# Understanding simple LRT test asymptotic using Taylor expansion?

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.

• 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. May 28, 2019 at 12:16
• How to get $ln(\hat \theta)=0$? May 28, 2019 at 13:23
• @user158565 $\hat{\theta}$ is the MLE. I meant $ln'(\hat{\theta})$ not $ln(\hat{\theta})$ sorry, typo May 28, 2019 at 14:35
• 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. Aug 30, 2022 at 11:13

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