Need help understanding this equality

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SUMMARY

The equality \(\sum_{i=1}^n (x_i - u)^2 = \sum_{i=1}^n (x_i - x_{avg})^2 + n(x_{avg} - u)^2\) holds true for a constant \(u\) and the average \(x_{avg} = \frac{\sum x_i}{n}\). The proof involves expanding both sides using the definition of \(x_{avg}\) and simplifying the expressions. Key steps include substituting \(\sum x_i\) with \(nx_{avg}\) and recognizing that the terms can be canceled to demonstrate equality. Understanding this relation is crucial for applications in probability theory.

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mkkrnfoo85
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Hello, I am really bothered by this because I can't seem to be able to prove the following is true.

\sum_{i=1}^n (x_i - u)^2 = \sum_{i=1}^n (x_i - x_{avg})^2 + n(x_{avg} - u)^2

for u = constant
and x_{avg} = \frac{\sum x_i}{n}

I just can't seem to grasp how I can prove that the right-side equals the left-side. Any push in the right direction would be extremely helpful.

Thanks in advance.

-Mark
 
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What've you tried doing? Have you tried expanding the RHS and using the definition of xavg? What do you get after doing this?
 
Your definition of "xavg" is wrong. You need
x_{avg}= \frac{\sum x_i}{n}
so \sum x_i= nx_{avg}

Multiply it out on the left: \sum (x_i^2- 2ux_i+ u^2)= \sum x_i^2- 2u\sum x_i+ u^2 \sum 1
Of course, \sum x_i= nx_{avg} and \sum 1= n so that is \sum x_i^2- 2nux_{avg}+ nu^2

Now do the same on the right and you should be able to cancel some things and arrive at the same answer.
 
oh that helped a lot. Thx, cristo and halls, and thanks for the correction. I got to the conclusion:

n(u^2-2ux_{avg}) == n(u^2-2ux_{avg})

=).

I had one more question. I need to know this relation to use in a proof in my 'probability' class. Is there an intuitive way to look at the equality to make it seem more obvious, and/or easier to memorize? Any takes on that would be helpful. Thanks again.

-Mark
 

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