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Log of Product |
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| Sep13-12, 11:16 AM | #1 |
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Log of Product
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? |
| Sep13-12, 11:21 AM | #2 |
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Admin
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| Sep13-12, 11:38 AM | #3 |
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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] |
| Sep13-12, 12:09 PM | #4 |
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Log of Product
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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