Why the Prediction Scale Matters
A model can produce a number that looks like a plausible prediction but is not on the scale of the response. For example, a logit-scale prediction of \(1.2\) is not a Bernoulli mean of \(1.2\): a Bernoulli mean is a probability and must lie between 0 and 1. Likewise, a log-scale prediction of \(0.7\) is not a positive response mean of \(0.7\). In both cases, the model first gives a value on the link scale, which must be converted before it is reported as a response mean.
This distinction is especially important when communicating predictions. The linear predictor is useful for fitting and interpreting the model on its specified scale, but the response mean answers a different question: what is the model's expected response at the stated covariate values? The conversion depends on the link function.
For example, in a Bernoulli model, \(\mu\) is the probability of a response of 1, so it must be between 0 and 1. In a log-link model, \(\mu\) is positive and is measured in the units of the modeled response mean. A predicted mean need not itself be an attainable response: for count data, for instance, a mean such as 2.4 events is possible even though an observed count is an integer.
Converting a Logit-Scale Prediction
For a Bernoulli response with mean \(\mu\), the logit link is \(g(\mu)=\log\{\mu/(1-\mu)\}\). It expresses the log of the odds, rather than the probability itself. Solving this equation for \(\mu\) gives the inverse-logit conversion.
Thus, a logit prediction \(\eta\) can be reported as a predicted probability only after applying this conversion. The intermediate value \(e^\eta\) is the predicted odds, not the probability. The inverse-logit result is always strictly between 0 and 1 for any finite \(\eta\), as required for a Bernoulli mean.
Worked Example: Turning a Logit Prediction Into a Probability
Suppose a hypothetical Bernoulli model gives a linear predictor of \(\eta=1.2\) for a particular covariate pattern. Find the predicted response mean and distinguish it from the predicted odds.
The inverse-logit formula gives \(\mu=e^{1.2}/(1+e^{1.2})\). Since \(e^{1.2}\approx3.3201\),
The predicted response mean is therefore about \(0.7685\), or a predicted probability of about 76.85% for response 1 at this covariate pattern. The value \(1.2\) is the log-odds prediction, and \(e^{1.2}\approx3.3201\) is the predicted odds. Neither is the probability.
A negative logit prediction is also valid. It corresponds to odds below 1 and a probability below 0.5. The direction of the conversion is monotone: larger \(\eta\) gives a larger predicted probability, but the numerical values generally differ. In particular, a positive logit prediction does not imply a probability equal to that positive number.
Worked Example: A Negative Logit Prediction
For a hypothetical Bernoulli outcome, suppose the model gives \(\eta=-0.8\). Convert this value to the response mean and check that the result is a valid probability.
Using the inverse-logit, \(\hat{\mu}=e^{-0.8}/(1+e^{-0.8})\). With \(e^{-0.8}\approx0.4493\),
The predicted probability of response 1 is about \(0.3100\), or 31.00%. The corresponding odds are about \(0.4493\), meaning the model predicts approximately 0.4493 occurrences of response 1 for each occurrence of response 0 in the odds ratio. The mean is within the required probability range \([0,1]\), while the link-scale value \(-0.8\) is not itself a probability.
Converting a Log-Scale Prediction
For a positive mean modeled with a log link, \(\eta=\log(\mu)\). Taking the exponential of both sides gives the response-scale prediction:
The exponential conversion guarantees a positive result for every finite \(\eta\). Unlike the logit case, the converted mean is not restricted to be at most 1. Its units are the units of the response mean: if the response is a count per clinic-day, for example, the predicted mean is in events per clinic-day. The exponentiated value is the mean, not an odds ratio.
Worked Example: Turning a Log Prediction Into a Mean
Suppose a hypothetical model for the mean number of events per clinic-day uses a log link and gives \(\eta=0.7\) at a specified covariate pattern. Find and interpret the response-scale prediction.
Since \(\mu=e^\eta\), substitute the given predictor:
The predicted mean is about 2.014 events per clinic-day. It is not a predicted count for one particular clinic-day; an observed count on a single day must be an integer, whereas the model's expected count can be noninteger. Reporting \(0.7\) as the expected number of events would confuse the log scale with the response scale.
