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Need help understanding this equality

  • Thread starter mkkrnfoo85
  • Start date
  • #1
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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.

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

for [tex]u = constant[/tex]
and [tex]x_{avg} = \frac{\sum x_i}{n}[/tex]

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
 
Last edited:

Answers and Replies

  • #2
cristo
Staff Emeritus
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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?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
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Your definition of "xavg" is wrong. You need
[tex]x_{avg}= \frac{\sum x_i}{n}[/tex]
so [itex]\sum x_i= nx_{avg}[/itex]

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

Now do the same on the right and you should be able to cancel some things and arrive at the same answer.
 
  • #4
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oh that helped a lot. Thx, cristo and halls, and thx for the correction. I got to the conclusion:

[tex]n(u^2-2ux_{avg}) == n(u^2-2ux_{avg})[/tex]

=).

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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