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

[Polymath89]

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?

 

[Borek]

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

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

 

[micromass]

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]

 

[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:

[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 occurring 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]

 

[CellCoree]

The \ln{v_{i}} term comes from the natural logarithm of the original expression. When taking the logarithm of a product, it can be rewritten as the sum of the logarithms of each term. In this case, the logarithm of the product \prod_{i=1}^m[\frac{1}{\sqrt{2\pi v}}\exp{(\frac{-u_{i}^2}{2v_{i}})}] can be written as \ln{\prod_{i=1}^m[\frac{1}{\sqrt{2\pi v}}\exp{(\frac{-u_{i}^2}{2v_{i}})}]}, which can then be expanded as \sum_{i=1}^m \ln{\frac{1}{\sqrt{2\pi v}}\exp{(\frac{-u_{i}^2}{2v_{i}})}}. The \ln{v_{i}} term comes from the logarithm of the term \frac{1}{\sqrt{2\pi v}} in the product.

It seems that in your calculation, you have missed the logarithm of this term and have only included the logarithm of the exponential term, resulting in a different expression. Make sure to include all terms when taking the logarithm of a product.