Regression On Standardized Variable

[Dreyfus]

Suppose Y and X are standardized and regression is run on the standardized variables.
What will be the relation b/w residuals ie the original error term ui and the error term ui* of new model?
I m getting ui* = (ui)/SD of Y.                                      (Where ui, ui* are all estimators nd SD stands for Standard deviation)
Someone please confirm...
Thanks

[kangkan]

me too :)

[Dreyfus]

OK thanks kangkan...one more question
If There are two regression models
i) LnY = ß1 + ß2*LnX + ui.         ( Ln stands for natural log)
ii) LnY = a1 + a2*LnwX + ui*    (w is some positive constant)
What will be the relation b/w intercept and slope coefficients?
I m getting ß2=a2 and a1 = ß1 - ß2*Lnw or a1= ß1 - a2*Lnw
Please cpnfirm

[Granpa Simpson]

I am getting something different..let me check may b i am committing a mistake somewhere..!!!

[Granpa Simpson]

Its ok..I got the same answers a2=b2=(dy/dx)*(x/y) and a1=b1-a2*logW.

[ani]

Hey...whch chapter d problems are concerned to???? Is it something stat related ??? As d regression term is dre...plzzz reply..I cud nt get d connection..

[Anakin Skywalker]

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[Anakin Skywalker]

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[SoniaKapoor]

pls explain the first one..hw did u find erroor term?thnx

[Dreyfus]

Sonia ...see the attached files of my workings,....IMG_20140616_212315.jpg IMG_20140616_212357.jpg

[SoniaKapoor]

Thnxx

[Shefali]

@vaibhav..
If There are two regression models
i) LnY = ß1 + ß2*LnX + ui.         ( Ln stands for natural log)
ii) LnY = a1 + a2*LnwX + ui*    (w is some positive constant)
What will be the relation b/w intercept and slope coefficients?
I m getting ß2=a2 and a1 = ß1 - ß2*Lnw or a1= ß1 - a2*Lnw

hw did you derive this...pls help

[Dreyfus]

@shefali...I hv attached my workings....IMG_20140617_002553.jpg

[Granpa Simpson]

let log Y=Y' and log X=X'
Now equation (1) can be written as,
Y'=ß1 + ß2*X'+ ui.
where ß2=Cov(X',Y')/Var(X'), ß1=mean(Y')-ß2*mean(X')
Now consider equation (2), it can be written as,
log Y=a1 + a2*(logX+logw)+ui*
or, Y'=a1 + a2*(X'+logw)+ui* (since logY=Y' and logX=X')
here a2=Cov(X'+log w,Y')/Var(X'+ log w)
Now Cov(X'+log w,Y)=Cov(X',Y')
And var(x'+log w)=var(x') (since variance is independent of change in origin).
thus a2=Cov(X',Y')/Var(X')=ß2.
Mean (X')=Mean(X')+log w.
therefore a1=mean(Y')-a2*{Mean(X')+log w},
a1=mean(Y')-a2*Mean(x')-a2*log w.
a1=mean(Y')-ß2*Mean(x')-ß2*log w. (since a2=ß2).
thus a1=ß1-ß2*log w, (since ß1=mean(Y')-ß2*mean(X')).

[Dreyfus]

Guys one more ques
If There are two regression models
i) expY = ß1 + ß2*expX + ui.         ( exp stands for natural exponent)
ii) expY' = a1 + a2*expX' + ui'    (Y'= Y+w1 nd X'= X+w2)
What will be the relation b/w intercept, slope coefficient and residual term?
I m getting these: a2 = ß2*e^(w1-w2)
                                 a1 = ß1*e^w1
                                  ui' = ui*e^w1
Please confirm.....

[Shefali]

@vaibhav.
In that pic after the eqn 3 b line how did you get next eqn and aftr that how is B2* and B2 equal?

[Dreyfus]

@shefali...actually its eq 2b.....I used the fact that (lnXiw -mean of lnxiw) = (lnxi + lnw - mean of lnXi -lnw)
Where mean of lnw is lnw as its a constant.......similary it applies to lnYiw also.....

[Shefali]

Thank You!