# CLT for random variables with positive support (e.g. exponential)

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: $$$$\sqrt{n}\left(1-\bar{X}_n\right) \rightarrow \mathcal{N}\left(0,\frac{1}{\lambda^2}\right)$$$$ in distribution. But doesn't the convergence to a standard normal implies a probability $$> 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.

• You can put a Gamma prior on $\lambda$, since it is the conjugate prior to the exponential distribution: en.wikipedia.org/wiki/Conjugate_prior Commented Jan 27, 2021 at 13:02
• Surely you can draw a sample where $\overline{X}_n$ has negative deviation from its expected value 1. So indeed you should have a positive probability for this happening. Commented Jan 27, 2021 at 13:17

You need to be careful with the order of quantifiers in understanding what (this version of) CLT is claiming. In particular, the order of $$x$$ (the point where you evaluate your CDF) and $$n$$ (sample size).
Convergence in distribution means that if you pick a value $$x$$ (yes, it can be negative), and consider the value of $$F_n(x) = P(\sqrt{n}(\overline{X}_n - 1) \le x)$$ at that point, then that value tends to $$\Phi(x)$$ as $$n \to \infty$$. Yes, that limit is positive (for all $$x$$).
This is indeed what happens. Let's be super concrete and consider $$x=-3$$. Can you have $$\sqrt{n}(\overline{X}_n-1) \le -3$$? Yes, consider for example $$n=10$$. Your sample mean $$\overline{X}_n$$ could be $$0.001$$, and then your $$\sqrt{n}(\overline{X}_n-1)$$ would be $$-3.159$$. There is a whole range of possible sample means that map to below $$-3$$, so indeed that event has a positive probability. Not a big one, but positive. It tends to $$\Phi(-3) \approx 0.0013$$ as $$n \to \infty$$. (Keeping $$x$$ fixed at $$-3$$.)
You can repeat the same experiment with some other value of $$x$$, say $$-10$$. Then you'll find that for $$n$$ large enough, the deviation of your sample mean, scaled by multiplying by $$\sqrt{n}$$, could indeed be that much negative.