An equality with summations

**datanewb** · August 16th 2012, 03:10 PM

I don't understand how the following is equivalent.

$n\Sigma(X_{i}-\overline{X})^2 = n\Sigma X_{i}^2 - (\Sigma X_{i})^2$

$\overline{X}$ is the average of all $X_{i}$ So, $\overline{X} = \Sigma X_{i}/n$

Could someone walk me through this. I've been staring at it for a long while now. I'll try to show my confusion below:

So, expanding the left side I get
$n\Sigma(X_{i}^2 - 2X_{i}\overline{X} + \overline{X}^2)$

Then substituting for $\overline{X}$ , I get

$n\Sigma(X_{i}^2 - 2X_{i}\Sigma X_{i}/n + (\Sigma X_{i}/n)^2$

Then multiplying it out, I get

$n\Sigma X_{i}^2 -\Sigma(2X_{i}\Sigma X_{i}) + \Sigma((\Sigma X_{i})^2/n)$

And I'll stop there, because I really don't know what taking the summation of a summation means???

But, I do think I'm not too far off track because I now see the first term from the right hand side of the initial equality.

**GJA** · August 16th 2012, 03:16 PM

Hi, datanewb.

You're definitely on the right track, nice work!

There is some confusion, because you've used the same index to represent both the original summation and the summation needed for $\overline{X}.$ Since the i index is used in the orginal summation, we would want to use a new dummy index, e.g. write

$\overline{X}=\frac{1}{n}\sum_{j}X_{j}$ .

Then the double summation is a summation over the indices i and j.

Does this help? I can write up more details if it's still confusing.

Good luck!

**datanewb** · August 16th 2012, 03:57 PM

Thank you, GJA, that definitely does help.

Rewriting the last line
$n \sum_{i}X_{i}^2 - \sum_{j}^n(2X_{j}\sum_{i}X_{i}) + \sum_{j}^n((\sum_{i}X_{i})^2/n^2)$

is equivalent to

$n \sum_{i}X_{i}^2 - \sum_{j}^n(2X_{j}\sum_{i}X_{i}) + ((\sum_{i}X_{i})^2/n)$

I think... okay, I need to think about this a little bit more. Hopefully I will solve it later tonight!

**GJA** · August 17th 2012, 07:32 AM

Glad it helped! I think expanding $\overline{X}$ a line or two later might simplify things. Good luck and nice job sticking with a tricky little problem like this.

**datanewb** · August 17th 2012, 09:44 AM

Okay, I finally solved it! Thank you for the advice and for giving me the space and time to figure it out!

I was simply too unfamiliar with algebraic manipulations of sigma notation to figure this out.

I had to realize the following:

$\sum_{i}^{n-1}(X_{i} + C) = \sum_{i}^{n-1}X_{i} + (n-i)C$

Secondly,
$\sum_{i}\frac{1}{n}X_{i} = \frac{1}{n}\sum_{i}X_{i}$

Finally, I had to realize that

$\sum_{i}(X_{i}\sum_{i}X_{i}) = \sum_{i}X_{i}\sum_{i}X_{i} = (\sum_{i}X_{i})^2$

Please, if any of this is wrong, correct me.

Using the above logic, I was able to see that

$n\sum_{i}X_{i}^2 - n\sum_{i}(2X_{i}\overline{X}) + n^2\overline{X}^2 =$

and substituting $\overline{X}$ with $\sum_{j}\frac{1}{n}X_{j}$

$n\sum_{i}X_{i}^2 - n\sum_{i}(2X_{i}\sum_{j}\frac{1}{n}X_{j}) + n^2(\sum_{j}\frac{1}{n}X_{j})^2 =$

$n\sum_{i}X_{i}^2 - 2\sum_{i}(X_{i}\sum_{j}X_{j}) + (\sum_{j}X_{j})^2 =$

and since $i\equiv{j}$

$n\sum_{i}X_{i} - (\sum_{i}X_{i})^2$

Which is exactly the right hand side of the equality I was trying to understand initially! Woohoo!

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