# Interpretation of quadratic term in Poisson regression

I fit a Poisson regression model where I used $$x$$ and $$x^2$$ to predict $$y$$.

Let's say the coefficients of $$x$$ and $$x^2$$ are $$\beta_1 = 0.8$$ and $$\beta_2 = -0.1$$.

Exponentiating the coefficient ($${\beta_1}$$) gives us the multiplicative factor by which the mean count changes when we increase $$x$$ by one unit: $$e^{\beta_1}$$ = 2.23. So increasing x by one unit changes the mean of $$y$$ by a factor of 2.23. This factor is constant over all $$x$$.

According to this answer, "$$e^{\beta_2}$$ would be called a ratio of ratio of rates comparing groups differing by 1 unit differing by 1 unit of $$X$$."

I tried this out with a toy example below with the parameters from above ($${\beta_1} = 0.8$$ and $${\beta_2} = -0.1$$) and indeed the ratio is not constant because of the nonlinear term but the ratio of the ratio is constant. But the ratio of ratios is not equal to $$e^{\beta_2}$$ but to $$e^{(2\beta_2)}$$. I think this relates to this part of the referenced answer above "But if you do a difference in differences for $$(E[Y|X=x+2] - E[Y|X=x+1]) - (E[Y|X=x+1] - E[Y|X=x]) = 2\beta_2$$. So basically the $$\beta_1$$ is the tangent slope of the quadratic curve at the origin, and $$\beta_2$$ is a quadratic slope."

With all this, I am not quite sure how to interpret a one-unit change of $$x$$ on the mean of $$y$$. Would it be correct to say that for each unit increase in $$x$$, the linear term $${\beta_1}$$ changes the mean of $$y$$ by a factor of 2.23 while the nonlinear term $${\beta_2}$$ leads to a decrease of this factor by $$e^{(2\beta_2)}$$ = 0.8187 for each increase in $$x$$. I don't think this is correct because the effect is dependent on the value of $$x$$.

Is there a way to express the effect on a one-unit increase in $$x$$ on the mean of $$y$$ that is independent of $$y$$ or is this not possible?

And if this is not possible, what is the correct way to specify a change from let's say $$x = 1$$ to $$x = 2$$ for this toy example? "Increasing $$x$$ from 1 to 2 increases the mean of $$y$$ by a factor of 1.6487"?

x predicted y ratio ratio of ratio
0 1
1 2.01375271 2.01375271
2 3.32011692 1.64872127 0.81873075
3 4.48168907 1.34985881 0.81873075
4 4.95303242 1.10517092 0.81873075
5 4.48168907 0.90483742 0.81873075

1. As you already stated, $$\exp(\beta)$$ for any first-order coefficient $$\beta$$ in a Poisson model can be interpreted as a relative rate of the Poisson process comparing groups differing by 1 in the corresponding variable being modeled.
2. In a linear model with identity link, like OLS, the interpretation of the quadratic term $$\beta_2$$ is readily seen as the rate of change of the effect, that is for every unit difference in $$X$$ the slope relating $$X$$ and $$Y$$ changes by a factor of $$\beta_2$$. Similarly, the linear term $$\beta_1$$ can be seen as the instantaneous rate of change relating $$X$$ and $$Y$$ when $$X=0$$. That is to say, since the shape of the predicted trend between $$X$$ and $$Y$$ is quadratic, $$\beta_1$$ is the slope of the tangent curve at $$X=0$$.
So, in your model with $$\beta_1 = 0.8$$ and $$\beta_2 = -0.1$$, comparing groups differing by 1 unit "near 0" (ignoring the issue of prediction at the means), has a relative rate ratio of 2.22, and groups differing by 1 unit have a ratio of relative rate ratios of 0.91
• Thanks! But groups differing by 1 unit have a ratio of relative ratios of 0.81 ($e^{(2\beta_2)}$) and not 0.91 ($e^{(\beta_2)}$), right? Since $e^{(2\beta_2)}$ is the second derivative of the relative rate of change with respect to $x$? Commented May 3, 2023 at 7:14
• Thanks for your patience! I did, and my result was $e^{2\beta_2}$... I also double-checked with WolframAlpha. $e^{2\beta_2}$ is also consistent with the ratio of ratios I calculated in the table for the toy example I provided. What am I missing? Commented May 4, 2023 at 19:04
• I also get $2\beta_2$, Showing some work, using $a,b$ for $\beta_1, \beta_2$ for simplicity in reading and typing: the crucial step is resolving: $ax +bx^2 + a(x+2) + b(x+2)^2 - 2\cdot (a(x+1) + b(x+1)^2)$, Here everytihng cancels except you get $4b$ from $b(x+2)^2$ and only $2b$ from $2\cdot (b(x+1)^2)$ Commented May 15, 2023 at 8:56