I see many facets of your question and in what follows will present my top 2 :-)
Facet 1 -- assessing the uncertainty in estimated coefficients
In logistic regression, assessing the uncertainty in the estimated coefficients is virtually the same as for least-squares regression .
In both logistic regression and least-squares regression, the regression coefficient table will include a column for the regression coefficients followed by a column of standard errors, then by a column of test statistics, and finally a column of p-values. The table below shows the example coefficient table output for the some regression problem (where the probability of buying some magazine for kids is estimated).
Note that the test statistics are labeled “z value” and the p-values are labeled “P(>|t|)” in the table above.
The standard errors can be used to construct confidence intervals for the regression coefficients. Roughly speaking, going plus or minus 2 times the standard error from the regression coefficient gives approximately a 95% confidence interval for the coefficient.
For example, for Residence Length, the regression coefficient is 0.024680. The next column gives the standard error of the regression coefficient which is 0.013800. Thus an approximate 95% confidence interval for the Residence Length regression coefficient is:
$$
0.024680 \pm 2 \times 0.013800 = 0.024680 \pm 0.0276\
$$
This means that the regression coefficient for Residence Length could be anywhere from -0.00292 to 0.05228 (with 95% confidence).
As is well known, we often use the odds-ratio, which is the exponential of the regression coefficient (i.e., $\exp(\beta)$ ), to help to interpret the meaning of the regression coefficient. The odds-ratio for the Residence Length coefficient, as shown in the coefficient table, is 1.0250. This means that there is a 2.5% increase in the odds of buying the magazine associated with each additional year of residence.
We can also compute the odds-ratios corresponding to the ends of the confidence interval. These odds-ratios will give us an equivalent confidence intervals for the odds. So continuing the example using Residence Length, the odds ratios corresponding to the ends of the confidence interval are
$$
\exp(-0.00292)=0.99708
$$
and
$$
\exp(0.05228)=1.05367.
$$
Thus, the interval [0.99708,1.05367] is an approximate 95% confidence interval for the odds ratio. This means that there could be anywhere from a 0.292% decrease to a 5.367% increase in the odds of buying the magazine associated with each additional year of residence.
Facet 2 -- modeling data with measurements uncertainty
Here again many options are known. Take a look at this to get started: https://pdfs.semanticscholar.org/0a7a/a5e3d407b24e2f5c24287bfdda20e573bd05.pdf
I will just cite the source:
in some datasets, the outcome of interest is measured with imperfect
sensitivity and specificity. It is well known that the
misclassification induced by such an imperfect diagnostic test will
lead to biased estimates of the odds ratios and their variances.
In this paper, the authors show that when the sensitivity and specificity
of a diagnostic test are known, it is straightforward to incorporate
this information into the fitting of logistic regression models.
An EM algorithm that produces unbiased estimates of the odds ratios and
their variances is described. The resulting odds ratio estimates tend
to be farther from the null but have greater variance than estimates
found by ignoring the imperfections of the test. The method can be
extended to the situation where the sensitivity and specificity differ
for different study subjects, i.e., nondifferential misclassification.
The method is useful even when the sensitivity and specificity are not
known, as a way to see the degree to which various assumptions about
sensitivity and specificity affect one's estimates.
Hope this helps!
