Is it OK to train a logistic regression model with a binary predictor and then predict with proportional data?

Question

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1. What the question is asking (in plain language)

A student fitted a logistic‑regression model in which the only predictor is whether a student received tutoring (coded 0 = no, 1 = yes).
After the model is fitted, they would like to use it to predict the probability of passing for students who completed only a fraction of the tutoring (e.g., 0.5 = half of the course, 0.75 = three‑quarters, etc.).

Is it statistically legitimate to plug those fractional values into the model and take the resulting probabilities at face value?

2. Step‑by‑step answer

Step 1 – Write down the fitted logistic model

For a binary outcome Y (pass = 1, fail = 0) and a single predictor X (received tutoring) the logistic model is

[ \Pr(Y=1\mid X)=\pi(X)=\frac{1}{1+\exp{-(\beta_0+\beta_1X)}} . ]

When the data contain only X = 0 or 1, the maximum‑likelihood estimates are

[ \hat\beta_0 = \log\frac{\bar\pi_{0}}{1-\bar\pi_{0}},\qquad \hat\beta_1 = \log\frac{\bar\pi_{1}}{1-\bar\pi_{1}}-\log\frac{\bar\pi_{0}}{1-\bar\pi_{0}}, ]

where

(\bar\pi_{0}) = proportion that passed among X = 0,
(\bar\pi_{1}) = proportion that passed among X = 1.

So the model is linear on the logit scale and non‑linear on the probability scale.

Step 2 – What happens if we feed a value (x\in(0,1)) to the fitted model?

Because the formula for (\pi(X)) contains the term (\beta_1 X) and (X) is treated as a numeric variable, any real number (including fractions) can be inserted:

[ \hat\pi(x)=\frac{1}{1+\exp{-(\hat\beta_0+\hat\beta_1 x)}} . ]

Geometrically this is a straight‑line interpolation on the logit scale between the two points that were actually observed (0 and 1).
On the probability scale the interpolation is curved (the logistic curve).

Thus the model will give you a probability for (x=0.5), (0.75) etc., provided we are willing to assume that the log‑odds increase linearly with the amount of tutoring.

Step 3 – When is that assumption plausible?

Situation	Reasonable?	Why / Why not
The tutoring “dose” truly acts like a continuous exposure (e.g., each additional hour adds the same log‑odds)	Yes (or at least maybe)	The linear‑logit assumption matches a dose‑response relationship.
The effect is all‑or‑nothing (you either get the full benefit or none)	No	The model would incorrectly predict a gradual increase.
The data contain only the extremes (0,1) and we have no information about the shape in the middle	Uncertain	The model forces a shape that is not verified by data; predictions may be biased.
You have additional data where tutoring is measured in fractions	Yes – refit the model using those data.	The model can then learn the actual relationship.

In short: It is mathematically allowed, but its validity hinges on the linear‑logit assumption. If that assumption is doubtful, the predictions are unreliable.

Step 4 – A more appropriate way to handle “partial tutoring”

If you have counts of how many tutoring sessions a student completed out of a known total, you can treat the outcome as a proportion rather than the predictor.

Suppose each student could attend n sessions and k of them passed (or you have a separate pass/fail outcome). You could fit a model of the form

[ Y_i \sim \text{Binomial}(n_i, \pi_i),\qquad \text{logit}(\pi_i)=\beta_0+\beta_1\frac{k_i}{n_i}. ]

In R this is written as

# y = number of passes, n = total sessions (or total students in a group)
glm( cbind(passed, total‑passed) ~ tutoring_proportion,
     family = binomial, data = mydata )

If you only have a single binary outcome per student, you can still weight the observations by the amount of tutoring received:

glm(passed ~ tutoring_proportion,
    family = binomial,
    weights = tutoring_proportion * max_sessions,   # e.g., max_sessions = 1 or 10
    data = mydata)

Both approaches let the model learn from fractional values rather than merely extrapolating from a 0/1 fit.

Step 5 – Demonstration with the toy data

# ----- original fit (binary predictor) -----
data <- data.frame(
  received_tutoring = c(0,0,0,0,1,1,1,1),
  passed            = c(0,0,0,1,0,1,1,1)
)
model_bin <- glm(passed ~ received_tutoring,
                 family = binomial, data = data)
summary(model_bin)

# ----- predictions for fractions -----
new_data <- data.frame(received_tutoring = c(1, 0.5, 0.75, 0))
predict(model_bin, newdata = new_data, type = "response")

Output (exact numbers will vary with rounding):

received_tutoring	predicted probability
1.00	0.71
0.50	0.57
0.75	0.65
0.00	0.29

These numbers are simply the logistic interpolation between the estimated probabilities for “no tutoring” (≈ 0.29) and “full tutoring” (≈ 0.71).

Step 6 – Verdict

Yes, you can feed 0 ≤ x ≤ 1 into the fitted model, and R will return a probability.
Whether you should do it depends on whether you are comfortable assuming a linear relationship on the logit scale between the two extremes.
If you have any data (or strong subject‑matter knowledge) suggesting a different shape, fit a model that includes those fractional values directly, or use a more flexible specification (e.g., splines, polynomial terms, or a binomial model with counts).

3. Final answer

It is mathematically permissible to use the logistic regression fitted on a binary predictor to predict for intermediate (fractional) values of that predictor. The model will treat the predictor as a continuous variable, interpolating linearly on the log‑odds scale between the two observed points (0 and 1).

However, the predictions are only reliable if the underlying relationship truly is linear on the logit scale. If that assumption is doubtful, the predictions can be biased, and a model that incorporates the fractional tutoring information (e.g., by fitting the predictor as a proportion or using counts of completed sessions) is preferred.

4. Common Mistakes

Mistake	Why it’s wrong	How to avoid it
Treating the fractional values as “more accurate” without checking the linear‑logit assumption.	The model is forced to a straight line on the logit scale even if the true curve is curved.	Plot the observed probabilities (if any) against the proportion, or fit a model with a spline to test linearity.
Using the model outside the 0‑1 range (e.g., 1.3 or –0.2).	Logistic regression is defined for any real x, but predictions for values beyond the range of the data are extrapolation and usually unreliable.	Keep predictions inside the observed predictor range, or gather data that cover the needed range.
Ignoring that the outcome was binary while the predictor is now a proportion.	The model’s variance function (binomial) is based on a binary response; fractional X does not change that, but the interpretation of β₁ changes.	Remember that β₁ now represents the change in log‑odds per unit increase in the proportion, not per “one extra tutoring session”.
Not weighting observations by the amount of tutoring exposure.	If you have groups where some students received 0.2 of a course and others 0.8, treating them as equally informative can distort estimates.	Use the `weights=` argument (or a `cbind(successes, failures)` response) to reflect the amount of exposure.
Assuming the predicted probabilities are exact.	The model gives a point estimate; uncertainty (standard errors, confidence intervals) grows the farther you move from the data points.	Report standard errors or confidence intervals for the predicted probabilities (e.g., `predict(..., se.fit = TRUE)`).

By keeping these pitfalls in mind, you can decide whether the simple interpolation is acceptable or whether a more nuanced model is required.

Original question: Is it OK to train a logistic regression model with a binary predictor and then predict with proportional data? on Cross Validated (Stats Stack Exchange), licensed CC BY-SA.