Intercept-Only Poisson Regression

Introduction

An intercept-only Poisson regression describes a set of independent count observations using one common conditional mean. With no predictors in the model, its fitted mean is the same for every observation. The likelihood calculation gives a simple result: when the observed counts have a positive total, that common fitted mean is their sample average.

The qualification matters. The log link restricts the model mean to be positive. If every observed count is zero, the likelihood has no maximum at any finite intercept; it approaches its supremum as the mean tends to zero. We derive both cases and work through the calculations.

Model and Likelihood

Let \(Y_1,\ldots,Y_n\) be independent Poisson responses, with observed values \(y_1,\ldots,y_n\). In an intercept-only model, each response has the same mean \(\mu\), and the log link writes that mean as \(\mu=\exp(\beta_0)\), where the intercept \(\beta_0\) can be any real number. Thus \(\mu>0\).

For a Poisson response, \(P(Y_i=y_i)=\exp(-\mu)\mu^{y_i}/y_i!\) for \(y_i=0,1,2,\ldots\). Independence means the joint likelihood is the product of these probabilities. Define the observed total \(S=\sum_{i=1}^n y_i\). Substituting \(\mu=\exp(\beta_0)\) gives

Taking logarithms separates the part that depends on the intercept from the factorial terms, which are constant as \(\beta_0\) varies:

Differentiating with respect to \(\beta_0\) yields the intercept likelihood equation. In this model the mean is \(\exp(\beta_0)\) for each of the \(n\) observations, so the equation balances the observed total against \(n\) times that common mean.

Why the Fitted Mean Is the Sample Average

Solving the equation requires \(S/n>0\). When that condition holds, exponentiating the solution gives \(\exp(\hat\beta_0)=S/n\). Since the sample average is \(\bar y=S/n\), the fitted mean for every observation equals \(\bar y\). This algebra identifies a stationary point; the next result verifies that it is in fact the unique maximum of the likelihood.

Proof. The first derivative is \(S-n\exp(\beta_0)\), and the second derivative is

Thus the log-likelihood is strictly concave, so it can have at most one stationary point, and any such point is a strict global maximum if the likelihood does not instead increase toward an endpoint. The derivative equation has the solution \(\exp(\beta_0)=S/n\), or \(\beta_0=\log(S/n)\), which is finite because \(S>0\). Also, as \(\beta_0\to-\infty\), the term \(S\beta_0\to-\infty\) while \(-n\exp(\beta_0)\to0\); hence \(\ell(\beta_0)\to-\infty\). As \(\beta_0\to+\infty\), the term \(-n\exp(\beta_0)\) dominates the linear term \(S\beta_0\), so again \(\ell(\beta_0)\to-\infty\). The stationary point is therefore the unique global maximum. At that point, \(\exp(\hat\beta_0)=S/n=\bar y\), as claimed. \(\square\)

The All-Zero Boundary Case

If \(S=0\), nonnegativity of the counts implies that every observation is zero. The derivative equation would require \(0-n\exp(\beta_0)=0\), which has no finite solution because \(\exp(\beta_0)>0\) for every real \(\beta_0\). This is not a failure of the likelihood calculation: it indicates that the likelihood’s supremum lies at the boundary \(\mu=0\), which the log-link parameterization cannot attain with a finite intercept.

Proof. When all observations are zero, the factorial terms are zero and the log-likelihood is \(\ell(\beta_0)=-n\exp(\beta_0)\). Its derivative is \(-n\exp(\beta_0)<0\), so the log-likelihood strictly decreases as \(\beta_0\) increases. For every finite \(\beta_0\), \(\ell(\beta_0)<0\), while \(\ell(\beta_0)\to0\) as \(\beta_0\to-\infty\). Thus the supremum is \(0\), but it is not attained at any finite intercept. Equivalently, the likelihood is \(\exp\{-n\exp(\beta_0)\}\), which approaches \(1\) but is less than \(1\) at every finite \(\beta_0\). \(\square\)

Common Mistake and Practical Meaning

A common mistake is to solve \(\exp(\hat\beta_0)=\bar y\) mechanically and report \(\hat\beta_0=\log(\bar y)\) for every sample. If \(\bar y>0\), this gives the finite maximum-likelihood estimate. If \(\bar y=0\), \(\log(\bar y)\) is not finite, and the model has no finite intercept estimate. The correction is to check the observed total before taking the logarithm.

For a positive sample average, the result also provides a useful computational check: the single fitted mean must equal the average of the observed counts. It follows from the intercept-only likelihood and the common-mean assumption, not from a rule that applies unchanged to every model with predictors.