Log of Product

  • Thread starter Polymath89
  • Start date
  • #1
Polymath89
27
0
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?
 

Answers and Replies

  • #2
Borek
Mentor
29,167
3,843
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).
 
  • #3
micromass
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
22,178
3,316
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]
 
  • #4
Polymath89
27
0
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]
 

Suggested for: Log of Product

Replies
1
Views
482
Replies
9
Views
661
  • Last Post
2
Replies
44
Views
2K
  • Last Post
Replies
13
Views
868
  • Last Post
Replies
1
Views
455
Replies
6
Views
463
  • Last Post
Replies
10
Views
805
Replies
16
Views
505
Replies
9
Views
554
Replies
7
Views
500
Top