0
$\begingroup$

I was wondering what the difference is between $Z_n=\frac{S_n-n\mu}{\sigma \sqrt{n}}=\frac{X_1+\ldots+X_n -n\mu}{\sigma \sqrt{n}}$ and $Z_n=\frac{X-n\mu}{\sigma/\sqrt{n}}$?

What is the benefit of one over the other?

$\endgroup$
2
  • 3
    $\begingroup$ I am not sure you have written a true equation? \begin{align} Z_n &=\frac{X_1+\ldots+X_n -n\mu}{\sigma \sqrt{n}}\\ & =\frac{n(\frac{X_1+\ldots+X_n}{n}-\mu)}{\sigma \sqrt{n}}\\ &= \frac{\bar{X}-\mu}{\sigma/\sqrt{n}} \end{align}. If you just have one X then why you need to normalize under $n\mu$? $\endgroup$
    – TPArrow
    Jul 29, 2015 at 6:25
  • $\begingroup$ Do you mean $\bar{X}$ rather than $X$ in your second one? $\endgroup$
    – Glen_b
    Jul 29, 2015 at 10:01

1 Answer 1

4
$\begingroup$

The Central Limit theorem refers to the limiting distribution of the sum of random variables, be it iid or simply independent. The equation is given in two equivalent forms, namely

$$Z=\frac{\sum_{i=1}^n X_i-n\mu}{\sqrt{n}\sigma}=\sqrt{n}\frac{\bar{X}-\mu}{\sigma} \xrightarrow{D} \mathcal{N}\left(0,1 \right)$$

The equations tell us that the distribution of the standardized version of either the sum or the sample mean converges to the distribution of a standard normal variable. By standardized we mean centering at the mean and dividing by the standard deviation so that the variance equals one. It is worth noting that this is not the case for random variables that do not have finite variance, e.g. samples from the Cauchy distribution.

$\endgroup$

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.