Interpretation of zero-truncated Poisson regression coefficients

Question

For non-truncated Poisson regression, a count random variable $Y$ is assumed to follow a Poisson distribution
$$ \mathbb{P} \left\lbrack Y = k \right\rbrack = e^{-\mu} \frac{\mu^k}{k!}, \; k = 0, 1, 2\ldots $$
with parameter $\mu > 0$. The expectation value of $Y$ is easily shown to be equal to $\mu$. If we want to model $Y$ with respect to some predictors $x \in \mathbb{R}^p$, we use a log link function and write
$$ \label{eq:link} \log \left( \mathbb{E} \left\lbrack Y \right\rbrack \right) = \log \mu = x^T \beta $$
with a vector $\beta \in \mathbb{R}^p$ of regression coefficients to be esimtated. Once one has computed the coefficient estimates $\hat{\beta} $, it is possible to interpret the value of one single regression coefficient estimate $ \hat{\beta}_j $ as follows, by virtue of the previous equation: if the value of the predictor $x_j$ increases by one, the other predictors kept fixed, then the expected count $\mu $ increases by a factor $e^{\beta_j}$.

In the context of zero-truncated Poisson regression, we consider a random variable $Z$ following the probability distribution
$$ \mathbb{P} \left\lbrack Z = k \right\rbrack = \frac{\lambda^k}{\left(e^\lambda -1\right) k!}, \; k = 1, 2, 3\ldots $$
with $ \lambda > 0$. When modeling $Z$ with respect to a set of predictors $x$, a log link is again chosen such that
$$ \log \lambda = x^T \beta $$
(see, for instance, the R family object ztpoisson for lme4::glm described at https://rdrr.io/rforge/countreg/man/ztpoisson.html). Here, the expectation value of $Z$ is not equal to the distribution parameter $\lambda $, but amounts to
$$ \mathbb{E} \left\lbrack Z \right\rbrack = \frac{\lambda}{1 - e^{-\lambda}} = \frac{e^{x^T \beta}}{1-e^{-e^{x^T \beta}}}. $$
This makes quantifying the effect of regression coefficient estimates $\hat{\beta_j}$ not as straightforward as for non-truncated Poisson regression.

I am wondering: Is there any intuitive way of quantifying the effect of a zero-truncated Poisson regression coefficient estimate analogous to non-truncated Poisson regression?

Answer 1

Unfortunately not. There is (to the best of my knowledge) no easy ceteris paribus interpretation of the coefficients' effect on the expectation of the zero-truncated response.

Sometimes the usual multiplicative effect on $\lambda$ has a natural interpretation. Namely, if the expectation of the underlying untruncated counts is of interest.

Moreover, it is clear that the multiplicative effect of a change in $x_j$ is at most $\text{e}^{\beta_j}$. For an observation with high $\lambda$ (i.e., sufficiently far away from the truncation point 0) the effect will become closer to $\text{e}^{\beta_j}$. In contrast, for an observation with low $\lambda$ (i.e., close to or even below the truncation point 0) the effect will be almost zero. But this, of course, depends on the entire regressor vector $x$ and not just the change in one of the regressors.