Suppose that for $t=1,\dots,T$, $x_t$ is an i.i.d. draw from a continuous distribution with p.d.f. $f(x_t;\theta_0)$.
Let $l(\theta;X):=\sum_{t=1}^T{\log{f(x_t;\theta)}}$ be the corresponding log-likelihood.
Let $\hat\theta:=\arg\max_\theta{l(\theta;X)}$ be the corresponding ML estimate.
Finally, define:
$$e:=l(\hat\theta;X)-\left.\left[\mathbb{E}l(\theta;X)\right]\right|_{\theta_0=\hat\theta, \theta=\hat\theta}$$
Now, for fairly standard distributions, $e=0$. E.g. for the normal distribution: $$l(\hat\theta;X)=-\frac{T}{2}\log{2\pi}-T\log\hat\sigma-\frac{T}{2}=\left.\left[\mathbb{E}l(\theta;X)\right]\right|_{\theta_0=\hat\theta, \theta=\hat\theta}.$$
Is this really a general property though? I suppose it is equivalent to the assertion that the log-likelihood evaluated at the ML estimate is only a function of the data via $\hat\theta$. This sounds a bit like $\hat\theta$ being a sufficient statistic, so perhaps it is only a property of distributions from the exponential family. (Side question: is this a theorem?)
My main questions then are the following:
Can you find a distribution $f$ for which $e\ne 0$?
Does one such $f$ exist for which the ML estimator admits a closed form solution?