13_AS_3_lec_a

# Compute the asymptotic standard error of the

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: equal to (I (β ))−1 , where I (β ) is the observed information matrix given by ˆ I (β ) = 22/45 − ∂ 2 ln L (β ) |ˆ ∂βi ∂βj β =β i ,j =1,··· ,p Actuarial Statistics – Module 3: Semi-parametric methods: Cox Regression Model Estimation of the regression parameters β Example I Given the partial likelihood: L= 1 2 3 23/45 Ce β , (e β + 2)(e β + 3) ˆ1 Show that the MLE of β is β = 2 log 6. Compute the asymptotic standard error of the estimator. Construct an approximate 95% conﬁdence interval for the parameter β . Solution: 1 log L = log C + β − log(e β + 2) − log(e β + 3) log eβ eβ Solving d d β L = 1 − e β +2 − e β +3 = 0 =⇒ β = 2 ˆ Noting that d log L | 1 < 0, β = 1 log 6. 2 dβ β = 2 log 6 2 1 2 log 6 Actuarial Statistics – Module 3: Semi-parametric methods: Cox Regression Model Estimation of the regression parameters β Example II 2 d 2 log L d β2 β 2e = − (e β +2)2 − 3e β (e β +3)2 2 ˆ I (β ) = [− d dlog L ]β =β The ˆ β2 ˜ asymptotic variance of β is ˆ ˆ [I (β )]−1 = [− 2e β ˆ (e β + Asymptotic standard error 3 − ˆ (e β + ˜ Var (β ) = 3)2 √ −1 = 2.02062 2.02062 = 1.4215 ˜ β is asymptotically normally distributed. A 95% conﬁdence interval for β is ˆ β ±z1−0.05/2 24/45 2)2 ˆ 3e β 1 ˜ var (β ) = log 6±1.96×1.4215 = (...
View Full Document

## This document was uploaded on 04/03/2014.

Ask a homework question - tutors are online