# Thread: What do you do it you can't increase X by one level while holding all other X constan

1. ## What do you do it you can't increase X by one level while holding all other X constan

Classical regression defines the slope as the change in Y for a one unit change in X holding all other X constant (or so I believe). What happens if its impossible to hold all other X constant when you change a given X. In some cases, substantively, if you change a specific X another X has to change, one is inherently linked to the other.

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

2. ## Re: What do you do it you can't increase X by one level while holding all other X con

hi,
this would mean that the X-es are correlated, so the model would be flawed. You would need to handle that,e.g. by regressing X1 on X2 and using X1 and the residuals instead if x1 and x2.

regards

3. ## Re: What do you do it you can't increase X by one level while holding all other X con

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?

4. ## The Following User Says Thank You to Dason For This Useful Post:

rogojel (03-22-2017)

5. ## Re: What do you do it you can't increase X by one level while holding all other X con

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?

6. ## Re: What do you do it you can't increase X by one level while holding all other X con

Well they could be mediated noetsi. Do you have an example for this question, so we can narrow down the responses? Is it possibly an output of both the predictor and outcome?

7. ## Re: What do you do it you can't increase X by one level while holding all other X con

Originally Posted by Dason
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?
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?

8. ## Re: What do you do it you can't increase X by one level while holding all other X con

Originally Posted by rogojel
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?
Collinearity isn't really something to be bothered by unless you're using the model in a way that's impacted by collinearity, and even then, there are some relatively easy fixes (but not always). If we're going to use the model for predictive purposes, multicollinearity is very rarely an issue in comparison to trying to make inferences on the betas, for example.

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.

9. ## Re: What do you do it you can't increase X by one level while holding all other X con

Originally Posted by rogojel
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?
Putting in a value for x1 but then something other than x1-squared for an x1*x1 term doesn't follow logical usage of the model.This kind of seems more like an end-user error rather than a flaw within the model. Someone using a perfect model but interpreting the slope as a correlation doesn't bring any fault on the model, it just indicates fault on the user.

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