in a regression, how do you handle measurements of proportions that fall outside (0,1)?

I saw some similar questions, but none of them seems to address precisely the problem I am going to describe.
If someone can please point me in the right direction, it'd be great.

In our work, we run some regressions that are interpreted using this equation:

$$P = \frac {10^{H \cdot LC}} {10^{H \cdot LC} + 10^{H \cdot X}}$$

$$P$$ is measured for different values of $$LC$$, say 8 values equally spaced between $$-11$$ and $$-4$$, and constants $$H$$ and $$X$$ are fitted. [$$H$$ is always positive, $$X$$ can take any real value].
This is done for several different 'items' to test, and each gets its own $$X$$ and $$H$$.

In other (cheaper) experiments, $$LC$$ is fixed to, say, $$-5$$, and $$P$$ is measured.
This too is done for several different items, but as the experiment is cheaper, many more items can be processed.
Our goal is then to use the $$P$$ value from this single measurement to calculate $$X$$ for each item, assuming an 'average' value of $$H$$, which is indeed most of the time not far from $$1$$.

And here's where we face a problem.

The items we most care about are those for which $$X$$ is as low as possible (a value of $$-9$$ is considered quite good).
As you can see from the equation, this corresponds to items for which $$P$$ gets close to 1.
Given the way $$P$$ is measured, it is subject to a rather large uncertainty, to the point that, although theoretically it should always be $$0 < P < 1$$, its measured value can easily fall outside $$(0,1)$$, in particular at the high end of the scale, so it can get up to $$1.1 - 1.2$$.

This of course stops us from using the inverse equation:

$$X = LC - \frac 1 H \cdot log_{10}( \frac P {1-P})$$

which requires $$P$$ to be strictly in $$(0,1)$$.

How would you address this issue?
Do you know of any literature or posts I could consult?

For completeness, I will mention that in other cases where we measured values that were not supposed to exceed $$1$$, but did due to measurement error, we found from experimental repeats that the error was log-normally distributed, so we applied Bayesian concepts to calculate the expected 'true' value from the 'measured' value.
Given that the distribution of true values was bounded, this in a way 'shrank' the interval back to where it should be.

EDIT adding R code for clarity and exemplification

We have no problem regressing $$P$$ vs $$LC$$. E.g.:

    if (length(find.package(package="FME", quiet=TRUE))==0)
install.packages("FME")
require(FME)

# 1. Regress P(LC)

# Simulate data

set.seed(012345)
N <- 8
X <- -7
H <- 0.9
LC <- rep((-11):(-4), each = 2)
P_true <- 10^(H*LC)/(10^(H*LC) + 10^(H*X))
P_meas <- rnorm(2*N, P_true, 0.05)
plot(P_meas ~ LC)

model.P.LC <- function(parms, LC) {
with(as.list(parms), {
10^(H*LC)/(10^(H*LC) + 10^(H*X))
})
}

modelCost.P.LC <- function(p) {
out <- model.P.LC(p, LC)
P_meas - out
}

start.P.LC <- c("H" = 1, "X" = -6)
fit.P.LC <- modFit(f = modelCost.P.LC, p = start.P.LC)

curve(model.P.LC(fit.P.LC$par,x), min(LC), max(LC), col = 2, lwd = 2, add = TRUE) summary(fit.P.LC) #Parameters: # Estimate Std. Error t value Pr(>|t|) #H 1.00997 0.11060 9.131 2.84e-07 *** #X -6.98042 0.04689 -148.853 < 2e-16 *** #--- #Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 # #Residual standard error: 0.04246 on 14 degrees of freedom # #Parameter correlation: # H X #H 1.00000 -0.01439 #X -0.01439 1.00000  Suppose instead we have measured many values of $$P$$, each for a different 'item', from experiments at a fixed value of $$LC$$. We assume $$H = 1$$, or some other suitable value, and we want to estimate $$X$$ for each item. Well, we can't do that for all items, because for some of them the error on $$P$$ causes it to be outside $$(0,1)$$, so the formula does not work.  # 2. Calculate X from P # Simulate data set.seed(012345) N <- 1000 X_true <- runif(N, -9, -4) H <- 1 LC <- -6 P_true <- 10^(H*LC)/(10^(H*LC) + 10^(H*X_true)) P_meas <- rnorm(N, P_true, 0.05) plot(P_meas ~ P_true, col = ifelse((P_meas > 0) & (P_meas < 1), "black", "red")) X_estimate <- LC - 1/H * log10(P_meas/(1-P_meas)) plot(X_estimate[!is.nan(X_estimate)] ~ X_true[!is.nan(X_estimate)]) abline(0, 1, col = "blue")  So I am wondering what a statistician would advise to do, to be able to 'use' the values of $$P$$ that do not fall within the allowed domain, in particular knowing that those close to $$1$$ are of particular interest to us, so we'd rather not throw away the data just because of some fluctuation in the signal. Any practical suggestion is very welcome. • If measured$P$can exceed$1$then modelling it as a proportion seems inappropriate. In particular you should not use a model where the error is inside the proportion calculation when you believe the error to be outside the proportion calcaulation Jan 28, 2021 at 8:53 • Thanks. But as you can see I am not modelling$P$, I am modelling$X$. Anyway, OK, this is what you say I should not do. What would you suggest to do then? Jan 28, 2021 at 9:42 • BTW,$P$is technically a ratio between two measurements, a 'maximal' response$E_{max}$and a measured response$E$. Given that$E$has random error, when the 'true'$E$is close to$E_{max}$, it can happen that its measured value exceeds$E_{max}$, and the ratio$P$is above$1$. I do not think this makes$P\$ not a proportion, does it? Jan 28, 2021 at 9:45
• @Henry Your conclusion about "inappropriate" does not follow and is counterproductive. It is reasonable to model a proportion as such, and to model its measurement as incorporating a measurement error. Nonlinear least squares methods are among the simplest such models. Indeed, we have (literally) dozens of threads with examples of fitting this model in the form $$P=\frac{1}{1+\exp(\log(10)H(X-LC))}+\varepsilon.$$ This is a submodel of the problem solved at stats.stackexchange.com/questions/478194.
– whuber
Jan 28, 2021 at 13:39
• Just reporting that the Bayesian approach works very well, superior to any other approach we have tried so far. Thanks again for your advice. Feb 5, 2021 at 12:32