Where does ln(v_i) come from in the log of this product?

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