Does the slope definition change then, or is there a technique that addresses this?

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Does the slope definition change then, or is there a technique that addresses this?

I think it wouldn't be appropriate to say that the model is flawed because the X's are correlated. Is any model with both X and X^2 in it automatically flawed?

I think I have an answer to this: if I have a model where I have the terms x and F(x) then the model also captures this dependency - if I input a value for x, F(x) will automatically have the right value and there is no way I can specify an inconsistent pair (x, F(x))

If I have a regression model that contains x1 and x2 which are correlated, this relationahip is not part of the model. If I input a new value for x1 I am still free to input any other value for x2, however unrealistic the combination(x1,x2). So, in this sense, the model would be flawed. Does this make sense?

Yepp, you are right!

This is an interesting question actually, why are we bothered by collinearity but not by x, x^2 type of models?

This is an interesting question actually, why are we bothered by collinearity but not by x, x^2 type of models?

To the OP: hlsmith got to it before I could. Another easy example (aside from square, cube,...) is a mediator/interaction, but it's not terribly important since we need to do a little algebra to interpret that slope at a fixed value of one of the variables (remember an interaction of X1*X2 can be rewritten as X3 if you want to see it more clearly).

E(Y) = b0 + b1x1 + b2x2 +b3x1*x2

Where x1 is some continuous variable and x2 is gender (1 if female, let's say; 0 for male).

The slope relating Y to X1 for males can be seen by plugging 0 in for x2. E(Y) = b0 + b1x1 + b2(0) + b3x1*(0) = b0 + b1x1...so the slope of Y with X1 is b1, (you can think of it like saying, holding x2 fixed at 0 (gender as a male))

The slope relating Y to X1 for females can be seen the same way but plugging in 1 for X2.... E(Y)= b0 + b1x1 + b2(1) + b3x1*(1) = b0 + b1x1 + b2 +b3x1 = (b0+b2) + (b1+b3)x1...so we can see the slope of y with x1 for females is b1+b3 holding gender (or anything else in our model, if it were bigger) fixed.

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hi Dason,

I think I have an answer to this: if I have a model where I have the terms x and F(x) then the model also captures this dependency - if I input a value for x, F(x) will automatically have the right value and there is no way I can specify an inconsistent pair (x, F(x))

If I have a regression model that contains x1 and x2 which are correlated, this relationahip is not part of the model. If I input a new value for x1 I am still free to input any other value for x2, however unrealistic the combination(x1,x2). So, in this sense, the model would be flawed. Does this make sense?

I think I have an answer to this: if I have a model where I have the terms x and F(x) then the model also captures this dependency - if I input a value for x, F(x) will automatically have the right value and there is no way I can specify an inconsistent pair (x, F(x))

If I have a regression model that contains x1 and x2 which are correlated, this relationahip is not part of the model. If I input a new value for x1 I am still free to input any other value for x2, however unrealistic the combination(x1,x2). So, in this sense, the model would be flawed. Does this make sense?