im new to Statistics so please just dont blast me if my question seems stupid for you :(

I studied the Central Limit theorem via the Khan Academy video, and know i know that

The central limit theorem states that the sampling distribution of the mean of any independent, random variable will be normal or nearly normal, if the sample size is large enough.

However, i listen to many people saying that

the sum of many independent random variables will tend towards a normal distribution

The fact is that im new able to reconcile this 2 definitions. Can you help me ?

  • 2
    $\begingroup$ Almost! In the second definition, it's the average that converges to a normal distribution. If you make that correction and think about it, these two definitions are actually saying the same thing! $\endgroup$ – CulDeVu Jan 19 '18 at 13:11

Suppose you have a random variable $X$ with finite mean $\mu$ and variance $\sigma^2$; it does not have to be normally distributed

Now suppose you take a sample size $n$ from this, so each value $X_1,X_2,\ldots, X_n$ is independently and identically distributed. You could consider the mean $\bar{X}_n = \frac1n \sum_1^n X_n$ which will have mean $\mu$ and variance $\frac{\sigma^2}{n}$, or the distribution of the distribution of the sum of these $S_n=\sum_1^n X_n$ which will have mean $n\mu$ and variance $n \sigma^2$

Informally the Central Limit Theorem says each of $\bar{X}_n$ and $S_n$ will be approximately normally distributed with their respective means and variances. More formally, as $n$ increases:

  • $\sqrt{{n}}\left(\bar{X}_n-\mu\right)$ converges in distribution to that of a normal distribution with mean $0$ and variance $\sigma^2$

  • $\frac{S_n-n\mu}{\sqrt{n}}$ converges in distribution to that of a standard normal distribution with mean $0$ and variance $\sigma^2$

These two statements are equivalent, and represent your two statements


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.