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 Answer 1


You can assemble two useful pieces of information to arrive at the interpretation of the linear and quadratic term in a Poisson GLM:

  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

  • $\begingroup$ 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 $x$? $\endgroup$
    – Ahorn
    Commented May 3, 2023 at 7:14
  • $\begingroup$ 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? $\endgroup$
    – Ahorn
    Commented May 4, 2023 at 19:04
  • $\begingroup$ @Ahorn you're not showing any work. $\endgroup$
    – AdamO
    Commented May 15, 2023 at 0:28
  • $\begingroup$ 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 $2\cdot (b(x+1)^2)$ $\endgroup$ Commented May 15, 2023 at 8:56

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