Two Different Questions: What Can Be Observed, and What Can the Mean Be?
A response distribution answers at least two distinct questions. First, which values can an individual response take? Second, which values can its expected value take? These questions are related, but their answers are not interchangeable. A model can have an integer-valued response and a non-integer mean, or a continuous response whose mean is restricted to a smaller range than the full set of possible observations.
This distinction matters when specifying a model and checking data. A response outside the model's support cannot have been generated by that response distribution. By contrast, a mean outside the model's allowed mean space indicates an invalid parameter value, not an unusual observation. The conditional mean and linear predictor are different quantities; the response distribution determines which response values are possible around that mean.
For example, a single Bernoulli response can only be 0 or 1, whereas its mean can be any probability between 0 and 1. A Poisson response can only be a nonnegative integer, whereas its mean can be any positive real number. A Normal response can be any real number, and its mean can also be any real number. The detailed exponential-family forms of these distributions are not needed to see this distinction.
Support and Mean Space in Exponential-Family Models
In the usual regular exponential-family setup, the response has a common support that does not change as the distribution parameter changes. One way to write the density or probability mass function is
Here \(\mathcal{S}\) is the support, \(\theta\) is a natural parameter, and \(\phi\) is a dispersion parameter when one is included. The formula assigns probability or density only to values in \(\mathcal{S}\). The set of allowed values of \(\theta\) and \(\phi\) is a separate parameter-space restriction. Through the mean function, those parameter restrictions determine the mean space. It is therefore possible for \(\mathcal{S}\) and the mean space to have different types of values.
A common support is an assumption of the regular setup, not a universal fact about every distribution that can be written in exponential-family-like form. If the support itself changes with a parameter, standard derivative arguments involving the normalizing integral may need additional boundary terms. Such cases require care rather than an automatic application of regular-family identities.
Why the Mean Is Related to the Support
For a discrete response taking values \(y_1,\ldots,y_k\) with probabilities \(q_1,\ldots,q_k\), where \(q_j\geq 0\) and \(\sum_{j=1}^k q_j=1\), the expectation is
This is a weighted average of support values, so it lies in their convex hull: the smallest interval containing those values. The same principle holds for a response with more support values, provided its expectation exists. For instance, if \(Y\geq 0\) with probability 1, then \(E(Y)\geq 0\). However, the convex-hull bound is not always the full mean space. Distributional restrictions can make some values in that hull unattainable, and the endpoints may be excluded when all support values must have positive probability.
Worked Example: A Bernoulli Response and a Fractional Mean
Suppose each hypothetical response \(Y\) records whether a one-time event occurred: \(Y=1\) means it occurred and \(Y=0\) means it did not. Let \(Y\) have a Bernoulli distribution with event probability \(p=0.35\). What values can one response take, and what is its mean?
The response support is \(\mathcal{S}=\{0,1\}\). The probabilities are \(P(Y=1)=0.35\) and \(P(Y=0)=0.65\), so both support values have positive probability. The mean is
Thus \(0.35\) is a valid mean but cannot be the value of one individual response. For a hypothetical sample of 12 responses with 4 event occurrences, the sample mean is \(4/12=0.3333\). That sample mean is an average of observed zeros and ones; it does not imply that any individual response was \(0.3333\). The parameter \(p=0.35\) is also not a claim that a response can take the fractional value 0.35.
Three Common Supports
The examples below emphasize the support-mean distinction rather than deriving each distribution's exponential-family representation. In each case, the response support follows from what is being measured and from the probability model.
| Response distribution | Possible individual responses | Mean space | Key distinction |
|---|---|---|---|
| Bernoulli | \(\{0,1\}\) | \([0,1]\); usually \(0<p<1\) for a nondegenerate model | The mean is a probability and need not be an observed outcome. |
| Poisson | \(\{0,1,2,\ldots\}\) | Positive real values for the usual nondegenerate model | The mean need not be an integer. |
| Normal | All real numbers | All real numbers | The response can be above or below its mean. |
In a Bernoulli model, the probability parameter \(p\) can reach 0 or 1 if degenerate distributions are allowed; otherwise a nondegenerate model uses \(0<p<1\). In a Poisson model, the mean parameter is positive, though a degenerate zero-count limit can be considered separately. In a Normal model with positive variance, the density is positive throughout the real line, whatever finite real mean is specified. These parameter details do not change the basic support distinction.
Worked Example: A Poisson Count With a Non-Integer Mean
Suppose a hypothetical count \(Y\), such as the number of events in a fixed observation window, follows a Poisson distribution with mean \(\lambda=2.4\). Its support is the nonnegative integers, so a single count of 2.4 is impossible. The mean 2.4 is nevertheless valid: it describes the expected count over repetitions, not a possible count in one window.
