# Covariance 's relationship with pure math and probabilty? [closed]

I've been looking up a lot of statistical books and cannot find out mathmatical insight behind it, but my math level wasn't allow me to read the mathmatical statistics books and get the math behind the definition, however, i still wonder if alien and human could came up the same definition of covariance with the universal mathmatical theory.

So, my question is, what is the Covariance 's relationship with pure math and probabilty?

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NARQ.    –  Did Jun 22 '12 at 19:35

## closed as not a real question by Did, t.b., Leonid Kovalev, Asaf Karagila, tomaszAug 18 '12 at 18:19

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Since $Cov(X,Y) = \dfrac{Var(X+Y) - Var(X) - Var(Y)}{2}$, it would be quite possible to work with something twice as large, rather like the pi v. tau suggestions.

There are more approachs: in particular correlation can be derived from covariance and variance, or covariance can be derived from correlation and variance, and so it would be possible to use correlation or its square for many practical statistical purposes without mentioning covariance explicitly.

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I don't know where that formula came from. Actually Cov(X,Y) = E[(X-mx)(Y-my)] where E(X)=mx and E(Y)=my. –  Michael Chernick Jun 22 '12 at 21:21
@did Regarding you r comment that this is not a real question, I think the OPs intended question was to find out what the covariance is and how it enters into probability. It is a statistical property of two random variables that when scaled represents correlation. It is important in probability and statistics. I don't see that it has a place in pure mathematics. I think the OP was just puzzled and fishing for some understanding about it. –  Michael Chernick Jun 22 '12 at 21:27
@Michael: My aim was to show that an alternative approach is possible which can be used instead of covariance, and so aliens might not start with the same definition you and most statistically educated humans do. –  Henry Jun 22 '12 at 22:03
Sorry Henry. I didn't get your point. I see below it you have a serious answer. I do see it though as an indirect way to get it since Var(X+Y) =VarX+VarY+2 Cov(X,Y). So it is true that Cov(X,Y) does = {Var(X+Y)-Var(X)-Var(Y)}/2. –  Michael Chernick Jun 22 '12 at 22:19