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.

Suppose that $X_1, \ldots ,X_n$ are i.i.d. sample with $E(X_1) = 0$ and $E(X_1^2) = \sigma^2 < \infty$. How can we prove that $l(\tau\sigma n^{-1/2})$ tends to $\chi^2_1(\tau^2)$?

Here $l(·)$ is the empirical likelihood ratio given by $l(\mu)=2\sum_{i=1}^{n}\log\{1+\lambda(X_i-\mu)\}$ for $\lambda$ satisfying $\sum_{i=1}^n \frac{X_i-\mu}{1+\lambda(X_i-\mu)}=0$.

share|improve this question
That "help" in the title would do nothing for you other than making it less informative. –  Gigili Mar 17 '13 at 17:27
¿¿¿$=2<1$??? ${}{}{}{}$ –  Pedro Tamaroff Mar 17 '13 at 17:34
@Gigili Look who's there! –  Pedro Tamaroff Mar 17 '13 at 17:34
Thanks a lot for the comments! The question has been corrected. –  XXX11235 Mar 17 '13 at 23:00

Your Answer


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

Browse other questions tagged or ask your own question.