For example, the probability of exactly five events is
The result is a probability for the valid support value 5. It is not a probability for observing the mean. To see how an average can be non-integer, consider three hypothetical counts \(0,2,5\). Their sample mean is
Every individual count is a nonnegative integer, while the average is not. This illustrates how averaging support values produces a value in the mean's range without making that value an individual response.
Worked Example: A Normal Response Has Unbounded Support
Suppose a hypothetical continuous measurement \(Y\), in millimetres, follows \(N(12.5,4)\). By the course convention, the second argument is the variance, so the standard deviation is \(\sqrt{4}=2\) mm. The support is all real numbers, and the mean is 12.5 mm. A value such as 14.8 mm is possible; so is a value below zero under this mathematical model, although that could be scientifically implausible for a measurement that must be positive.
For an illustration of probability over an interval, calculate the probability that \(Y\) lies between 14 mm and 16 mm. Standardizing gives
where \(Z\sim N(0,1)\). Therefore,
Under the stated model, about 18.66% of measurements lie strictly between 14 and 16 mm. The event has the same probability if either endpoint is included, because a continuous distribution assigns probability zero to any single point. The calculation concerns an interval of possible responses; it does not restrict the mean to that interval.
Checking the Assumptions and Diagnosing Mismatch
Before using a response distribution, check that its support is consistent with how observations were recorded. The distributional assumption is about the response represented by the model, not merely the label given to a column in a data file. For example, a code of 9 used to mean “missing” is not a valid ninth category or count; it must be treated according to the data-collection convention rather than as a response value.
- Check impossible values. A negative count is outside Poisson support, and a Bernoulli response of 2 is outside Bernoulli support. Such values often signal coding, measurement, or model-specification problems.
- Check the scale and units. A response recorded as a percentage on a 0-to-100 scale and one recorded as a proportion on a 0-to-1 scale have different numerical supports. Confirm which representation is being modeled.
- Check whether the model permits implausible values. A Normal model permits every real value. If a measurement must be positive, substantial probability assigned to negative values may indicate a poor fit to the measurement scale, even though the observed data contain no negative values.
- Do not use the observed minimum and maximum as the support. A finite sample usually covers only part of the values permitted by its distribution. Seeing no large counts, for example, does not mean those counts are outside Poisson support.
An observed value outside the assumed support is a direct incompatibility that should be investigated. A model that permits values outside the scientifically meaningful range is a different kind of concern: it may still approximate the observed range well, but it can yield implausible predictions or probabilities. Alternative approaches include choosing a response distribution with an appropriate support or, when justified by the sampling process, using a model that explicitly conditions on a restricted range. A transformation can also change the modeled scale, but it does not erase the need to check what values are possible on the original scale.
Why Support Constraints Matter
Support constraints are foundational because a model's probability statements only apply to values it permits. If the response coding allows values that the chosen distribution rules out, calculations may be mathematically defined but scientifically irrelevant or impossible. If the distribution allows values that the process cannot produce, its predictions can be misleading at the boundaries. Conversely, the fact that a mean is not itself an attainable response does not make it invalid: means summarize distributions, and summaries need not share the response's support.
The principal alternative is not to choose between “support” and “mean.” Both must be checked. First verify that the support matches the possible observations; then verify that the mean parameter is allowed and interpretable for the response scale. In a GLMM, this distinction remains important when the mean depends on covariates and random effects: the conditional mean still has to lie in the distribution's mean space, and possible conditional responses still have to lie in its support.
Check Your Understanding
Answer each question, then compare your reasoning with the brief solutions.
- A Bernoulli response has event probability 0.62. What are its support and mean? Can one individual response equal 0.62?
- A Poisson response has mean 3.7. Is a single response of 3.7 possible? Is a sample mean of 3.7 possible?
- A Normal response has mean \(-2\) and positive variance. Is 0 in its support? Is the mean itself an allowed response value?
- Explain why observing only the values 0, 1, and 2 in a finite Poisson sample does not establish that the support is \(\{0,1,2\}\).
- Suppose a purported Bernoulli data set contains the response value 3. Is this an unusual but valid outcome under the model, or a support mismatch?
Solutions. (1) The support is \(\{0,1\}\), and the mean is \(0.62\); no single response can equal \(0.62\). (2) A single response of 3.7 is impossible because Poisson responses are nonnegative integers. A sample mean of 3.7 is possible because averages of integer counts need not be integers. (3) Yes: a Normal distribution with positive variance has support over all real numbers. The mean \(-2\) is also an allowed real-valued response, though a particular observation equals it with probability zero. (4) A finite sample may fail to include some values that remain possible under the model; observed extrema do not define distributional support. (5) The value 3 is outside Bernoulli support, so it indicates a coding, data, or model mismatch to investigate, not a valid Bernoulli outcome.