A Practical Scale Check
A reliable reporting check is to name both the scale and the quantity before interpreting a number. Ask: “Is this the linear predictor, or has the inverse link been applied?” Then check whether the result has the response's range and units. For a Bernoulli mean, the answer must be a probability between 0 and 1. For a positive mean with a log link, the answer must be positive and expressed in the response's units.
The link determines the conversion; the same numerical value of \(\eta\) can correspond to different response means under different links. Nor should a reader infer the link from a number alone. The model specification must identify it. An identity link is a special case: its inverse is itself, so the linear predictor and mean have the same numerical value. That does not make the distinction unnecessary; it means the conversion happens to leave the value unchanged.
| Model link | Linear predictor represents | Response-mean conversion | Response-scale check |
|---|---|---|---|
| Logit | Log-odds | \(\mu=e^\eta/(1+e^\eta)\) | Probability between 0 and 1 |
| Log | Log of the mean | \(\mu=e^\eta\) | Positive, in response units |
| Identity | The mean itself | \(\mu=\eta\) | Check that the value is meaningful for the response |
Determine whether the value is a linear predictor \(\eta\) or a response-scale prediction.
Use the link specified by the model, not the apparent size or sign of the prediction.
Use the inverse-logit for a logit predictor or exponentiate a log-link predictor.
Confirm the transformed value is allowed for the response mean and interpret it in the response's units.
Common Mistakes and Model Checks
The most common error is to report \(\eta\) directly as the response mean. Under a logit link, this can produce a value outside \([0,1]\), an immediate warning that the interpretation is wrong. Under a log link, the error may be less obvious because \(\eta\) can be positive or negative while the mean must be positive. A negative log-scale predictor is perfectly valid; exponentiating it gives a positive mean below 1.
- Wrong reasoning: “The model predicts \(\eta=1.2\), so the event probability is 1.2.” Correction: For a logit link, convert \(1.2\) to \(e^{1.2}/(1+e^{1.2})\approx0.7685\).
- Wrong reasoning: “The log-link predictor is \(0.7\), so the expected count is 0.7.” Correction: Exponentiate to obtain \(e^{0.7}\approx2.0138\) expected events per clinic-day.
- Wrong reasoning: “Exponentiating a logit prediction gives its probability.” Correction: \(e^\eta\) gives odds under the logit link; the probability is \(e^\eta/(1+e^\eta)\).
- Wrong reasoning: “Any predicted mean must be an attainable individual response.” Correction: A mean is an expectation, so it can be noninteger for a count response.
When auditing a report or software output, verify which scale the output uses. Some outputs present linear predictors and others present response-scale fitted means; labels such as “estimate” or “prediction” alone may not settle the question. Compare the numerical range with the model's response and link, and check the units stated in the interpretation. If the scale cannot be established from the output, do not silently treat the value as a probability or mean.
These conversions apply to predictions at a specified covariate pattern. They do not, by themselves, answer every question about a population-wide average or about uncertainty around a prediction. The central task here is narrower: identify the scale of the model's prediction and apply the appropriate inverse link before describing it as a response mean.
Check Your Understanding
For each question, identify the scale before interpreting the number. Solutions follow the questions.
- A Bernoulli model has a logit prediction of \(\eta=0\). What is the predicted probability?
- A Bernoulli model has a logit prediction of \(\eta=2\). Is 2 the predicted probability? Compute the probability.
- A positive-mean model has a log-link prediction of \(\eta=-1.1\). Find the response-scale mean and state why a negative predictor is not a problem.
- A log-link model predicts \(\eta=1.5\) for a response measured in milligrams. What quantity, including units, should be reported as the predicted mean?
Solutions. (1) The inverse-logit gives \(e^0/(1+e^0)=1/2=0.5\), a predicted probability of 50%. (2) No. The predictor is log-odds; the probability is \(e^2/(1+e^2)=7.3891/8.3891\approx0.8808\), or about 88.08%. (3) The mean is \(e^{-1.1}\approx0.3329\) in the response's units. A negative value is allowed on the log scale; exponentiation gives a positive mean. (4) Exponentiate: \(e^{1.5}\approx4.482\). Report a predicted mean of about 4.482 milligrams, not 1.5 milligrams.