Asymptotic distribution of MLE of iid exponentials? If $Y_1, ..., Y_n$ are iid exponentially distributed with mean $\zeta$ then the Fisher information is $n/\zeta^2$, the MLE estimator is $\sum Y_i /n$, and the variance of $Y_i$ is $\zeta^2$. 
By a theorem of Cramër, the MLE is asymptotically normal with mean $\zeta$ and variance equal to $1/n$ times the inverse Fisher information, that is $\zeta^2 /n^2$. 
Yet by CLT, the MLE is asymptotically normal with mean $\zeta$ and variance equal to $1/n$ times the variance of $X_i$, that is $\zeta^2/n$. 
Which one is it? Why don't they agree? Did I make a mistake?
 A: First of all, the Fisher information in this case (as defined in the Wikipedia article) is $\frac{n}{\zeta^2}$. This can be derived easily enough.
From the CLT we have that $\frac{\bar Y_n - \zeta}{SE(\bar Y_n)} \rightarrow_d \mathcal N(0,1)$. 
$Var(\bar Y_n) = \frac{1}{n^2} \sum_{i=1}^n Var(Y_i) = \zeta^2/n$ (although we could have just used $I^{-1}(\zeta)$ instead) so
$$
\bar Y_n \rightarrow_d \mathcal N(\zeta, \zeta^2/n).
$$
As for the Cramer theorem, I'm not sure what you mean. Could you clarify which theorem you are referring to?
If you are referring to the Cramer-Rao bound, this tells us that $Var(\hat \zeta) \geq I^{-1}(\zeta) = \frac{\zeta^2}{n}$. This is the same variance in the CLT so there is no disagreement here; indeed, this says that asymptotically we are doing as well as we can.
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Here's the derivation that $I(\zeta) = \frac{n}{\zeta^2}$. By $Y_i$ iid we have
$$
L(\zeta ; \vec Y) = \prod_i \frac{1}{\zeta} \exp (\frac{-Y_i}{\zeta}) = {1 \over \zeta^n} \exp (-\frac{1}{\zeta} \sum_i Y_i)
$$
$$
\implies l(\zeta; \vec Y) := \log L(\zeta; \vec Y) = -\frac{1}{\zeta} \sum_i Y_i - n \log \zeta.
$$
Taking two derivatives we find that 
$$
l''(\zeta; \vec Y) = \frac{-2 \sum_i Y_i}{\zeta^3} + \frac{n}{\zeta^2}
$$
and therefore
$$
I(\zeta) = \mathbb E_Y(-l''(\zeta; \vec Y)) = \frac{n}{\zeta^2}.
$$
A: The Fisher information, defined as 
$$\int_\mathbb{R} \frac{[f'(u)]^2}{f(u)} du,$$
is independent of the sample size $n$. I didn't include any parameter $\theta$ here since the above definition applies to more general setting such as nonparametric inference. 

For your case, based on your description, $Y_1, \ldots, Y_n$ i.i.d. $\sim \text{Exp}(\zeta)$, where $\zeta$ is a scale parameter, thus the density of $Y_1$ is
$$f_Y(y; \zeta) = \frac{1}{\zeta}e^{-y/\zeta}, y > 0.$$
Hence by definition (once again, the standard definition of Fisher information doesn't depend on sample), the Fisher information is 
$$I(\zeta) = \int_0^\infty \left(\frac{1}{\zeta^3} - \frac{2y}{\zeta^4} + \frac{y^2}{\zeta^5}\right) e^{-y/\zeta}dy = \frac{1}{\zeta^2}$$
Since $\bar{Y} = \dfrac{1}{n}\sum_{i = 1}^n Y_i$ is MLE, based on MLE asymptotic theory (may correspond to your "Cramer theorem"), we have
$$\sqrt{n}(\bar{Y} - \zeta) \Rightarrow N(0, I(\zeta)^{-1}) = N(0, \zeta^2).$$
On the other hand, since $\text{Var}(Y_1) = \zeta^2$, it follows by classical CLT that
$$\sqrt{n}(\bar{Y} - \zeta) \Rightarrow N(0, \zeta^2).$$
So these two results agree very well.
In fact, Cramer's theorem is deduced from CLT (under some regularity conditions), so logically they are bounded to give the identical limiting distributions. 
