# Quadratic term of standardized predictor in logistic regression

A random intercept logistic regression is performed to assess the association between $$Y$$: Disease (Yes/No) and Standardized Predictor($$X_1$$) adjusting for control variables ($$X_2$$, $$X_3$$) based on clustered survey data. A $$X_1^2$$ term is considered in the model to explore the nonlinear relationship. Results:

               coef     p-value
intercept     0.240     <0.001
X1        0.285     <0.01
I(X1)^2  -0.084     <0.01
X2        0.114     <0.05
X3        0.210     0.345


I'm trying to interpret the $$X_1$$ and $$X_1^2$$ as follows: "A unit increase in $$X_1$$ (standardized) is associated with $$exp(0.285)$$ higher odds of disease when $$X_1$$ (standardized) is at its mean, each additional level of $$X_1$$ is associated with $$exp(-0.084)$$ decrease in the likelihood of disease." Is this appropriate? Does anybody have any thoughts on this?

Thank you.

A few things:

1) If you have a regression of the form $$y = \beta_{0} + \beta_{1}x_{1} + \beta_{2}x^2_{1}$$, then taking a derivative of $$y$$ with respect to $$x_{1}$$ will return $$dy/dx_{1} = \beta_{1} + 2\times\beta_{2}x_{1}$$. So $$\beta_{1}$$ should be interpreted as the expected change in $$y$$ from a one-unit change in $$x_{1}$$ when $$x_{1} = 0$$. Even though you standardized your $$x_{1}$$, the $$0$$ part does not change, so you should interpret the coefficient on the linear part of $$x_{1}$$ as the effect of the variable when it is at $$0$$.

2) If you look again at the derivative formula above ($$dy/dx$$), you'll see that each additional one-unit increase in $$x_{1}$$ is associated not with a $$\beta_{2}$$ change in $$y$$, but with a $$\beta_{1} + 2\times\beta_{2}x_{1}$$ change in $$y$$. In other words, the change in $$y$$ is not constant, but varies depending on what the value of $$x_{1}$$ is. So you should not interpret the two coefficients separately, but simply say what each additional one-unit increase in $$x_{1}$$ does using values of both $$\beta_{1}$$ and $$\beta_{2}$$.

3) You might as well calculate and report the values of odds by exponentiating the coefficients. If you do that with $$0.285$$, you will get $$1.330$$, and you can then say that a one-unit change in $$x_{1}$$ is associated with a $$33\%$$ increase in odds of having the disease (when $$x_{1}$$ is at $$0$$). Exponentiating $$-0.084$$ will give you a number smaller than $$1$$ and you can use it together with the $$33\%$$ value to say what an additional one-unit change in $$x_{1}$$ is associated with.

4) You should use "on average" and "holding other independent variables constant" when discussing the effect of an explanatory variable on the dependent variable.

• Thank you very much. For $X_1^2$ do I need to interpret this as exp(0.285+(2*(-0.084)))= 1.124, that means additional one-unit change in x1 is associated with a 12% increase in odds of having the disease?
– JRK
May 21, 2019 at 10:50
• Sorry, I should have been more clear. $exp(\beta_{1} + 2 \times \beta_{2}x_{1})$ is the odds ratio for a one-unit change in $x_{1}$. You still need to keep the $x_{1}$ in the expression. So an additional one-unit change in $x_{1}$ is associated with a $exp(0.285 - 2 \times 0.084x_{1})$ change in odds of having the disease. May 22, 2019 at 0:26
• Thank you, its clear now.
– JRK
May 22, 2019 at 2:04