I am trying to understand the proof that the LRT test for

$$H_0: \theta = \theta_0 \quad vs \quad H_1: \theta \neq \theta_0$$

is asymptotically $\chi_1^2$. I am reading the proof presented in Casella Berger (2002, Ch. 10).

They first use a taylor expansion to write the log-likelihood under the null as

$$\ell(\theta_0) = \ell(\hat{\theta}) + (\theta_0-\hat{\theta})\ell'(\hat{\theta}) + \frac{1}{2} (\theta_0-\hat{\theta})^2 \ell''(\hat{\theta})^2 +R_n$$

and then substitute this into the test statistic:

$$-2 \log \lambda (X) = -2 \log \frac{\mathcal{L}(\theta_0)}{\mathcal{L}(\hat{\theta})}=-2\left(\ell(\theta_0) - \ell(\hat{\theta})\right).$$

After doing this, I end up with

$$-2\log \lambda (X) = -2\ell(\hat{\theta})-2\left(\theta_0 - \hat{\theta}\right)\ell'(\hat{\theta}) - (\theta_0 - \hat{\theta})^2\ell''(\hat{\theta})-(-2\ell(\hat{\theta}))$$

$$=- (\theta_0 - \hat{\theta})^2\ell''(\hat{\theta})$$

since $\ell'(\hat{\theta})=0$ as it is the MLE.

However Casella & Berger arrive at

$$-2\log \lambda(X) = \frac{(\theta_0-\theta)^2}{-\ell''(\theta)}$$

I don't understand how the observed fisher information is in the denominator, when it is multiplied by $(\theta_0 - \hat{\theta})$ in the Taylor expansion.

Also, wouldn't we require it to be in the numerator anyway as $\sqrt{n}(\theta-\theta_0) \to \mathcal{N}(0, I^{-1}(\theta))$? The proof in question can be found on page 489 here.

  • $\begingroup$ For what is worth, I had a quick look and obtained the same result, it seems it would be in the numerator. I am curious to see if somebody have an explanation. $\endgroup$
    – matteo
    May 28, 2019 at 12:16
  • 1
    $\begingroup$ How to get $ln(\hat \theta)=0$? $\endgroup$
    – user158565
    May 28, 2019 at 13:23
  • $\begingroup$ @user158565 $\hat{\theta}$ is the MLE. I meant $ln'(\hat{\theta})$ not $ln(\hat{\theta})$ sorry, typo $\endgroup$
    – Xiaomi
    May 28, 2019 at 14:35
  • $\begingroup$ I am reading this book recently. I got the same derivation as yours. I believe there is a typo in the book here. How is the $\chi^2_1$ then arrived at? We can go back to Theorem 10.1.12 of the same book, where it indicates that $\sqrt{n}(\hat\theta-\theta)\to n(0,1/I(\theta_0))$. So your guess on this distribution is correct. $\endgroup$
    – Wenxu
    Aug 30, 2022 at 11:13

1 Answer 1


it can be seen by using the properties of logarithms to bring an exponent in front:

$$ ln''(\hat{\theta}) = ln''(\hat{\theta})^{(-1)(-1)} = -\ln''(\hat{\theta})^{-1}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.