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Log of Product

  1. Sep 13, 2012 #1
    I have a problem taking the log of this expression [tex]\prod_{i=1}^m[\frac{1}{\sqrt{2\pi v}}\exp{(\frac{-u_{i}^2}{2v_{i}})}][/tex]

    Now I would get [tex]\ln({\frac{1}{\sqrt{2\pi v}}})(\sum_{i=1}^m{\frac{-u_{i}^2}{v_{i}}})[/tex]

    The author gets, by ignoring the constant multiplicative factors, [tex]\sum_{i=1}^m (-\ln{v_{i}}-\frac{u_{i}^2}{v_{i}})[/tex]

    Can anybody tell me where the [itex]\ln{v_{i}}[/itex] comes from and what I have done wrong?
     
  2. jcsd
  3. Sep 13, 2012 #2

    Borek

    User Avatar

    Staff: Mentor

    You are missing m (not that it answers your question).
     
  4. Sep 13, 2012 #3
    Do you mean

    [tex]\prod_{i=1}^m[\frac{1}{\sqrt{2\pi v}}\exp{(\frac{-u_{i}^2}{2v_{i}})}][/tex]

    or

    [tex]\prod_{i=1}^m[\frac{1}{\sqrt{2\pi v_i}}\exp{(\frac{-u_{i}^2}{2v_{i}})}][/tex]
     
  5. Sep 13, 2012 #4
    I'm sorry, I just noticed the difference in the terms, first the author uses v as a constant, so he starts with this term:

    [tex]\prod_{i=1}^m[\frac{1}{\sqrt{2\pi v}}\exp{(\frac{-u_{i}^2}{2v})}][/tex]

    and then he gets, by ignoring the constant multiplicative factors:

    [tex]\sum_{i=1}^m (-\ln{v}-\frac{u_{i}^2}{v})[/tex]

    Then he replaces v with [itex]v_{i}[/itex], so [tex]\prod_{i=1}^m[\frac{1}{\sqrt{2\pi v_i}}\exp{(\frac{-u_{i}^2}{2v_{i}})}][/tex]

    and gets [tex]\sum_{i=1}^m (-\ln{v_{i}}-\frac{u_{i}^2}{v_{i}})[/tex]

    To put all of this in perspective, the author tries to estimate parameters of a GARCH(1,1) model and the first part(with v as a constant) is supposed to be an example of a Maximum Likelihood Estimation, where he estimates the variance v of a random variable X from m observations on X when the underlying distribution is normal with zero mean. Then the first term is just the likelihood of the m observations occuring in that order.
    For the second part with [itex]v_{i}[/itex], he uses MLE to estimate the parameters of the GARCH model. [itex]v_{i}[/itex] is the variance for day i and he assumes that the probability distribution of [itex]u_{i}[/itex] conditional on the variance is normal. Then he gets [tex]\prod_{i=1}^m[\frac{1}{\sqrt{2\pi v_i}}\exp{(\frac{-u_{i}^2}{2v_{i}})}][/tex]

    and [tex]\sum_{i=1}^m (-\ln{v_{i}}-\frac{u_{i}^2}{v_{i}})[/tex]
     
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