What a Reciprocal-Link Coefficient Describes
A regression coefficient is interpreted through the link function, not directly through the response mean. In Gamma regression with a reciprocal-mean link, the linear predictor describes the inverse conditional mean. A coefficient therefore gives an additive change on the inverse-mean scale. To obtain a change in the mean itself, first update the linear predictor and then take its reciprocal.
This distinction matters when reporting results. A coefficient of a given size does not imply that the conditional mean changes by that same amount, or by a fixed percentage. The corresponding mean change depends on the starting predictor value.
The reciprocal-mean link \(g(\mu)=1/\mu\) is distinct from the canonical negative-inverse link \(g(\mu)=-1/\mu\). This tutorial uses the reciprocal-mean link: the sign in the model equation is positive. With that convention, a positive coefficient increases the inverse mean as its covariate increases, and thus decreases the mean, with other predictors held fixed.
Interpreting Coefficients and Contrasts
Write \(x_j\) for one covariate and \(\beta_j\) for its coefficient. Compare two predictor vectors that differ only in \(x_j\), from \(x_j=a\) to \(x_j=b\). The linear predictor changes by \(\beta_j(b-a)\). Since the link equates this predictor to the inverse conditional mean, the coefficient's direct interpretation is on that inverse scale.
The resulting mean-scale contrast follows by taking reciprocals. If \(\eta_0\) and \(\eta_1\) are the two linear predictors, both positive, then \(\mu_0=1/\eta_0\) and \(\mu_1=1/\eta_1\). The exact mean difference is
For a change \(\Delta\) in \(x_j\), \(\eta_1-\eta_0=\beta_j\Delta\). Thus the mean difference depends on both the coefficient contrast and the two predictor values. It is not generally constant across covariate profiles.
Worked Example: A One-Unit Increase in a Continuous Predictor
Suppose a Gamma regression for a positive clinical measurement has \(x\) as a continuous predictor and
Compare \(x=2\) with \(x=4\). At \(x=2\), the predictor is \(0.20+0.05(2)=0.30\), so the fitted conditional mean is \(\mu(2)=1/0.30=10/3\), approximately \(3.333\). At \(x=4\), the predictor is \(0.20+0.05(4)=0.40\), and \(\mu(4)=1/0.40=2.5\).
The covariate increases by 2 units. The inverse mean increases by \(0.05(2)=0.10\), from \(0.30\) to \(0.40\). The mean changes by \(2.5-10/3=-5/6\), approximately \(-0.833\). This is not \(-0.10\): \(0.10\) is the increase in the inverse mean, not the decrease in the mean.
Exact Mean-Scale Change and Local Effect
Two useful calculations translate the inverse-mean interpretation to the response-mean scale. The first gives an exact contrast between two predictor settings. The second gives the instantaneous rate of change when a continuous covariate varies.
Proof. Since \(\eta_0=1/\mu_0\), the new predictor is
Taking its reciprocal gives \(\mu_1=\mu_0/(1+\beta_j\Delta\mu_0)\). For the difference, subtract the two reciprocals and use \(\eta_1-\eta_0=\beta_j\Delta\):
Because \(\mu_0\mu_1=1/(\eta_0\eta_1)\), this is also \(-\beta_j\Delta\mu_0\mu_1\). Substitution of \(\eta_0=1/\mu_0\) into the expression for \(\mu_1\) also shows that its denominator \(1+\beta_j\Delta\mu_0\) must be positive under the stated condition \(\eta_1>0\). \(\square\)
The local effect follows by differentiating the mean as a function of the covariate. This derivative describes a rate of change, not the effect of a finite jump.
To derive this, write \(\mu(x)=1/\eta(x)\). With other covariates fixed, \(\partial\eta(x)/\partial x_j=\beta_j\). The chain rule gives
Since \(\mu(x)=1/\eta(x)\), this equals \(-\beta_j\mu(x)^2\). For a positive coefficient, the local mean effect is negative; for a negative coefficient, it is positive. Its magnitude depends on the current mean. \(\square\)
Worked Example: A Binary Predictor
Suppose a treatment indicator \(T\) is coded 0 for the reference group and 1 for the treatment group, and the fitted model is
For \(T=0\), the predictor is \(0.25\), so the fitted mean is \(\mu(0)=1/0.25=4\). For \(T=1\), the predictor is \(0.35\), so \(\mu(1)=1/0.35=20/7\), approximately \(2.857\). The coefficient \(0.10\) says the treatment group's inverse mean is \(0.10\) larger than the reference group's inverse mean.
The mean difference is \(20/7-4=20/7-28/7=-8/7\), approximately \(-1.143\). The ratio of means is \((20/7)/4=5/7\), approximately \(0.714\). Neither the difference nor the ratio is the coefficient. They are consequences of the two inverse means in this particular comparison.
Calculating Predictions at Specified Values
For a numerical prediction, substitute the specified covariate values into the full linear predictor, check that the resulting predictor is positive, and take its reciprocal. A coefficient cannot generally be converted into a mean prediction on its own: the intercept and the other covariate contributions determine the starting inverse mean.
Worked Example: Comparing Two Continuous-Predictor Values
Consider the model
Compare \(x=0\) with \(x=5\). At \(x=0\), the predictor is \(0.50\), which is positive, and the fitted mean is \(\mu(0)=1/0.50=2\). At \(x=5\), the predictor is \(0.50-0.04(5)=0.30\), also positive, and the fitted mean is \(\mu(5)=1/0.30=10/3\), approximately \(3.333\).
The coefficient is \(-0.04\), so the five-unit increase changes the inverse mean by \((-0.04)(5)=-0.20\), from \(0.50\) down to \(0.30\). The conditional mean therefore increases by \(10/3-2=4/3\), approximately \(1.333\). The sign is consistent with the local derivative: \(-\beta\mu^2\) is positive when \(\beta=-0.04\). The size of the finite mean increase is not obtained by multiplying \(0.04\) by five; that multiplication gives the inverse-mean decrease.
Common Mistake and Practical Interpretation
A common mistake is to say, “The coefficient is \(0.10\), so increasing the predictor by one unit increases the mean by \(0.10\).” This treats the coefficient as though the model used the identity link. Under the reciprocal-mean link, a one-unit increase changes \(1/\mu\) by \(0.10\); the mean is then found by taking the reciprocal of the updated inverse mean.
A second mistake is to report a fixed percent change as though the coefficient were a log-link coefficient. Under a log link, a coefficient has a direct interpretation through a mean ratio. Here, the reciprocal link instead gives an additive contrast in inverse means. Mean differences and ratios can still be calculated for specified covariate values, but they depend on those values.
When reporting a result, name the scale explicitly: for example, “holding the other covariates fixed, a one-unit increase in \(x_j\) raises the inverse conditional mean by \(\beta_j\).” If a mean-scale statement is needed, give the predictor settings and calculate both fitted means. This keeps the coefficient interpretation separate from the derived predictions.
Check Your Understanding
For each question, distinguish the inverse-mean scale from the mean scale.
- In a reciprocal-mean Gamma model, what quantity changes additively when \(x_j\) increases by \(\Delta\), holding other covariates fixed?
- If the old predictor is \(0.40\) and a covariate change increases it by \(0.10\), calculate the old and new fitted means and their difference.
- For a positive coefficient on a continuous covariate, what is the sign of the local effect on the mean, assuming the predictor is positive?
- Why does a reciprocal-link coefficient not, by itself, give a constant mean difference for a one-unit covariate increase?
- For \(1/\mu(x)=0.60-0.05x\), calculate the fitted means at \(x=2\) and \(x=4\), and state the change in the inverse mean.