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If we write the regression equation like so:

y = B x + C

$y = B x + C$

But if both sides are standardized, then we have:

(y - my)/sy = B ( x - mx )/sx + C

$\dfrac{y - \bar y}{\sigma_y} = B \dfrac{x - \bar x}{\sigma_x} + C$

Now solving for y$y$ by multiplying by sy$\sigma_y$ then adding my$\bar y$:

y = B ((x - mx) sy)/sx + C sy + my

y = B sy/sx + C sy + my - (B sy mx)/sx

$y = B \dfrac{x - \bar x}{\sigma_x} \sigma_y + C \sigma_y + \bar y$

$y = B \dfrac{\sigma_y}{\sigma_x} x - B \dfrac{\sigma_y}{\sigma_x} \bar x + C \sigma_y + \bar y$

Therefore the metric coefficients are:

B' = B sy/sx

C' = C sy + my - sum((B sy mx)/sx)

$B' = B \dfrac{\sigma_y}{\sigma_x}$

$C' = B \dfrac{\sigma_y}{\sigma_x} \bar x + C \sigma_y + \bar y$

In the above notation, the multiplication operator is implied.

R test code:

coef(m)[1]*sd(d0$y)+mean(d0$y)-  
  (coef(m)['x1']*sd(d0$y)*mean(d0[['x1']])/sd(d0[['x1']]) +
   coef(m)['x2']*sd(d0$y)*mean(d0[['x2']])/sd(d0[['x2']]))

Gives 1 as desired.

If we write the regression equation like so:

y = B x + C

But if both sides are standardized, then we have:

(y - my)/sy = B ( x - mx )/sx + C

Now solving for y by multiplying by sy then adding my:

y = B ((x - mx) sy)/sx + C sy + my

y = B sy/sx + C sy + my - (B sy mx)/sx

Therefore the metric coefficients are:

B' = B sy/sx

C' = C sy + my - sum((B sy mx)/sx)

In the above notation, the multiplication operator is implied.

R test code:

coef(m)[1]*sd(d0$y)+mean(d0$y)-  
  (coef(m)['x1']*sd(d0$y)*mean(d0[['x1']])/sd(d0[['x1']]) +
   coef(m)['x2']*sd(d0$y)*mean(d0[['x2']])/sd(d0[['x2']]))

Gives 1 as desired.

If we write the regression equation like so:

$y = B x + C$

But if both sides are standardized, then we have:

$\dfrac{y - \bar y}{\sigma_y} = B \dfrac{x - \bar x}{\sigma_x} + C$

Now solving for $y$ by multiplying by $\sigma_y$ then adding $\bar y$:

$y = B \dfrac{x - \bar x}{\sigma_x} \sigma_y + C \sigma_y + \bar y$

$y = B \dfrac{\sigma_y}{\sigma_x} x - B \dfrac{\sigma_y}{\sigma_x} \bar x + C \sigma_y + \bar y$

Therefore the metric coefficients are:

$B' = B \dfrac{\sigma_y}{\sigma_x}$

$C' = B \dfrac{\sigma_y}{\sigma_x} \bar x + C \sigma_y + \bar y$

In the above notation, the multiplication operator is implied.

R test code:

coef(m)[1]*sd(d0$y)+mean(d0$y)-  
  (coef(m)['x1']*sd(d0$y)*mean(d0[['x1']])/sd(d0[['x1']]) +
   coef(m)['x2']*sd(d0$y)*mean(d0[['x2']])/sd(d0[['x2']]))

Gives 1 as desired.

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Source Link
Chris
  • 1.3k
  • 10
  • 31

If we write the regression equation like so:

y = B x + C

But if both sides are standardized, then we have:

(y - my)/sy = B ( x - mx )/sx + C

Now solving for y by multiplying by sy then adding my:

y = B ((x - mx) sy)/sx + C sy + my

y = B sy/sx + C sy + my - (B sy mx)/sx

Therefore the metric coefficients are:

B' = B sy/sx

C' = C sy + my - sum((B sy mx)/sx)

In the above notation, the multiplication operator is implied.

R test code:

coef(m)[1]*sd(d0$y)+mean(d0$y)-  
  (coef(m)['x1']*sd(d0$y)*mean(d0[['x1']])/sd(d0[['x1']]) +
   coef(m)['x2']*sd(d0$y)*mean(d0[['x2']])/sd(d0[['x2']]))

Gives 1 as desired.

If we write the regression equation like so:

y = B x + C

But if both sides are standardized, then we have:

(y - my)/sy = B ( x - mx )/sx + C

Now solving for y by multiplying by sy then adding my:

y = B ((x - mx) sy)/sx + C sy + my

y = B sy/sx + C sy + my - (B sy mx)/sx

Therefore the metric coefficients are:

B' = B sy/sx

C' = C sy + my - sum((B sy mx)/sx)

In the above notation, the multiplication operator is implied.

If we write the regression equation like so:

y = B x + C

But if both sides are standardized, then we have:

(y - my)/sy = B ( x - mx )/sx + C

Now solving for y by multiplying by sy then adding my:

y = B ((x - mx) sy)/sx + C sy + my

y = B sy/sx + C sy + my - (B sy mx)/sx

Therefore the metric coefficients are:

B' = B sy/sx

C' = C sy + my - sum((B sy mx)/sx)

In the above notation, the multiplication operator is implied.

R test code:

coef(m)[1]*sd(d0$y)+mean(d0$y)-  
  (coef(m)['x1']*sd(d0$y)*mean(d0[['x1']])/sd(d0[['x1']]) +
   coef(m)['x2']*sd(d0$y)*mean(d0[['x2']])/sd(d0[['x2']]))

Gives 1 as desired.

Source Link
Chris
  • 1.3k
  • 10
  • 31

If we write the regression equation like so:

y = B x + C

But if both sides are standardized, then we have:

(y - my)/sy = B ( x - mx )/sx + C

Now solving for y by multiplying by sy then adding my:

y = B ((x - mx) sy)/sx + C sy + my

y = B sy/sx + C sy + my - (B sy mx)/sx

Therefore the metric coefficients are:

B' = B sy/sx

C' = C sy + my - sum((B sy mx)/sx)

In the above notation, the multiplication operator is implied.