# Prove that $l(\tau\sigma n^{(-1/2)})$ tends to $\chi^2_1(\tau^2)$

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 $\sum_{i=1}^n \frac{X_i-\mu}{1+\lambda(X_i-\mu)}=0$.

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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