Why a Gaussian Link Can Restrict the Mean
A Gaussian response can have any real-valued mean. This does not mean every link function can be used with every possible mean, however: a nonidentity link may be undefined for some means, or its inverse may only produce means in a restricted range. The distinction matters when specifying a model and when checking fitted values or predictions.
For example, a Gaussian model with an identity link can represent a negative mean, while a log link represents only positive means. The observed Gaussian response can still be negative under a log link; the restriction applies to its conditional mean, not to each observation. The earlier material on Gaussian GLMs establishes the Gaussian mean and variance framework. Here we focus on the compatibility between the mean domain, the link, and the linear predictor.
For a Gaussian response, \(Y\sim N(\mu,\sigma^2)\), the density is defined for every real \(\mu\), and the possible values of \(Y\) are all real numbers. Thus the unrestricted Gaussian mean domain is \(\mathbb R\). With the identity link, \(g(\mu)=\mu\), every real predictor gives a valid mean. For a nonidentity link, we must check its domain and choose a branch if the function is not one-to-one.
Domains of Three Nonidentity Links
In this tutorial, the log link is \(g(\mu)=\log\mu\), the reciprocal link is \(g(\mu)=1/\mu\), and the square link is \(g(\mu)=\mu^2\). For reciprocal and square links, we use the positive-mean branch unless stated otherwise. These conventions make each link one-to-one on its chosen mean domain.
| Link | Chosen mean domain | Predictor domain | Inverse link |
|---|---|---|---|
| Log: \(g(\mu)=\log\mu\) | \(\mu>0\) | \(\eta\in\mathbb R\) | \(\mu=\exp(\eta)\) |
| Reciprocal: \(g(\mu)=1/\mu\) | \(\mu>0\) | \(\eta>0\) | \(\mu=1/\eta\) |
| Square: \(g(\mu)=\mu^2\) | \(\mu>0\) | \(\eta>0\) | \(\mu=\sqrt{\eta}\) |
The log link maps positive means onto the entire real line: very small positive means have very negative predictors, and large means have large positive predictors. By contrast, the reciprocal and square links on the positive-mean branch have only positive predictors. Assigning a negative predictor to either of these links cannot produce a positive mean on that branch.
Worked Example: Checking Three Predictors
Suppose a model gives the linear predictor \(\eta=-1, 0,\) or \(4\) at three covariate values. Determine whether each predictor gives a valid mean under each link, using the positive-mean branches.
Log link. All three predictors are allowed, and the corresponding means are \(\exp(-1)\), \(\exp(0)\), and \(\exp(4)\). These are approximately \(0.368\), \(1\), and \(54.598\), all positive.
Reciprocal link. The predictor must be strictly positive. At \(\eta=-1\) and \(\eta=0\), the link is invalid on the chosen branch: the inverse \(1/\eta\) is negative in the first case and undefined in the second. At \(\eta=4\), the mean is \(1/4=0.25\), which is valid.
Square link. The predictor must also be strictly positive. At \(\eta=-1\), there is no real square root; at \(\eta=0\), the inverse gives mean zero, outside the chosen domain \(\mu>0\). At \(\eta=4\), the mean is \(\sqrt{4}=2\), which is valid.
The Gaussian distribution itself would permit any of these means, including a negative one. The restrictions in the last two checks come from the selected link branches.
Results for Predictor and Coefficient Constraints
In a regression model, the predictor at observation \(i\) is \(\eta_i=x_i^\mathsf{T}\beta\), where \(x_i\) is its design vector and \(\beta\) is the coefficient vector. Thus a link’s domain condition becomes a condition on the predictors generated by the design and coefficients.
Proof. For the log link, the inverse is \(\mu_i=\exp(\eta_i)\). The exponential is defined and strictly positive for every real \(\eta_i\), so each finite real predictor produces a mean in the chosen domain \((0,\infty)\). Conversely, every \(\mu_i>0\) has the unique predictor \(\eta_i=\log\mu_i\), which is real. For the reciprocal link, the inverse on the positive branch is \(\mu_i=1/\eta_i\). This is defined and positive exactly when \(\eta_i>0\). For the square link, the inverse is \(\mu_i=\sqrt{\eta_i}\), which is real and strictly positive exactly when \(\eta_i>0\). Applying these conditions to each observation proves the result. \(\square\)
The equivalence in Result 1 concerns model validity at the design points being considered. If predictions are requested at new covariate values, their predictors must satisfy the same link-specific conditions. A model valid on the observed design can therefore produce an invalid predictor when extrapolated.
Proof. The \(i\)th component of \(X\beta\) is \(\eta_i=x_i^\mathsf{T}\beta\). By Result 1, a reciprocal or square model is valid at all design points precisely when \(x_i^\mathsf{T}\beta>0\) for each \(i=1,\ldots,n\), which is the stated set. For each row \(x_i^\mathsf{T}\), the condition \(x_i^\mathsf{T}\beta>0\) defines an open half-space if \(x_i\ne 0\), and defines the empty set if \(x_i=0\). The full feasible set is a finite intersection of these sets. It is therefore open when nonempty, and is convex because each half-space is convex; if a zero row makes it empty, it is still either empty or an open convex set as stated. For the log link, each finite \(\beta\) gives finite real predictors \(X\beta\), and the exponential of each predictor is positive. No sign constraint on \(\beta\) is required. \(\square\)
This result gives a practical way to diagnose restrictions before fitting. For a reciprocal or square link, the coefficient constraints are strict linear inequalities. They may be feasible or infeasible for a given design. If they are infeasible, no coefficient vector makes the specified model valid at every design point on the chosen branch.
