Estimating Confidence Intevals

#1
Hi,

I have used a loglikelihood function to estimate 24 parameters. How do I construct a confidence interval given the Hessian (avaliated on the estimatives) and the gradiente(avaliated on the estimatives). Do I need both?

What I saw is that, given that MLE is assymptoticaly normal I could construct using:

X +- 1.96*[-(hessian)^-1] ... Thats ok when I have one parameter. How do I do it for the 24 ones, using the VAR-COV matrix?

tks!
 

BGM

TS Contributor
#2
If you satisfy these regularity conditions

http://en.wikipedia.org/wiki/Maximum_likelihood#Asymptotic_normality

then it follows that the MLE \( \hat{\theta} \) jointly follows a multivariate normal distribution asymptotically:

\( \sqrt{n}(\hat{\theta} - \theta_0) \sim \mathcal{N}_p(0, I^{-1}) \)

where \( p \) is the dimension of the vector, \( \theta_0 \) is the true parameter, \( I \) is the information matrix.

Now we invoke another well-known result:
(see e.g. Theorem 7 in http://www2.econ.iastate.edu/classes/econ671/hallam/documents/QUAD_NORM.pdf)

\( [\sqrt{n}(\hat{\theta} - \theta_0)]^T (I^{-1})^{-1} [\sqrt{n}(\hat{\theta} - \theta_0)]
= n(\hat{\theta} - \theta_0)^T I (\hat{\theta} - \theta_0) \sim \chi^2(p) \)

Once you can provide a consistent estimate of the information matrix \( \hat{I} \), then the following set

\( \{\theta_0: n(\hat{\theta} - \theta_0)^T \hat{I} (\hat{\theta} - \theta_0) \leq \chi^2_{1 - \alpha}(p) \} \)

forms a \( 1 - \alpha \) confidence region for \( \theta_0 \). In high dimension like this it will be a generalization of the ellipsoid.
 

Dason

Ambassador to the humans
#3
Are you asking about making a confidence interval for a single parameter in the 24 parameter set (and maybe you'll do this for all 24 parameters) or are you asking about making a confidence interval for a linear combination of the parameters (so maybe something like a confidence interval for \(\theta_1 - \theta_2\))