[R] How to find inverse of glm model?

[Luigi Marongiu]

Thank you. This topic is more complicated than anticipated. Best regards, Luigi

On Tue, Apr 15, 2025 at 11:09 PM Andrew Robinson <apro using unimelb.edu.au> wrote:
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> A statistical (off-topic!) point to consider: when the GLM was fitted, you conditioned on x and let y be the random variable. Therefore the model supports predictions of y conditional on x. You’re now seeking to make predictions of x conditional on y. That’s not the same thing, even in OLS.
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> It might not matter for your application but it’s probably worth thinking about. Simulation experiments might shed some light on that.
>
> Cheers, Andrew
>
> --
> Andrew Robinson
> Chief Executive Officer, CEBRA and Professor of Biosecurity,
> School/s of BioSciences and Mathematics & Statistics
> University of Melbourne, VIC 3010 Australia
> Tel: (+61) 0403 138 955
> Email: apro using unimelb.edu.au
> Website: https://researchers.ms.unimelb.edu.au/~apro@unimelb/
>
> I acknowledge the Traditional Owners of the land I inhabit, and pay my respects to their Elders.
>
> On 16 Apr 2025 at 1:01 AM +1000, Gregg Powell via R-help <r-help using r-project.org>, wrote:
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>
> Take a look at this Luigi...
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> # The model is: logit(p) = β₀ + β₁*Cycles
> # Where p is the probability (or normalized value in your case)
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> # The inverse function would be: Cycles = (logit⁻¹(p) - β₀)/β₁
> # Where logit⁻¹ is the inverse logit function (also called the expit >function)
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> # Extract coefficients from your model
> intercept <- coef(b_model)[1]
> slope <- coef(b_model)[2]
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> # Define the inverse function
> inverse_predict <- function(p) {
> # p is the probability or normalized value you want to find the >cycles for
> # Inverse logit: log(p/(1-p)) which is the logit function
> logit_p <- log(p/(1-p))
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> # Solve for Cycles: (logit(p) - intercept)/slope
> cycles <- (logit_p - intercept)/slope
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> return(cycles)
> }
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> # Example: What cycle would give a normalized value of 0.5?
> inverse_predict(0.5)
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> This function takes a probability (normalized value between 0 and 1) and returns the cycle value that would produce this probability according to your model.
> Also:
> This works because GLM with binomial family uses the logit link function by default
> The inverse function will return values outside your original data range if needed
> This won't work for p=0 or p=1 exactly (you'd get -Inf or Inf), so you might want to add checks
>
> All the best,
> Gregg
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> On Tuesday, April 15th, 2025 at 5:57 AM, Luigi Marongiu <marongiu.luigi using gmail.com> wrote:
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> I have fitted a glm model to some data; how can I find the inverse
> function of this model? Since I don't know the y=f(x) implemented by
> glm (this works under the hood), I can't define a f⁻¹(y).
> Is there an R function that can find the inverse of a glm model?
> Thank you.
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> The working example is:
> `V = c(120.64, 66.14, 34.87, 27.11, 8.87, -5.8, 4.52, -7.16, -17.39, -14.29, -20.26, -14.99, -21.05, -20.64, -8.03, -21.56, -1.28, 15.01, 75.26, 191.76, 455.09, 985.96, 1825.59, 2908.08, 3993.18, 5059.94, 6071.93, 6986.32, 7796.01, 8502.25, 9111.46, 9638.01, 10077.19, 10452.02, 10751.81, 11017.49, 11240.37, 11427.47, 11570.07, 11684.96, 11781.77, 11863.35, 11927.44, 11980.81, 12021.88, 12058.35, 12100.63, 12133.57, 12148.89, 12137.09) df = data.frame(Cycles = 1:35, Values = V[1:35]) M = max(df$Values) df$Norm = df$Values/M df$Norm[df$Norm<0] = 0 b_model = glm(Norm ~ Cycles, data=df, family=binomial) plot(Norm ~ Cycles, df, main="Normalized view", xlab=expression(bold("Time")), ylab=expression(bold("Signal (normalized)")), type="p", col="cyan") lines(b_model$fitted.values ~ df$Cycles, col="blue", lwd=2)`
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-- 
Best regards,
Luigi