2πkD, where D is the determinant of the covariance matrix, i.e. = √(σ12σ22 - ρ4).
    Note e.g. that the peak pdf value is 1/2πD
    If it's right, there must be an easier way.
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