Worked Example: Testing a Reciprocal Model on a Design
Consider the predictor \(\eta(x)=\beta_0+\beta_1x\) at \(x=0,1,2\). For the positive-mean reciprocal link, find a coefficient choice that is valid at all three points, and check a choice that fails.
The constraints are \(\beta_0>0\), \(\beta_0+\beta_1>0\), and \(\beta_0+2\beta_1>0\). Choose \(\beta_0=1\) and \(\beta_1=0.5\). The three predictors are \(1\), \(1.5\), and \(2\), all strictly positive. The fitted means are their reciprocals: \(1\), \(2/3\), and \(1/2\). Each is positive and finite, so this choice is valid.
Now choose \(\beta_0=1\) and \(\beta_1=-0.75\). The predictors are \(1\), \(0.25\), and \(-0.5\). The last violates the strict positivity constraint; applying \(1/\eta\) would give \(-2\), which is not on the chosen positive-mean branch. The coefficient choice is therefore invalid for this design, even though it gives valid means at the first two points.
Branches, Boundary Values, and Gaussian Observations
The restrictions above depend on branch choices. The reciprocal function is defined for \(\mu\ne0\); on the negative-mean branch \(\mu<0\), its predictor is also negative and the inverse remains \(1/\eta\). Using both positive and negative means together is possible only with care: \(\mu=0\) is excluded, and \(\eta=0\) is excluded, so the full domains are disconnected. The square function takes the same value at \(\mu\) and \(-\mu\), so it is not one-to-one over all nonzero real means. Its inverse must select either \(+\sqrt{\eta}\) or \(-\sqrt{\eta}\), and in either case \(\eta>0\).
The boundary is not included on the positive branches. For the reciprocal link, \(\eta=0\) has no finite reciprocal mean; for the square link, \(\eta=0\) gives \(\mu=0\), which is outside the chosen domain. A value arbitrarily close to zero but positive may be valid, but it can imply a very large reciprocal mean or a very small square-link mean. Validity alone does not guarantee numerically stable fitting.
Finally, a positive conditional mean does not require all observed responses to be positive. If \(Y\sim N(\mu,\sigma^2)\) with \(\mu>0\), the Gaussian distribution still assigns positive probability to negative values. The link constrains \(E[Y\mid x]\), not the support of \(Y\). Confusing those two restrictions leads to incorrectly rejecting a Gaussian model whenever a negative observation appears.
Worked Example: A Positive Mean Does Not Prevent Negative Responses
Let \(Y\sim N(1,4)\), so its mean is \(1\) and its standard deviation is \(2\). This mean could arise from a log-link predictor \(\eta=\log(1)=0\). The response distribution is nevertheless not restricted to positive values.
Standardizing the event \(Y<0\) gives \(P(Y<0)=P((Y-1)/2<-1/2)=\Phi(-0.5)\), where \(\Phi\) is the standard normal distribution function. Numerically, \(\Phi(-0.5)\approx0.3085\). Thus the model has a positive mean but assigns about \(30.85\%\) probability to negative responses. That is consistent with a Gaussian response; only the fitted mean must satisfy the log-link domain.
Common Mistakes and Practical Checks
- Restricting the observations instead of the means. A positive-mean link does not turn a Gaussian response into a positive-valued random variable. Check the fitted conditional means, not the signs of individual observations.
- Assuming every real predictor is allowed. This is true for the log link, but not for the positive-mean reciprocal or square links. Verify \(\eta_i>0\) at each design point.
- Including zero at a strict boundary. The reciprocal inverse is undefined at zero, and the positive square-root branch gives zero there, not a strictly positive mean.
- Using the square link without selecting a branch. Since \(\mu^2=(-\mu)^2\), the square link does not uniquely determine a mean on the full real line. State the mean branch explicitly.
- Checking only observed covariate values. A model may meet its domain restrictions in the sample and fail them at a prediction point. Check each predictor where a fitted mean will be reported.
Check Your Understanding
For each question, distinguish the Gaussian response distribution from the restrictions imposed by the link.
- For the positive-mean log link, what is the mean when \(\eta=-2\), and is this predictor valid?
- Under the positive-mean reciprocal link, explain why \(\eta=0\) is invalid even though the Gaussian family permits a zero mean.
- For the positive-mean square link, what condition must hold for \(\eta\), and what is the inverse link?
- A reciprocal model has predictors \(0.8\), \(1.1\), and \(-0.2\) at three design points. Is it valid at all three? Explain.
- Can a Gaussian response have a negative observed value when its conditional mean is positive? Explain briefly.
- Why must a branch be specified for the square link if Gaussian means are allowed to be negative?