In our Probability and statistics materials, I run into this equation:
$X_i$ is one result (I hope)
$n$ is a count of the results
$\overline{X_n}$ is sample mean of the results
It is connected with point estimations, we are trying to estimate variance, and this adjustment should help. But I don't understand how we can do it.
I expect it has something to do with this adjustment:
$(a-b)^2=a^2-2ab+b^2$
But I just don't see how (especially the minus sign confuses me). I understand that the $(\overline{X_n})^2$ was removed from the sum. But where is the $2ab$ part?
So, How can the adjustment be accurate?
It's actually pretty simple but the brackets in your picture might be a bit confusing. For simplicity reasons I'll only look at the sum.
We know that $$\sum^{n}_{i=1}(X_i - \bar{X_{n}})^2=(\sum^{n}_{i=1}X_i^2)-2\bar{X_n}\sum^{n}_{i=1}X_i+\sum^{n}_{i=1}\bar{X_n}^2$$
Which in turn reduces to $$(\sum^{n}_{i=1}X_i^2)-2\bar{X_n}\sum^{n}_{i=1}X_i+\sum^{n}_{i=1}\bar{X_n}^2=(\sum^{n}_{i=1}X_i^2)-2\bar{X_n}n\bar{X_{n}}+n\bar{X_n}^2$$
Because ofcourse we know that $\frac{1}{n}\sum^{n}_{i=1}X_i=\bar{X_n}$ so $\sum^{n}_{i=1}X_i=n\bar{X_n}$. Here is now the thing we want because $$(\sum^{n}_{i=1}X_i^2)-2\bar{X_n}n\bar{X_{n}}+n\bar{X_n}^2=(\sum^{n}_{i=1}X_i^2)-n\bar{X_n}^2$$
Put the constant of $\frac{1}{1-n}$ that we left out for simplicity, back in front, and you're done!