Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ and $Y$ be random variables

The correlation between $X$ and $Y$ is:

$\frac{\text{Cov}(X,Y)}{\sigma_X \sigma_Y}=\frac{E[(X-\mu_X)(Y-\mu_Y)]}{\sigma_X \sigma_Y}$

So, correlation is suppose to measure the strength of a linear relationship.

But, given the above formula, I don't understand how it is going to work.

Let's assume that $X$ and $Y$ are perfectly correlated i.e. 1. Then, based on the formula, all I know is $Cov(X,Y)$ will some big number. How is it that dividing by the product of the standard deviation of $X$ and $Y$ helps to measure the strength of its linear relationship?

I also tried to draw a graph of $Y=X$ to get a geometric understanding of the formula but that hasn't help yet. My guess is that the correlation is suppose to be the slope of the line Y=X but the way the formula is defined does not make it obvious to me.

share|improve this question
    
Erm what...? The size of covariance partly reflect how the variables are. If X=Y, then X and Y perfectly correlated, if var (X) = 0.00000000000001, then what is covariance between X and Y and how is that a very big number? Dividing by standard deviation scales it. –  Lost1 Jan 5 at 22:23
    
@Lost1 If $X=Y$, then $Cov(X,Y)=Var(X)$. $Var(X)$ can be a big number, can't it? –  mauna Jan 5 at 22:46
    
I just gave an example when it is very small. The point is it can be anything.... You wrote 'it will be a big number'. This is complete bs. –  Lost1 Jan 6 at 0:21
    
However the correlation is always 1, regardless what the variance is. That is why you look at correlation. It is always between -1 and 1 and independent of scales. If you look at relationship between height and weight. Using weight in kg and lb will give different covariance against height in m or feet, but correlation is always the same independent of choice of units. –  Lost1 Jan 6 at 0:23
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.