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Seda
Oct25-10, 06:59 PM
I'm trying to prove that abs(r) <= 1.

(Ill apologize up front that Im not sure on how to write all equations properly in this forum, but Ill try to make it clear)

Note that this is all sample statistics, not population, which is why I'm using r and not rho.

I know that I have to use the Cauchy-Schwartz inequality, and I can use that without proving that.

I have:

r= Cov(x,y)/(sxsy)

Therefore by Cauchy-Schwartz:

abs(r) <= (var(x)var(y))/(sxsy)

And since variance is the deviation squared

abs(r) <= (sx2sy2)/(sxsy)

leaving me with

abs(r) <= (sxsy)

Instead of the "1" I want.

My guess my error is somewhere in utlizing the cauchy schwartz but im not sure..

statdad
Oct26-10, 08:13 AM
Start this way.


cov(x,y) = \int \int (x - \overline x)(y - \overline y) h(x,y) \, dx dy


or, if you prefer


cov(x,y) = E[(x-\overline x)(y - \overline y)]


Now apply cauchy-schwartz.