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Correlation cannot be more extreme than +1/-1, proof using Cauchy-Schwarz inequality

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The correlation coefficient cannot exceed an absolute value of 1

This is well-known. But why is that the case? How can we proof it? This post gives one explanation using the Cauchy-Schwarz inequality.

Here’s one version of the definition of correlation:

\[ r = \frac{\sum(\Delta x \Delta y)}{\sqrt{\sum \Delta x^2} \sqrt{\sum \Delta y^2}} \]

where \(\Delta x\) and \(\Delta y\) are the differences of \(x_i\) and \(\bar{x}\), that is: \(\Delta x_i = x_i - \bar{x}\), and similarly for \(\Delta y_i\).

For the ease of notation, let’s proceed with the understanding that \(x\) stands for the differences, ie \(\Delta x\) (and similarly for \(y\)):

\[ r = \frac{\sum(xy)}{\sqrt{\sum x^2} \sqrt{\sum y^2}} \]

Now, we conjecture that

\[ r = \frac{\sum(xy)}{\sqrt{\sum x^2} \sqrt{\sum y^2}} \le 1 \]

Let’s multiply the equation by the denominator of the LHS:

\[ \sum(xy) \le \sqrt{\sum x^2} \cdot \sqrt{\sum y^2} \]

The Cauchy Schwarz inequality states that

\[ \big| \langle x,y\rangle \big |\leq ||x|| \cdot ||y|| \]

In words, the inner product \(\langle x,y\rangle\) (in its positive variant, ie \(>0\)) is smaller or equal to the product of the vector norms.

Stated differently:

\[ \sum xy \le \sqrt{\sum x^2} \cdot \sqrt{\sum y^2} \]

Which is what we wanted to proof in the first place.

Here’s a quite nice intuition on the Cauchy Schwarz inequality.