Logistic Regression with (Normal) Distributions for Independent Variables

Consider the logistic regression where $Y_i \in {0,1}$ are dependent variable observations and $X_i \in \mathbb{R}$ are the independent variables.

However we do not observe the $X_i$ themselves. Instead we observe some vector of parameters $\boldsymbol{\mu}_i$ and we know the distribution $F$ s.t. $F(X_i|\boldsymbol{\mu}_i)$.

How can we perform a logistic regression where the $X_i$ are not themselves observed? For $F$ a general distribution, one way may be to sample for each $i$, many values from $F(X_i|\boldsymbol{\mu}_i)$ and put them all into the regression model as observations.

Can we do things more efficiently if we know for example that $F$ is a normal distribution?

I think you can also go for a maximum likelihood approach considering the $$x_i$$ are latent variables over which you marginalize a likelihood.
Let's say the likelihood of your usual logistic regression, if you oberved the $$x$$ values, is $$\mathcal{L}(\beta, x, y)$$ where $$\beta$$ is the vector of parameters (typically, $$\mathcal{L}(\beta, x, y) = (\frac{1}{1 + e^{-\beta x}})^y (\frac{1}{1 + e^{\beta x}})^{1 - y}$$).
Then the likelihood only observing $$\mu$$ and $$y$$ is $$\mathcal{L}(\beta, y, \mu) = \mathbb{E}_{X \sim F_{\mu}}[\mathcal{L}(\beta, y, X)]$$ And the total likelihood is just the product of the likelihoods over all observed $$(y_i, \mu_i)$$.
Unfortunately, these expectations may be untractable (maybe for a simple normal distribution it is not, but it is not obvious to me...), so you can estimate them by Monte Carlo. For instance, sample $$x_i \sim F_{\mu_i}$$ and take empirical mean of $$\mathcal{L}(\beta, y_i, x_i)$$. I don't think that this is equivalent to simulating data according to $$F_{\mu_i}$$ and put them into the model, but it would be nice to see the links...
Another way would be to go with an E-M algorithm (where $$x_i$$ are the latent variables) to maximize this likelihood, this would vertainly be more computationally efficient.