Why is correlation coefficient -1 to +1

stunner5000pt
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Homework Statement


Why is the correlation coefficient between -1 and +1?


Homework Equations


we know correlation coefficient
[tex]\rho = \frac{E[xy]-E[x]E[y]}{\sqrt{\sigma_{x} \sigma_{y}}}[/tex]

OR

[tex]r = \frac{\sum ^n _{i=1}(X_i - \bar{X})(Y_i - \bar{Y})}{\sqrt{\sum ^n _{i=1}(X_i - \bar{X})^2} \sqrt{\sum ^n _{i=1}(Y_i - \bar{Y})^2}}[/tex]

The Attempt at a Solution


Is there a way to prove this analytically? Perhaps we can use the second formula and prove by induction the bottom is greater than the top? Or perhaps equal??

I tried using the expected value formula for the first version with rho - i couldn't really use that properly. Canyou please suggest an approach? Can this even be done analytically? Or would it just have to be explained?

Thanks for your help!
 
on Phys.org
stunner5000pt said:

Homework Statement


Why is the correlation coefficient between -1 and +1?


Homework Equations


we know correlation coefficient
[tex]\rho = \frac{E[xy]-E[x]E[y]}{\sqrt{\sigma_{x} \sigma_{y}}}[/tex]

OR

[tex]r = \frac{\sum ^n _{i=1}(X_i - \bar{X})(Y_i - \bar{Y})}{\sqrt{\sum ^n _{i=1}(X_i - \bar{X})^2} \sqrt{\sum ^n _{i=1}(Y_i - \bar{Y})^2}}[/tex]

The Attempt at a Solution


Is there a way to prove this analytically? Perhaps we can use the second formula and prove by induction the bottom is greater than the top? Or perhaps equal??

I tried using the expected value formula for the first version with rho - i couldn't really use that properly. Canyou please suggest an approach? Can this even be done analytically? Or would it just have to be explained?

Thanks for your help!

Your formula for ##\rho## is incorrect; it should be
[tex]\rho \equiv \frac{E(X-EX)(Y-EY)}{\sigma_X \sigma_Y} = \frac{E(XY) - EX EY}{\sigma_X \sigma_Y}[/tex] You should not have ##\sqrt{ \;\; }## in the denominator.
 
stunner5000pt said:

Homework Statement


Why is the correlation coefficient between -1 and +1?


Homework Equations


we know correlation coefficient
[tex]\rho = \frac{E[xy]-E[x]E[y]}{\sqrt{\sigma_{x} \sigma_{y}}}[/tex]

OR

[tex]r = \frac{\sum ^n _{i=1}(X_i - \bar{X})(Y_i - \bar{Y})}{\sqrt{\sum ^n _{i=1}(X_i - \bar{X})^2} \sqrt{\sum ^n _{i=1}(Y_i - \bar{Y})^2}}[/tex]

The Attempt at a Solution


Is there a way to prove this analytically? Perhaps we can use the second formula and prove by induction the bottom is greater than the top? Or perhaps equal??

I tried using the expected value formula for the first version with rho - i couldn't really use that properly. Canyou please suggest an approach? Can this even be done analytically? Or would it just have to be explained?

Thanks for your help!


Since
[tex]0 \leq \text{Var}(aX+bY) = a^2 \sigma_X^2 + 2 a b\, \text{Cov}(X,Y) + b^2 \sigma_Y^2,[/tex]
for all ##a,b##, and since ##\text{Cov}(X,Y) = \sigma_X \sigma_Y \, \rho##, the matrix
[tex]M = \pmatrix{\sigma_X^2 & \sigma_X \sigma_Y \, \rho\\<br /> \sigma_X \sigma_Y \, \rho & \sigma_Y^2}[/tex]
must be positive semidefinite. Apply the standard tests for semidefinitness.
 
stunner5000pt said:
Why is the correlation coefficient between -1 and +1?
This is a direct consequence of the Cauchy-Schwarz inequality, a very important result which shows up in many forms throughout mathematics. There is a proof on the Wiki page in the context of an abstract inner-product space. It's a good exercise to go through the proof as written, and then restate the result by translating the concepts of inner product and norm into the language of probability.
 

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