Re: Correlation of X with XY ?
From: Charles Knapp (nowhere_at_nomailspam.com)
Date: 09/27/04
- Next message: Charles Knapp: "Re: Correlation of X with XY ?"
- Previous message: Charles Knapp: "Re: Correlation of X with XY ?"
- In reply to: Ray Koopman: "Re: Correlation of X with XY ?"
- Next in thread: Richard Ulrich: "Re: Correlation of X with XY ?"
- Messages sorted by: [ date ] [ thread ]
Date: Sun, 26 Sep 2004 23:34:41 -0400
"Ray Koopman" <koopman@sfu.ca> wrote in message
news:1096241534.783487.155680@h37g2000oda.googlegroups.com...
> Charles Knapp wrote:
> > "Ray Koopman" <koopman@sfu.ca> wrote in message
> > news:1096227018.840862.156010@h37g2000oda.googlegroups.com...
> >> Charles Knapp wrote:
> >>> But I have a computer program which uses the built in (BASIC)
> >>> random number generator which is a "rectangular" distribution with
> >>> Mean=.5 and std. Deviation = .2889 (the range is (0<x<1).
> >>> This gives a correlation between X and XY of about .6 .... clearly
> >>> not ZERO.
> >>> Is this because XY is "nonlinear" meaning that the corellation
> >>> with X will depend critically on the "mean" and the "variance"
> >>> whatever X and Y that you happen to use?
> >>
> >> Corr[x,xy] = Cov[x,xy]/(SD[x] SD[y]).
> >
> > There seems to be something wrong with this, shouldn't it read:
> >
> > Corr[x,xy] = Cov[x,xy]/(SD[x]SD[xy]) ????
> >
> > this is not identical to your equation above ????
> >
> >>
> >> Cov[x,xy] = E[x^2 y] - E[x] E[xy].
> >>
> >> If x and y are independent then
> >>
> >> E[x^2 y] = E[x^2] E[y] and E[xy] = E[x] E[y].
> >>
> >> Substituting gives
> >>
> >> Cov[x,xy] = E[x^2] E[y] - E[x]^2 E[y]
> >> = Var[x] E[y],
> >
> > There seems to be something wrong with this too
> > because:
> >
> > E[x^2] =/= E[x]^2 ????
> >
> > i.e. Var[x] =/= (Mean[x])^2 ????
> >
> >>
> >> and Corr[x,xy] = SD[x] E[y] / SD[y].
> >
> > Therefore, I don't think this is correct ????
> > But I think you know how to produce the correct answer
> > if you correct your mistakes.
> >
> > Can you give me the corrected expression for:
> >
> > Corr[x,xy] = ????????????????
>
> You're right, SD[y] in the denominator should be SD[xy],
> which complicates matters. However, the numerator is correct.
>
> Corr[x,xy] = Cov[x,xy]/(SD[x] SD[xy]).
>
> Cov[x,xy] = E[x^2 y] - E[x] E[xy].
>
> Var[xy] = E[x^2 y^2] - E[xy]^2.
>
> If x and y are independent then
>
> E[x^2 y] = E[x^2] E[y], E[xy] = E[x] E[y],
>
> and E[x^2 y^2] = E[x^2] E[y^2].
>
> Substituting gives
>
> Cov[x,xy] = E[x^2] E[y] - E[x](E[x] E[y])
>
> = (E[x^2] - E[x]^2) E[y]
>
> = Var[x] E[y],
OK, the above eqn looks correct
>
> Var[xy] = E[x^2] E[y^2] - (E[x] E[y])^2
Ok, the above eqn. is correct
>
> = (Var[x] + E[x]^2)(Var[y] + E[y]^2) - E[x]^2 E[y]^2
No, this eqn is not correct. It is not equal to the line above it.
>
> = Var[x] Var[y] + Var[x] E[y]^2 + E[x]^2 Var[y],
>
> SD[x] E[y]
> and Corr[x,xy] = ----------------------------------------------------.
> Sqrt(Var[x] Var[y] + Var[x] E[y]^2 + E[x]^2 Var[y])
Boy..... we're sure a long way apart..... I get:
E{X^2}*E{Y}-E{X}*E{X}*E{Y}
Corr(x,xy) = ------------------- -------------------------------------
sqrt( Var{X}*( E{X^2}*E{Y^2}-E{X}^2*E{Y}^2 ) )
?????????????
- Next message: Charles Knapp: "Re: Correlation of X with XY ?"
- Previous message: Charles Knapp: "Re: Correlation of X with XY ?"
- In reply to: Ray Koopman: "Re: Correlation of X with XY ?"
- Next in thread: Richard Ulrich: "Re: Correlation of X with XY ?"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|
|
