# Why is it that when you add normally distributed random variables the variance gets larger but in the Central Limit Theorem it gets smaller?

When you add two independent normal distributions the resulting distributions' variance is the sum of the variances i.e. it gets larger. However, the Central Limit Theorem states that

when independent random variables are added, their properly normalized sum tends toward a normal distribution (informally a bell curve) even if the original variables themselves are not normally distributed. https://en.wikipedia.org/wiki/Central_limit_theorem

and also says that as you add more variables the variance gets smaller.

Why does the variance increase in one case but decrease in the other? Is it because in the CLT case the random variables come from the same underlying distribution?

• The CLT deals with sample mean $\bar X,$ for which the standard error is $SD(\bar X) = \sigma/\sqrt{n}.$ – BruceET Nov 23 '20 at 19:45
• True, but how's this a comment and not an answer? – Lewian Nov 23 '20 at 21:16

Let $$X_1, X_2\overset{iid}{\sim}F_X(x)$$ with $$\mu$$ and $$\sigma^2$$ defined and finite.

$$var(X_1+X_2) = var(X_1) + var(X_2) = 2\sigma^2$$

But...

$$var(\bar{X}) = var\bigg(\dfrac{X_1 + X_2}{2}\bigg) = \dfrac{1}{4}var(X_1 + X_2) = \dfrac{1}{2}\sigma^2$$

(Remember that variance is not linear; the constant pulls out of the variance operator as a squared value.)

Using more than two $$iid$$ variables results in that familiar $$var(\bar{X}) = \dfrac{\sigma^2}{n}$$.

$$var(\bar{X}) = var\bigg(\dfrac{1}{n}\sum_{i=1}^n X_i\bigg) = \dfrac{1}{n^2}var\bigg(\sum_{i=1}^n X_i\bigg) = \dfrac{1}{n^2} n\sigma^2 = \dfrac{\sigma^2}{n}$$