This is what I think the technical statement of CLT is:

If we consider $\overline{X}_{n}$ coming from a sample of $\mathcal n$ independent and identically distributed random variables $X_{i}$ with mean $\mu$ and variance $\sigma^{2}$ then as n $\rightarrow \infty$

$\sqrt n \left( \left(\frac{1}{n} \ \sum_{i=1}^{n} X_{i} \right) - \mu \right) \approx N(0,\sigma^{2})$

Question: I've read textbooks which claim $n \ge 30$ without offering any support for the claim, why $n \ge 30$?

  • 1
    $\begingroup$ The rule that $n \ge 30$ is a rough rule-of-thumb. The Statistics Stackexchange has some good questions on this topic, see this or this. $\endgroup$ – George V. Williams Jan 28 '15 at 19:41
  • 1
    $\begingroup$ It is a very rough rule of thumb that only gives a reasonable approximation when the the distribution of the $X_i$ is "nice." If the $X_i$ have binomial distribution $p=0.01$, the approximation will be very poor at $n=30$. $\endgroup$ – André Nicolas Jan 28 '15 at 19:41

There are quantitative versions of the CLT which provide an estimate of the rate of convergence. One is given at http://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem. This says that the sequence of CDFs of the rescaled sample means converges uniformly to the appropriate normal CDF, provided the third moment of the variables exists.

Assuming the summands have finite third absolute moment, we get $|F_n - \Phi| \leq \frac{C \rho}{\sigma^3 \sqrt{n}}$ where $\sigma$ is the standard deviation of the summands and $\rho$ is the absolute third moment of the summands. Here $C$ does not depend on the variables, and has been shown to be less than $\frac{1}{2}$. So the practical convergence rate depends on how big $\rho$ is relative to $\sigma$.

Let's try this with the binomial. The standard deviation is $(p(1-p))^{1/2}$. The third absolute moment is $p (1-p)^3 + (1-p) p^3 = p (1-p) ((1-p)^2+p^2)=p (1-p) (1-2p+2p^2)$. The factor of $(1-2p+2p^2)$ is between $1/2$ and $1$, so we can essentially ignore that. The important thing is that $\frac{\rho}{\sigma^3}$ is on the order of $(p (1-p))^{-1/2}$. So for the binomial, the theorem estimates the error as being at most $\frac{1}{2 p^{1/2} (1-p)^{1/2} \sqrt{n}}$. Making this small could require $n$ to be quite large, if $p$ is close to $0$ or $1$.

A quantitative example: take $p=0.01$, then $\frac{1}{2 p^{1/2} (1-p)^{1/2}}$ is a little larger than $5$. In this case, the theorem only guarantees that the error is less than $0.1$ uniformly if $n>2500$. In reality this is probably rather blunt (indeed, a sharp argument for this special case can be found at http://en.wikipedia.org/wiki/De_Moivre%E2%80%93Laplace_theorem), but it gives an idea of how slow the convergence rate could be in the worst cases.

  • $\begingroup$ Fantastic answer. The assumption of a finite third moment, not just first or second, for a convergence rate is often forgotten. $\endgroup$ – Paul Jan 28 '15 at 20:42

$n \ge 30$ is Rule-of-Thumb. Basically to make distribution less skewed, uni-modal, and to make it look more like Normal Distribution.

The number of variables can be less than 30, for example if, population distribution is normal, or etc.

It really depends on population or sampling distribution,


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.