Understanding simple LRT test asymptotic using Taylor expansion?

Question

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.

Answer 1

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

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