I was trying to understand asymptotic normality of the posterior better, and came across a confusing point. So let's say we have a likelihood, $$L(\theta | X) = \prod_{i=1}^n p(X_i | \theta), $$ so the log-likelihood is $$J(\theta) = \log L = \Sigma_{i=1}^n \log(p(X_i | \theta)).$$
$J$ is itself a sum of random variables, so the log-likelihood J will be asymptotically normal, by the central limit theorem.
But we can also show the likelihood is asymptotically normal through a Taylor expansion. Let $\hat{\theta}$ be the mle. So we have
$$J(\theta) = J(\hat{\theta}) + \nabla J \cdot (\theta-\hat{\theta}) + \frac{1}{2}(\theta-\hat{\theta})H(\theta-\hat{\theta}). $$ Since $\hat{\theta}$ is the mle, we know $\nabla J = 0$, and $I(\theta)=-H$ so this reduces to
$$J(\theta) = \log(L) = J(\hat{\theta}) - \frac{1}{2}(\theta-\hat{\theta})I(\theta)(\theta-\hat{\theta})\tag 1$$
Now exponentiating (1), we get
$$e^J = L = ke^{-\frac{1}{2}(\theta-\hat{\theta})I(\theta)(\theta-\hat{\theta})}, $$ which is also asymptotically normal, with $L \sim N(\hat{\theta},I(\theta)^{-1})$.
Am I making a mistake here?