Required number of random numbers for using Central Limit Theorem

I wanted to know how many i.i.d random variables have to be summed in order to be able to use Central Limit Theorem. I know it varies depending on the distribution, but does there exist any number $N$, that for any $n\geq N$, we can say sum of any $n$ random distribution has a negligible difference with Normal distribution? Can you introduce a reference for that to me?

Moreover, is it true that CLT cannot be applied for the sum of iid random variables with heavy tailed distributions such as Pareto and Zipf?

• Certainly the CLT establishes that when you have an infinite number of values they will converge to the normal distribution. The CLT doesn't prove that any finite number will so converge. Sep 6, 2015 at 18:13
• Possibly of interest: Regression when the OLS residuals are not normally distributed. Sep 6, 2015 at 19:02

There will never be such a number. Convergence will always depend on properties of your random variables. Having said that, there are some results you may want to use. There is Berry Eseen (note there is a bound for the constant in the theorem $C < 0.4748$).

So, if $\mathbb{E}[X_i] = 0, 0 < \mathbb{E}[X_i^2] = \sigma^2 < \infty$ and $\mathbb{E}[|X_i|^3] = \rho < \infty$, then the maximal difference between the cumulative distribution of $\frac{X_1 +...+ X_n}{\sigma \sqrt{n}}$ and that of a standard normal is smaller than $\frac{0.4748\rho}{\sigma^3 \sqrt{n}}$ for every $x\in \mathbb{R}$.

Then you might say this doesn't make much sense, all this bounding stuff. In which case I would suggest you first ask yourself - "what exactly am I looking for?". You are certainly not the first to ask this kind of question and the answer is going to depend on how you phrase the question.

Classic CLT holds if the distribution has two moments (check wikipedia). For a Pareto distribution, the existence of the integrals corresponding to the moments depends on $\alpha$ (again, see wikipedia).

In theory, there is no $N$, such that $\forall n\geqslant N$, sum of any $n$ random distribution has a negligible difference with Normal distribution. Proof: for any $N$, make you i.i.d a Bernoulli distributed with $p=\frac 1N$.

In practise, sampling distribution can be very normal-like even for a very heavy-tailed distributions. That is why t-test is known to be robust, even when dealing with heavy-tailed distributions, when you have large sample sizes. In my application (mobile gaming), 10s of thousands of users are enough to make sampling distribution approach normal, for practical applications (e.g. I can use normal approximation, instead of costly brute-force sampling).

If you have a particular application, you can run a simulation, by sampling from your distribution/sample and test sampling distribution for normality.