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:
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?
Is it common to use Truncated Normals for observation distribution in GLM literature? I couldn't find any example
Can I make it work by assuming $\mathbf{E} (y\mid x) \simeq \mu$ and specify $\mu = ax + b$?
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.