Assume a Poisson random effect model (with a random intercept) $$\log E(Y_{it}|b_i,x_i)=\beta_0+\beta_1x_i+b_i$$ $$b_i \sim N(0,\tau^2)$$

How to derive the population-averaged effect of one unit change in $x_i$?


1 Answer 1


Start with your first expression:

$$\log E(Y_{i}|b_i,x_i)=\beta_0+\beta_1x_i+b_i$$

The effect of a one-unit change in $x_i$ is easy to write: $$\log E(Y|x_i+1,b_i) - \log E(Y|x_i,b_i) = \beta_0+\beta_1(x_i+1) + b_i - (\beta_0+\beta_1x_i+b_i) = \beta_1$$

Exponentiating both sides gives us:

$${E(Y|x_i+1,b_i) \over E(Y|x_i,b_i)} = e^{\beta_1}$$

So a one-unit increase in $x_i$ causes an $e^{\beta_1}-1$ percent change regardless of $x_i$ and $b_i$, implying that the population-averaged effect of a one-unit increase in $x_i$ also causes an $e^{\beta_1}-1$ percent change in $E(Y)$.


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