# Estimating Confidence Intevals

#### celiofeltrinjr

##### New Member
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
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

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 $$\theta_1 - \theta_2$$)