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Statistics Sum of Squares x*y Proof

  1. Sep 13, 2014 #1
    1. The problem statement, all variables and given/known data

    Prove that

    [tex]\sum[(x_{i} - \overline{x})(y_{i} - \overline{y})] = \sum[(x_{i} - \overline{x})y_{i}][/tex]


    2. Relevant equations
    None.


    3. The attempt at a solution
    I tried using the fact that the sum of the mean values is just the mean value, because the sum of a constant is simply a constant, and I expanded out the first sum, but I didn't end up anywhere.

    [tex] \sum[x_{i}y_{i}] - \overline{y}\sum(x_{i}) - \overline{x}\sum(y_{i}) + \overline(x)\overline(y) [/tex]

    I have no idea where to go from here, so some inspiration would be much appreciated.
     
  2. jcsd
  3. Sep 13, 2014 #2

    Ray Vickson

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    I assume you have n values ##x_1, \ldots, x_n## and ##y_1, \ldots,y_n##. In that case your last term above, ##\bar{x} \bar{y}## is incorrect. Can You see why? You can also do other simplifications, but if I say more I will essentially be doing the question for you.
     
  4. Sep 13, 2014 #3

    LCKurtz

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    Notice that your first and third terms give what is on the right side of the equation. So you are left with showing$$
    -\sum x_i\bar y + \sum \bar y \bar x = 0$$You're pretty close...
     
  5. Sep 13, 2014 #4
    Ok, so I actually did end up getting somewhere and didn't realize it =)

    EDIT:

    I've done proof

    [tex] \sum[(x_{i}-\overline{x})y_{i}] + \sum[\overline{y}(\overline{x} - x_{i})][/tex]

    Playing around in excel has taught me that [tex]\sum[\overline{y}(\overline{x} - x_{i})][/tex] is actually just equal to zero, because your subtracting an array of x values from the mean and then just multiplying them by a constant, [tex]\overline{y}[/tex], then adding them, which nets you zero.
     
    Last edited: Sep 13, 2014
  6. Sep 13, 2014 #5

    Ray Vickson

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    Try writing out a few small examples, such as for n = 2 or n = 3. These are small enough that you can write down everything explicitly and see exactly what is going on.
     
  7. Sep 13, 2014 #6
    I've figured out why, but I'm not sure how to explain it symbolically, do you think I would need to elaborate more or is my edited explanation good enough in your view?

    Thanks for the help!
     
  8. Sep 13, 2014 #7

    LCKurtz

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    ##\bar x## and ##\sum x_i## are related to each other. How?
     
  9. Sep 13, 2014 #8
    I'm a little bit rusty on my operation of sums, so excuse me if this is incorrect, but can I do this?

    [tex]

    [(\overline{y})/n][\sum x_{i}] - \overline{y}[\sum x_{i}]

    = [(\overline{y}^2)/n][\sum(x_{i} - x_{i})]

    [/tex]

    Then what tends to zero appears to be obvious
     
  10. Sep 13, 2014 #9

    LCKurtz

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    I don't follow that at all. Why don't you just answer my question at the top?
     
  11. Sep 13, 2014 #10
    They're related by

    [tex]

    \overline{x} = [\sum x_{i}]/n

    [/tex]
     
  12. Sep 13, 2014 #11

    LCKurtz

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    So you could use ##\sum x_i = n\bar x##. Do you see how you might use that to get$$
    \sum \bar y(\bar x - x_i) = 0\text{?}$$
     
  13. Sep 13, 2014 #12
    Wow, it appears my brain took a nap today.

    Thank you for your help lol, I appreciate it.
     
  14. Sep 13, 2014 #13

    LCKurtz

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    So amuse me and show the finishing steps...
     
  15. Sep 13, 2014 #14
    Sorry, I'm very slow at typing this latex code..

    [tex]
    \overline{y}*[\sum(\overline{x} - x_{i})] = 0
    [/tex]

    [tex]
    [\sum(\overline{x} - x_{i})] = 0
    [/tex]

    [tex]
    \sum(\overline{x}) - \sum(x_{i}) = 0
    [/tex]

    [tex]
    n(\overline{x}) - n(\overline{x}) = 0
    [/tex]

    LHS = RHS

    used

    [tex]
    \sum(c) = n*c
    [/tex]

    Where n is the upper bound and 1 is the lower bound of the sum
     
  16. Sep 13, 2014 #15

    LCKurtz

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    Good. Thanks for finishing it up.
     
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