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Rearranging a formula (Multivariate Gaussian function)

  1. Dec 16, 2012 #1
    1. The problem statement, all variables and given/known data
    See image, p(y|θ) is the Likelihood function which has to be rearranged in the form of equation (3). θ is a vector variable.
    1zobx9u.png

    2. Relevant equations
    None?

    3. The attempt at a solution
    I first expanded the exponent in the original function, equation (2).

    [tex](b-A\theta)^T(b-A\theta)=b^Tb - b^TA\theta - \theta^T A^T b + \theta^T A^T A \theta[/tex]

    Now suppose I can write the function equivalently as

    [tex]C \exp\left( -\frac{1}{2} (\theta - \theta_0)^T L (\theta - \theta_0) \right)[/tex]

    where C represents the same constant multiplying with exp in equation (2). In this case, the exponential parts must be the same. So:

    [tex]b^Tb - b^TA\theta - \theta^T A^T b + \theta^T A^T A \theta = \theta^TL\theta - \theta^TL\theta_0 - \theta_0^TL\theta + \theta_0^TL\theta_0 [/tex]

    If this this expression is true for all θ, then all coefficients of (...)θ , θT(...) , θT(...)θ and the constants must match.

    So L = ATA and θ0=L-1ATb.

    Now I don't know how to get L0. (equation (5) ). I only have this condition left from my assumption about the constants above:

    bTb=θT00

    There just don't seem to be enough terms in either exponent to allow the exponential part of L0. I don't think I can add and subtract anything either...

    Any ideas?
     
    Last edited: Dec 16, 2012
  2. jcsd
  3. Dec 16, 2012 #2

    Ray Vickson

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    Try to show that
    [tex] (b-A\theta)^T(b-A\theta) = (\theta - \theta_0)^T L (\theta - \theta_0) + K[/tex] for some matrix L and some constant K, and for ##\theta_0## as given in the question.
     
  4. Dec 16, 2012 #3
    Ok.

    [tex](b-A\theta)^T(b-A\theta)=b^Tb - b^TA\theta - \theta^T A^T b + \theta^T A^T A \theta[/tex]
    [tex]= b^Tb - (L L^{-1}A^T b)^T\theta - \theta^T L( L^{-1}A^T b) + \theta^T L \theta[/tex]
    [tex]= b^Tb + \theta_0^T L \theta - \theta^T L \theta_0 + \theta^T L \theta[/tex]
    [tex]= b^Tb + (\theta-\theta_0)^T L \theta - \theta^T L \theta_0[/tex]
    [tex]= b^Tb + (\theta-\theta_0)^T L \theta - (\theta-\theta_0)^T L \theta_0 -\theta_0^T L \theta_0 [/tex]
    [tex]= b^Tb + (\theta-\theta_0)^T L (\theta-\theta_0)-\theta_0^T L \theta_0 [/tex]

    I used the fact that LT=L=ATA.

    On the basis of this it seems K = bTb - θ0T0

    I still can't see the origin of L0 here though..
     
  5. Dec 20, 2012 #4
    Bump. I haven't been able to find any leads. Anyone have an idea?
     
  6. Dec 20, 2012 #5

    pasmith

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    I'm somewhat confused by the fact that if [itex]L = A^TA[/itex] and [itex]\theta_0 = L^{-1}A^Tb[/itex] then [itex]L^{-1} = A^{-1}(A^T)^{-1}[/itex] so that
    [tex]A\theta_0 = AL^{-1}A^Tb = A(A^{-1}(A^T)^{-1})A^Tb = b[/tex]
    which means that [itex]b - A\theta_0 = b - b = 0[/itex]. So I'm at a loss to explain why the author has included the exponential in [itex]\mathcal{L}_0[/itex], when its value appears to be [itex]\exp(0) = 1[/itex].

    On the other hand, I was able to show (using the additional fact that [itex]L[/itex] is symmetric so that [itex](L^T)^{-1} = (L^{-1})^T = L^{-1}[/itex]) that
    [tex](\theta - \theta_0)^TL(\theta - \theta_0) = (A\theta - b)^T(A\theta - b)
    = (b - A\theta)^T(b - A\theta)[/tex]
    It's just a case of expanding the left hand side, substituting the definitions of [itex]L[/itex] and [itex]\theta_0[/itex] and simplifying.
     
  7. Dec 21, 2012 #6

    pasmith

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    Ignore that: A is not square, so cannot be invertible.

    The aim is to show that [itex](b - A\theta)^T(b - A\theta) = (b - A\theta_0)^T(b - A\theta_0) + (\theta - \theta_0)^T(\theta - \theta_0)[/itex]

    Now the left hand side is
    [tex]b^Tb - b^TA\theta - \theta^TA^Tb + \theta^TA^TA\theta[/tex]

    The right hand side is
    [tex]b^Tb - b^TA\theta_0 - \theta_0^TA^Tb + \theta_0^TA^TA\theta_0
    + \theta^TL\theta - \theta^TL\theta_0 - \theta_0^TL\theta + \theta_0^TL\theta_0\\
    = b^Tb - (b^TA\theta_0 + \theta_0^TA^Tb) + 2\theta_0^TL\theta_0
    + \theta^TL\theta - \theta^TL\theta_0 - \theta_0^TL\theta \\
    = b^Tb - 2b^TAL^{-1}A^Tb + 2b^TAL^{-1}A^Tb
    + \theta^TA^TA\theta - \theta^TA^Tb - b^TA\theta \\
    = b^Tb -b^TA\theta - \theta^TA^Tb + \theta^TA^TA\theta
    [/tex]
    as required.
     
    Last edited: Dec 21, 2012
  8. Dec 22, 2012 #7
    Edit: there seems to be a mistake between the second and third lines. You go from -(bT0 + θ0TATb) to -2bT(....)+2bT(....)

    The plus sign should be minus, and I'm not sure of where the factor of 2 comes from. Anyway I will investigate your approach.
     
    Last edited: Dec 22, 2012
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