I'm trying to setup a Bayesian GLM with Truncated Normal, $\mathcal{N}_+(\mu, \sigma, 0)$ truncated at $0$.

I want to specify $\mathbf{E} (y\mid x) = ax + b$ but it looks like $\mathbf{E} (y\mid x)$ does not have a nice formula in terms of the parameters.

My questions are:

  1. Is it possible to use $\mathcal{N}_+(\mu, \sigma, 0)$ for observation distribution in a GLM and be able to specify $\mathbf{E} (y\mid x)$ as the parametric family of curve that you're trying to fit?

  2. Is it common to use Truncated Normals for observation distribution in GLM literature? I couldn't find any example

  3. Can I make it work by assuming $\mathbf{E} (y\mid x) \simeq \mu$ and specify $\mu = ax + b$?

  4. In GLMs, we need to specify $\mathbf{E} (y\mid x)$ as the parametric family that we're trying to fit because intuitively we "expect" the observed data to behave like that family. Is this explanation correct?

Thank you so much! I'm new to GLMs, so any help would be much appreciated!

EDIT: Thanks much for the detailed answer Achim. Not sure this helps, but in 4. I wanted to say that -- suppose we choose observation distribution, i.e, likelihood, $D$ to model our response, our goal would be to specify the expected value of $y$ as the parametric function that we're trying to fit.

For instance, if we're fitting a straight line, we specify $E(y|x) = b^Tx$. But we could also make this complex -- $E(y|x) = \sigma(x)$ the sigmoid function.

Now, if $D$ is as simple as $N(\mu, 1)$, you could just specify $\mu = E(y|x) = b^Tx$ and the model will fit straight line to the data.

In case $D$ is $\gamma(\alpha, 2)$, the Gamma distribution in shape and scale, you could simply specify $\alpha / 2 = E(y|x) = b^Tx$.

The issue with Truncated Normal $TN(\mu, \sigma)$ as $D$ is that -- there's no parameter for $E(y|x)$. The best that could be done, as you mentioned, is, $\mu = E(y|x) = b^Tx$ and specify $\sigma$ in terms of $x$ too.

In all these cases, we're trying to specify $E(y|x)$ in terms of the parametric function $g(x)$ that we "expect" best explains the data we have. Not sure if I make sense though.


1 Answer 1


General remarks

The truncated normal model (just like the censored normal model) does not belong to the GLM (generalized linear model) family in the classical sense (a la McCullagh & Nelder, 1989, doi:10.1201/9780203753736). In GLMs the dispersion parameter (usually denoted $\phi$) can be treated as a nuisance parameter and the regression coefficients $\beta$ that affect the expectation $\mu$ via $g(\mu) = x^\top \beta$ can be estimated without the dispersion.

So the truncated normal model is similar in spirit but also different because $\mu$ and $\sigma$ are not orthogonal and have to be estimated simultaneously. Misspecification of one parameter will lead to inconsistent estimation of the other. That's why for these models not only $\mu$ is often specified as a function of regressors but also $\sigma$, often in the framework of GAMLSS (generalized additive models of location, scale, and shape).


  1. No, the model does not belong to the GLM family. The expectation is $\mathrm{E}(y | x) = \mu + \sigma \cdot \frac{\phi(\mu/\sigma)}{\Phi(\mu/\sigma)}$ where $\phi(\cdot)$ and $\Phi(\cdot)$ are the probability density and cumulative distribution function of the standard normal distribution, respectively. This intrinsically depends on both $\mu$ and $\sigma$.
  2. The model is commonly used in various contexts but it is not a GLM. In econometrics it is often used as one component of the two-part Cragg model, consisting of a binary "hurdle" for 0 vs. greater than 0 and a zero-truncated model. My lecture notes on this might be a useful starting point.
  3. The assumption that the mean is approximately linear really only makes sense if the observations are far enough from the truncation point $0$. But the closer the observations get towards $0$, the more effect the truncation will have. This means that the marginal effects will go from $\beta$ when you are far away to virtually $0$ when you are close to the truncation point. The figure below (from my notes linked above) shows the relationship between the untruncated mean $\mu$ (dotted) and the truncated mean $\mu + \phi(\mu)/\Phi(\mu)$ (solid, assuming $\sigma = 1$). For $\mu$ greater than two standard deviations the truncation has almost no effect, but for $\mu$ around $0$ or below the slope is almost zero.

Relationship between untruncated mean, probability of truncation, and truncated mean

  1. I'm not sure what exactly you mean by this. I would say that we assume a response distribution that fits our observations and then link the parameters of that distribution to the available regressors. And for some distributional families (GLM) it turns out that you can separate effects on the location and the scale nicely (like in the linear regression model).


For fitting truncated regression models where also the scale parameter can be modeled by regressors, there are various packages available in R:

Of course, there are other packages as well, including general model estimation frameworks, notably brms for Bayesian regression models.

  • $\begingroup$ Thank you so much for the detailed explanation. Even though it's not a GLM, it gives me relief that it is used as an observational distribution by others. I wanted to confirm this so that I know I'm not doing something grossly incorrect. $\endgroup$
    – chesslad
    Commented Jun 8, 2023 at 19:36
  • $\begingroup$ I have made an edit to explain 4. in detail. It would help me a lot if you could clarify whether my understanding is correct. $\endgroup$
    – chesslad
    Commented Jun 8, 2023 at 20:05
  • 1
    $\begingroup$ Thanks for the additional explanations. I'm not sure, though, whether I have much to add. While it is generally a good strategy to model the mean directly through a linear predictor, plus a link function, this is not possible here due to the dependency on both parameters of the latent distribution. Also a linear link will in general not work well due necessarily non-constant marginal effects. $\endgroup$ Commented Jun 8, 2023 at 23:15

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