I have a bunch of iid $\{X_i\}$ with $X_i \sim \exp(\lambda)$ - let's say $\lambda = 1$. Now, classic version of CLT tells me: \begin{equation} \sqrt{n}\left(1-\bar{X}_n\right) \rightarrow \mathcal{N}\left(0,\frac{1}{\lambda^2}\right) \end{equation} in distribution. But doesn't the convergence to a standard normal implies a probability $> 0$ of negative draws of the sample mean? - Which cannot be drawn as the exponential distribution has a positive support.

I am curious. And for a practical reason as well, because as in my real world example I want to apply Baysian estimate for the mean of a random variable with non-negative support. My assumption is that I have iid $\{X_i\}$ and a known variance $\sigma^2$. I want to estimate the mean of the RV. Now, the Likelihood is normal and the informative prior as well.

But if feels weird to apply a prior with negative support on to estimate the mean of a random variable with positive support. And I guess the situation is not changed by applying a Normal-inverse-gamma distribution as prior.