Proving a Sum Identity for Variance Estimator

Question

I'm trying to prove the following identity:

for a set of numbers: $$ x_1, ... ,x_n $$

$$\frac{1}{n(n-1)} \sum_{i=1}^n (x_i- \bar x)^2 = \bar x^2 - \frac{1}{n(n-1)} \sum_{i \neq j} x_ix_j $$

but I cant manage to prove it.

I found this identity:

$$ \frac{1}{n(n-1)} \sum_{i=1}^n (x_i- \bar x)^2 = \frac{1}{n(n-1)}((\sum_{i=1}^n x_i^2 ) - n\bar x) $$

but couldnt advance more.

Thanks!

Answer 1

Note that $$ \sum_{i=1}^n (x_i-\bar{x})^2=\sum_{i=1}^nx_i^2-n\bar{x}^2, $$ and $$ n^2\bar{x}^2=\left(\sum_{i=1}^nx_i\right)^2=\sum_{i=1}^nx_i^2+\sum_{i\ne j}x_ix_j. $$

Answer 2

Try these to warm up:

What is $\sum x_i$? Write it out more explicitly using "..."

What is $(\sum x_i)^2$? Write it out the same way.

Can you split that last sum into the terms $x_ix_j$ where $i=j$ and terms where $i \neq j$?

Can you expand out $\sum (x_i - \bar x)^2$?