- #1
Seda
- 71
- 0
I'm trying to prove that abs(r) <= 1.
(Ill apologize up front that I am 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 I am not sure..
(Ill apologize up front that I am 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 I am not sure..