Solving the Mystery of \ln{v_{i}} in an Expression

[Polymath89]

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

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

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

Can anybody tell me where the \ln{v_{i}} comes from and what I have done wrong?

 

[Borek]

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

You are missing m (not that it answers your question).

 

[micromass]

Do you mean

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

or

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

 

[Polymath89]

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:

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

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

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

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

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

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 occurring in that order.
For the second part with v_{i}, he uses MLE to estimate the parameters of the GARCH model. v_{i} is the variance for day i and he assumes that the probability distribution of u_{i} conditional on the variance is normal. Then he gets \prod_{i=1}^m[\frac{1}{\sqrt{2\pi v_i}}\exp{(\frac{-u_{i}^2}{2v_{i}})}]

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