I was looking into Central Limit Theorems and how a CLT is derived and I found this source quite helpful. The only thing I am having trouble to comprehend is the transformation of the formula

$$\frac{\overline X_n - E[\overline X_n]}{\sqrt{Var[\overline X_n]}}\overset{d}{\rightarrow}Z$$

in the case of iid variables. With iid variables, the formula becomes

$$\sqrt{n}\frac{\overline X_n - E[X_i]}{\sqrt{Var[X_i]}}\overset{d}{\rightarrow}Z$$

because of

$$E[\overline X_n]=E[X_i]\quad \text{ and }\quad Var[\overline X_n]= \frac{Var[X_i]}{n}.$$ How can the two assumptions regarding the expected value and the variance be derived fromm iid variables and why is the variance of a single observation i divided by the number of observations n the same as varaince of the sample mean?


1 Answer 1


If you have IID random variables $X_1, X_2, ..., X_n$, then the average $\bar{X}_n = \frac{1}{n}\sum_{i=1}^n X_i$ has expectation $E(\bar{X}_n) = \frac{1}{n}\sum_{i=1}^n E(X_i) = E(X_1)$. This previous fact has nothing to do with the fact that we have independence though! This is just from the identical distribution of the $X_i$'s and linearity of expectation.

We use the independence assumption when we calculate $Var(\bar{X}_n) = Var \left(\frac{1}{n}\sum_{i=1}^n X_i\right) = \frac{1}{n^2}\sum_{i=1}^n Var(X_i) = \frac{1}{n^2} \cdot n Var(X_i) = \frac{Var(X_1)}{n}.$

Notice in both cases the $X_1$ in the final expressions can be replaced by any of the $X_i$'s, since they are all identically distributed.

  • $\begingroup$ Ah I see, thanks a lot for the explanations! $\endgroup$
    – Phil
    Nov 21, 2020 at 23:23

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