Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

http://dl.dropbox.com/u/5681270/Screen%20Shot%202012-03-25%20at%204.42.10%20PM.png

This is a solution to 1 of my tutorial questions, anybody got any clue as to the step:

$\frac{1}{nI(\sigma^2)} = \frac{2\sigma^2}{n}$ ?

Thanks.

share|improve this question

1 Answer 1

up vote 0 down vote accepted

If you compute the second derivative of the log likelihood function or equivalently the first derivative of the score vector (since the score vector by definition is the first derivative of the log likelihood function) you get $$\frac{ds}{d\sigma^2}=\frac{n}{2\sigma^4}-\frac{1}{\sigma^6}\sum_{i=1}^n{(x_i-u_i)^2}$$ Now taking expectations yields:

$$E(\frac{ds}{d\sigma^2})=\frac{n}{2\sigma^4}-\frac{n}{\sigma^6}E\frac{\sum_{i=1}^n{(x_i-u_i)^2}}{n}=\frac{n}{2\sigma^4}-\frac{n}{\sigma^4}=-\frac{n}{2\sigma^4}$$

Lastly by the fisher information matrix equality we have

$$I(\sigma^2)=-E(\frac{ds}{d\sigma^2})=\frac{n}{2\sigma^4}$$

Remark1: As you may be aware of, we are assuming that $\mu$ is known above. As is standard I also denote the score vector by $s()$, i.e. $s(\sigma^2)=\frac{dl}{d\sigma^2}$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.