# Asymptotic distribution of the MLE of an exponential via the CLT

Let $$X_1,..,X_n$$ i.i.d from an exponential distribution : \begin{align*} f(x) = \lambda e^{-\lambda x} , x > 0 \end{align*} I've computed the MLE and the Fisher information number : \begin{align*} \lambda_{MLE} &= \frac{1}{\bar{X}} \\ I(\lambda) &= \frac{1}{\lambda^2} \end{align*} I therefore obtain the following asymptotic distribution for $$\lambda_{MLE}$$ : \begin{align} \sqrt{n} (\lambda_{MLE} - \lambda_0) \xrightarrow{D} \mathcal{N}(0,\lambda^2) \end{align} I wish to check that my result is correct by using the CLT theorem and the Delta Method. CLT : \begin{align} \sqrt{n} (\bar{X}-\mu) \xrightarrow{D} \mathcal{N}(0,1/\lambda^2) \end{align} Because $$\lambda^2$$ is the variance of an exponential random variable.

The delta method states that : \begin{align} \sqrt{n}(g(\hat{\theta}) -g(\theta)) \xrightarrow{D} \mathcal{N}(0,(g'(\theta))^2 \sigma^2(\theta) ) \end{align} With in this case : \begin{align} g(x) &= \frac{1}{x} \\ \frac{dg(x)}{dx} &= \frac{-1}{x^2} \end{align} Therefore : \begin{align} \sqrt{n}(\lambda_{MLE} - \lambda_0) \xrightarrow{D} \mathcal{N}(0,\Big[\frac{-1}{\lambda^2}\Big]^2 \frac{1}{\lambda^2}) \end{align} Which would yields an asymptotic variance of $$1/\lambda^6$$ which cannot be true. What am I doing wrong ?

When you use delta method, you should use $$\theta = \frac{1}{\lambda}$$ but not $$\theta = \lambda$$. In this case
\begin{align} \sqrt{n}(\lambda_{MLE} - \lambda_0) \xrightarrow{D} \mathcal{N}(0,\Big[\frac{-1}{\theta^2}\Big]^2 \frac{1}{\lambda^2}) \end{align}