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Thread: Properties of Summations

  1. #1
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    Properties of Summations

    Hello,

    I'm working through a proof of the coefficient of linear regression (r) from its verbose form to its concise one. I realize the concept is statistics, but the the proof seems more algebra and possibly calculus-based.

    Verbose Form:
    $\displaystyle
    r = \frac{n\sum xy - \sum x \sum y}{\sqrt{n\sum x^2 - (\sum x)^2} \sqrt{n\sum y^2 - (\sum y)^2}}
    $

    Concise Form:
    $\displaystyle
    r = \frac{\sum (z_{x} z_{y})}{n-1}
    $

    It seems easiest to work backward from the Concise Form to the Verbose one, and to do so I'm using the following definitions:

    $\displaystyle
    z_{x} = \frac{x - \bar{x}}{s_{x}}
    $
    $\displaystyle
    \bar{x} = \frac{\sum x}{n}
    $

    $\displaystyle
    s_{x} = \sqrt{\frac{n\sum x^2 - (\sum x)^2}{n(n-1)}}
    $

    Doing basic substitution in the Verbose Form I get (didn't substitute for $\displaystyle s_{x}$ or $\displaystyle s_{y}$ to keep some semblance of readability):
    $\displaystyle
    r = \frac{\sum (\frac{x - \frac{\sum x}{n}}{s_{x}} * \frac{y - \frac{\sum y}{n}}{s_{y}})}{n-1}
    $

    I'm kind of stuck on what to do with the numerator, which seems to result in distributing a summation to other summations. Since x and y are two "paired" sets n will be the same for all summations and also a constant.

    To simplify my question:

    Am I able to distribute the $\displaystyle \sum$ like so?:
    $\displaystyle \sum (\frac{x - \frac{\sum x}{n}}{s_{x}})$ -> $\displaystyle \sum(\frac{\frac{nx - \sum x}{n}}{s_{x}})$ -> $\displaystyle \frac{\frac {n\sum x - \sum \sum x}{n}}{\sum s_{x}}$

    If so, what would the term $\displaystyle \sum \sum x$ resolve to?


    Note: I don't want anyone to solve the proof here, I'm just trying to understand how I might be able to resolve the summations. I'd like to work through the proof myself to understand how it works.

    Whew, okay first time using LaTeX that took a lot out of me...
    Last edited by jstandard; Apr 22nd 2010 at 03:04 PM.
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  2. #2
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    Quote Originally Posted by jstandard View Post
    Hello,

    I'm working through a proof of the coefficient of linear regression (r) from its verbose form to its concise one. I realize the concept is statistics, but the the proof seems more algebra and possibly calculus-based.

    Verbose Form:
    $\displaystyle
    r = \frac{n\sum xy - \sum x \sum y}{\sqrt{n\sum x^2 - (\sum x)^2} \sqrt{n\sum y^2 - (\sum y)^2}}
    $

    Concise Form:
    $\displaystyle
    r = \frac{\sum (z_{x} z_{y})}{n-1}
    $

    It seems easiest to work backward from the Concise Form to the Verbose one, and to do so I'm using the following definitions:

    $\displaystyle
    z_{x} = \frac{x - \bar{x}}{s_{x}}
    $
    $\displaystyle
    \bar{x} = \frac{\sum x}{n}
    $

    $\displaystyle
    s_{x} = \sqrt{\frac{n\sum x^2 - (\sum x)^2}{n(n-1)}}
    $

    Doing basic substitution in the Verbose Form I get (didn't substitute for $\displaystyle s_{x}$ or $\displaystyle s_{y}$ to keep some semblance of readability):
    $\displaystyle
    r = \frac{\sum (\frac{x - \frac{\sum x}{n}}{s_{x}} * \frac{y - \frac{\sum y}{n}}{s_{y}})}{n-1}
    $

    I'm kind of stuck on what to do with the numerator, which seems to result in distributing a summation to other summations. Since x and y are two "paired" sets n will be the same for all summations and also a constant.

    To simplify my question:

    Am I able to distribute the $\displaystyle \sum$ like so?:
    $\displaystyle \sum (\frac{x - \frac{\sum x}{n}}{s_{x}})$ -> $\displaystyle \sum(\frac{\frac{nx - \sum x}{n}}{s_{x}})$ -> $\displaystyle \frac{\frac {n\sum x - \sum \sum x}{n}}{\sum s_{x}}$

    If so, what would the term $\displaystyle \sum \sum x$ resolve to?


    Note: I don't want anyone to solve the proof here, I'm just trying to understand how I might be able to resolve the summations. I'd like to work through the proof myself to understand how it works.

    Whew, okay first time using LaTeX that took a lot out of me...
    For your double sum, what is the indexing?
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  3. #3
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    It actually doesn't list the indexing in the formula definition (at least in the text I'm using), just $\displaystyle \sum$

    I would guess it would have to be: $\displaystyle \sum_{i=1}^{n} \sum_{i=1}^{n} x_{i}$ since all of the calculations are done for 2 sets of values, both sets having n number of values (since they're paired).

    I should mention, I'm not even certain if my formula progression is correct and I'm able to "distribute" the summation in that manner.
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  4. #4
    Senior Member
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    The sum $\displaystyle \sum_{i=1}^nx_i$ has only n as a free variable. Therefore the sum $\displaystyle \sum_1^n\sum_{i=1}^n x_i$ is just $\displaystyle n \sum_{i=1}^n x_i$. But I don't think that's what this sum actually is. Note that you may NOT (!!) distibute a sum across a quotient vis. $\displaystyle \sum \frac ab \neq {\sum a \over \sum b}$, which it looks like you've done.

    I'm confused about what the summation is over in $\displaystyle s_x$. There do not appear to be any free variables under the summation sign.
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  5. #5
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    maddas,

    Thanks, that makes sense about the not being able to distribute the sum across a fraction.

    For the summation in the denom, $\displaystyle s_{x} = \sqrt{\frac{n\sum_{i=1}^n x_i^2 - (\sum_{i=1}^n x_i)^2}{n(n-1)}}$

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  6. #6
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    Ah, I see. What text are you working out of?
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  7. #7
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    The text is: Elementary Statistics 11th ed, Mario F. Triola.

    All of the summations in the book are listed simply as $\displaystyle \sum$ without any indexing.

    The original formulas are the ones I have above in the "Concise" and "Verbose" forms, where I'm trying to track the proof from "Concise" to "Verbose" using whatever manipulations possible.

    I should mention I spoke with a tutor at school who also came to the conclusion that the double summation term resolves to $\displaystyle n \sum_{i=1}^n x_i$, but that presents a problem because then you'd have:
    $\displaystyle n\sum_{i=1}^n x_i - \sum _{i=1}^n \sum _{i=1}^n x_i$ -> $\displaystyle n \sum_{i=1}^n x_i - n \sum_{i=1}^n x_i$ -> = 0. Eliminating the terms you need in the numerator of the "Verbose" form of the equation.

    Ha, sorry, I realize the "medium" of communication for this isn't necessarily the greatest, so let me know if I'm not doing a good job explaining things. I'm somewhat in new territory here.
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