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10-30-2010   1 links from elsewhere to this Post. Click to view. #1 (permalink)
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Default [Statistics] Why can't |correlation| be above one?

We're not exactly expected to know this for our course/it's not explained in the textbook, but I feel like I can't grasp the true meaning of correlation until I find an answer for this.

So... correlation r = (n-1)^(-1)Σ[(x-xbar)/sx*(y-ybar)/sy]

I get how Σ(x-xbar/sx) would just be 0 because of adding all z scores together, how multiplying x-xbar/sx and y-ybar/sy has to do with the least-square regression idea, and how dividing n-1 is just with relationship to the sample; however, why |r|< 1 I have no idea. It's also really close to s^2 = (n-1)^(-1)Σ(x-xbar)^2, but I'm not sure if that's just by chance. If anyone can prove this algebraically or with logical deduction, I'd really appreciate it.
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11-04-2010   #2 (permalink)
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The sample correlation coefficient is defined as r = (n - 1)^{-1} Σ [(x_i - xbar)/s_x] [(y_i - ybar)/s_y].

Using s_x = sqrt[(n - 1)^{-1} Σ (x_i - xbar)^{2}] and s_y = sqrt[(n - 1)^{-1} Σ (y_i - ybar)^{2}], we can rewrite r as

(1) r = Σ [(x_i - xbar)/sqrt(Σ (x_i - xbar)^{2})] [(y_i - ybar)/sqrt(Σ (y_i - ybar)^{2})]

The triangle inequality states that the absolute value of a sum is never greater than the sum of the absolute values; in its simplest form, if a and b are real numbers, then |a + b| ≤ |a| + |b|. In this case, we get

(2) |r| = |Σ [(x_i - xbar)/sqrt(Σ (x_i - xbar)^{2})] [(y_i - ybar)/sqrt(Σ (y_i - ybar)^{2})]| ≤ Σ |(x_i - xbar)/sqrt(Σ (x_i - xbar)^{2})| |(y_i - ybar)/sqrt(Σ (y_i - ybar)^{2})|.

Note that the terms |(x_i - xbar)/sqrt(Σ (x_i - xbar)^{2})| and |(y_i - ybar)/sqrt(Σ (y_i - ybar)^{2})| are normalized distances; that is,
(3) Σ |(x_i - xbar)/sqrt(Σ (x_i - xbar)^{2})|^{2} = Σ |(y_i - ybar)/sqrt(Σ (y_i - ybar)^{2})|^{2} = 1.

Now, for any real numbers a and b, (a - b)^{2} = a^{2} - 2ab + b^{2} ≥ 0. Shifting terms around, this tells us that ab ≤ (a^{2} + b^{2})/2.

Hence, for all i, we have the following inequality:
(4) |(x_i - xbar)/sqrt(Σ (x_i - xbar)^{2})| |(y_i - ybar)/sqrt(Σ (y_i - ybar)^{2})| ≤ 1/2 * |(x_i - xbar)/sqrt(Σ (x_i - xbar)^{2})|^{2} + 1/2 * |(y_i - ybar)/sqrt(Σ (y_i - ybar)^{2})|^{2}

And because this is true for each term, it is also true for the sum:
(5) Σ |(x_i - xbar)/sqrt(Σ (x_i - xbar)^{2})| |(y_i - ybar)/sqrt(Σ (y_i - ybar)^{2})| ≤ 1/2 * Σ |(x_i - xbar)/sqrt(Σ (x_i - xbar)^{2})|^{2} + 1/2 * Σ |(y_i - ybar)/sqrt(Σ (y_i - ybar)^{2})|^{2}

But, from (3), each of the sums on the right-hand side of this last inequality is 1, making the right-hand side 1/2 * 1 + 1/2 * 1 = 1. So we have
(6) Σ |(x_i - xbar)/sqrt(Σ (x_i - xbar)^{2})| |(y_i - ybar)/sqrt(Σ (y_i - ybar)^{2})| ≤ 1.

Combining (2) and (6), we get
(7) |r| = |Σ [(x_i - xbar)/sqrt(Σ (x_i - xbar)^{2})] [(y_i - ybar)/sqrt(Σ (y_i - ybar)^{2})]| ≤ Σ |(x_i - xbar)/sqrt(Σ (x_i - xbar)^{2})| |(y_i - ybar)/sqrt(Σ (y_i - ybar)^{2})| ≤ 1.
 
11-04-2010   #3 (permalink)
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Wow math genius here. It took me awhile to finish reading it. Thanks for explaining it ;3 I didn't think it would get answered at all. I'm pretty sure I understand everything except this part:

"Note that the terms |(x_i - xbar)/sqrt(Σ (x_i - xbar)^{2})| and |(y_i - ybar)/sqrt(Σ (y_i - ybar)^{2})| are normalized distances; that is,
(3) Σ |(x_i - xbar)/sqrt(Σ (x_i - xbar)^{2})|^{2} = Σ |(y_i - ybar)/sqrt(Σ (y_i - ybar)^{2})|^{2} = 1."

Normalized distances and getting 1 when you square the inside...how does that work? O.o
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11-04-2010   #4 (permalink)
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Let's start with a simpler example. Take the standard xy-coordinate system and draw a line segment from the origin to an arbitrary point (x, y). The length of this segment will be sqrt(x^2 + y^2). Now, suppose that we want the coordinates of the point on this segment that is a distance of 1 from the origin. The coordinates of this second point would have to be (x / sqrt(x^2 + y^2), y / sqrt(x^2 + y^2)). One can verify that this second point does have a distance of 1 from the origin.

The same principle applies for (3), the difference being that we are operating in n-dimensional space instead of 2-dimensional space. In this case, view sqrt(Σ (x_i - xbar)^2) as the distance the point (x_1, x_2, ..., x_n) is from the point (xbar, xbar, ..., xbar), or, if you prefer, the distance the point (x_1 - xbar, x_2 - xbar, ..., x_n - xbar) is from the origin.
 

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