# Homework Help: Why is correlation coefficient -1 to +1

1. Mar 23, 2014

### stunner5000pt

1. The problem statement, all variables and given/known data
Why is the correlation coefficient between -1 and +1?

2. Relevant equations
we know correlation coefficient
$$\rho = \frac{E[xy]-E[x]E[y]}{\sqrt{\sigma_{x} \sigma_{y}}}$$

OR

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

3. 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 couldnt really use that properly. Canyou please suggest an approach? Can this even be done analytically? Or would it just have to be explained?

2. Mar 24, 2014

### Ray Vickson

Your formula for $\rho$ is incorrect; it should be
$$\rho \equiv \frac{E(X-EX)(Y-EY)}{\sigma_X \sigma_Y} = \frac{E(XY) - EX EY}{\sigma_X \sigma_Y}$$ You should not have $\sqrt{ \;\; }$ in the denominator.

3. Mar 24, 2014

### Ray Vickson

Since
$$0 \leq \text{Var}(aX+bY) = a^2 \sigma_X^2 + 2 a b\, \text{Cov}(X,Y) + b^2 \sigma_Y^2,$$
for all $a,b$, and since $\text{Cov}(X,Y) = \sigma_X \sigma_Y \, \rho$, the matrix
$$M = \pmatrix{\sigma_X^2 & \sigma_X \sigma_Y \, \rho\\ \sigma_X \sigma_Y \, \rho & \sigma_Y^2}$$
must be positive semidefinite. Apply the standard tests for semidefinitness.

4. Mar 24, 2014

### jbunniii

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.