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.


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – JRK
    Commented May 21, 2019 at 10:50
  • 1
    $\begingroup$ 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. $\endgroup$
    – AlexK
    Commented May 22, 2019 at 0:26

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