# Variance formulae

I know that the variance formula is $$\sigma^2 = \frac{ \left( x_1 - \bar{x} \right) ^2 + \left( x_2 - \bar{x} \right) ^2 + \dots + \left( x_n - \bar{x} \right) ^2 }{n}$$ Where $\sigma^2$ is the variance; $x_1,\ x_2,\ \dots,\ x_n$ are the statistical data, and $n$ is the number of data.

My question is: how can I expand that formula to get this equivalent one:

$$\sigma^2 = \frac{x_1^2 + x_2^2 + \dots + x^2_n}{n} -\bar{x}^2$$

?

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Hi JnxF. Wellcome to MathStackExchange. I think you may be confusing statistical data $x_1, ...,x_n$ with random variables $X_1,...,X_n$ that have zero correlation $\sigma(X_1,X_2)=0,...,\sigma(X_{n-1},X_{n})=0$. – MathOverview Dec 16 '12 at 14:39
Thanks, but I have not much "mathematical level", I'm doing high school, and, by the moment, I don't know so much statistic. – JnxF Dec 16 '12 at 14:42

Each of the $(x_i - \bar x)^2$ terms expands into $x_i^2 - 2x_i\bar x + \bar x^2$.

Since the sum of all $x_i$ is also equal to $n \bar x$ (by definition), we get that the sum of all the $2x_i\bar x$ is actually $2n\bar x^2$.

Then, the equation becomes: $\displaystyle \sigma^2 = \frac{x_1^2 + x_2^2 + \cdots + x_n^2 - 2n\bar x^2 + n\bar x^2}{n}$

which is $\displaystyle \frac{x_1^2 + x_2^2 + \cdots + x_n^2 - n\bar x^2}{n}$ or $\displaystyle \frac{x_1^2 + x_2^2 + \cdots + x_n^2}{n} - \bar x^2$, as required

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Thanks, really clear explanation. – JnxF Dec 16 '12 at 14:37
The same trick works for finding the expected value of X^3, X^4, etc. as well. – Joe Z. Dec 25 '12 at 16:08
Sorry, I don't understand you. Can you explain it a bit more, please? Do you want to mean that is also possible to expand $(x_i - \bar{x})^n$ for any integer? – JnxF Dec 25 '12 at 21:35
Yes, and the result will still be $(x_1^k + x_2^k + \cdots + x_n^k)/n - \bar x^k$. – Joe Z. Dec 25 '12 at 23:07

Since $$\bar{x}=\frac1n\sum_{i=1}^nx_i$$ we have that \begin{align} \sigma^2 &= \frac{1}{n}\sum_{i=1}^n \left( x_i - \bar{x} \right) ^2\\ &=\frac1n\sum_{i=1}^nx_i^2-2x_i\bar{x}+\bar{x}^2\\ &=\frac1n\sum_{i=1}^nx_i^2-2\bar{x}\frac1n\sum_{i=1}^nx_i+\frac1n\sum_{i=1}^n\bar{x}^2\\ &=\frac1n\sum_{i=1}^nx_i^2-2\bar{x}\bar{x}+\frac nn\bar{x}^2\\ &=\frac1n\sum_{i=1}^nx_i^2-\bar{x}^2 \end{align}

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I hope you don't mind that I edited your answer to use the align environment to, in my opinion, improve readability. – robjohn Dec 16 '12 at 14:50
@robjohn No problem mr. robjohn. It is much better now – Nameless Dec 16 '12 at 14:51

Expand square terms and use the definition $\bar{x} = (x_1 + x_2 + .. + x_n)/n$

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