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Derivative of Log Likelihood Function

  1. Nov 15, 2015 #1
    So looking through my notes I cant seem to understand how to get from one step to the next. I have attached a screenshot of the 2 lines I'm very confused about. Thanks.

    BTW: The equations are for the log likelihood in a mixture of gaussians model

    EDIT: To elaborate I am particularly confused about how they get numerator term π_{k} N(x_{n}|μ_{k}, Σ). I can't seem to understand how they are differentiating this to obtain that. I understand how they obtain the denominator term from differentiating the log but thats about all. To differentiate the multivariate gaussian I would think the log function needs to be used to break up the internal terms. Although I cant put this intuition together.

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    Last edited: Nov 15, 2015
  2. jcsd
  3. Nov 15, 2015 #2


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    I think it's because ##\Sigma_k## appears both inside and outside (as an inverse) the exponent in the cdf function ##\mathscr{N}##.
    $$\frac{\partial}{\partial \Sigma_k}\mathscr{N}(\mu,\Sigma_k)=
    \frac{\partial}{\partial \Sigma_k}\left[C\Sigma_k{}^{-1}\exp[f(\mu,\Sigma_k)]\right]$$
    for known constant ##C## and function ##f##.
    By the product rule, this is then equal to
    $$C\exp[f(\mu,\Sigma_k)]\left[\frac{\partial}{\partial \Sigma_k}\Sigma_k{}^{-1}+\Sigma_k{}^{-1}\frac{\partial}{\partial \Sigma_k}f(\mu,\Sigma_k)]\right]$$

    There will be some messy algebra involved.

    You might find it easier to first work through the univariate case, differentiating wrt ##\sigma## and seeing if you can obtain an analogous expression. If that works out, it shouldn't be too hard to extend it to the multivar case.
  4. Nov 15, 2015 #3
    I don't understand how you got $$C\Sigma_{k}^{-1}$$ In the multivariate gaussian we have $$\frac{1}{|\Sigma_{k}|}$$ How did you convert that determinant into an inverse ? Maybe you meant the same thing but forgot the determinant sign ?
    Last edited: Nov 15, 2015
  5. Nov 15, 2015 #4


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    I didn't. What I wrote is only broadly indicative of the structure. I didn't look up the multivariate Gaussian formula. With your correction that line becomes:

    $$C\exp[f(\mu,\Sigma_k)]\left[\frac{\partial}{\partial \Sigma_k}|\Sigma_k|^{-1}+|\Sigma_k|^{-1}\frac{\partial}{\partial \Sigma_k}f(\mu,\Sigma_k)]\right]$$
    which is
    $$C\exp[f(\mu,\Sigma_k)]\left[-|\Sigma_k|^{-2}\frac{\partial |\Sigma_k|}{\partial \Sigma_k}+|\Sigma_k|^{-1}\frac{\partial}{\partial \Sigma_k}f(\mu,\Sigma_k)]\right]$$

    I think if you work through the univariate case first it'll become much clearer.
  6. Nov 15, 2015 #5
    Okay so rewriting with exponents of -1/2 (for the gaussian) and repeating the operation we would have:
    $$C\exp[f(\mu,\Sigma_k)]\left[-\frac{1}{2}|\Sigma_k|^{\frac{-3}{2}}\frac{\partial |\Sigma_k|}{\partial \Sigma_k}+|\Sigma_k|^{\frac{-1}{2}}\frac{\partial}{\partial \Sigma_k}f(\mu,\Sigma_k)]\right]$$
    So the problem becomes the extra $$|\Sigma_k|^{-1}$$ that gets left over after we factor out $$|\Sigma_k|^{\frac{-1}{2}}$$ Any ideas ?
  7. Nov 16, 2015 #6
    So I think I resolved my troubles using a few properties outlined in the matrix cookbook.
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