Tutorials › Generalized Linear Models › Scaling Continuous Predictors

Home  /  Tutorials  /  Count Effects and Predictions

Count Effects and Predictions · Tutorial 469 of 981

Scaling Continuous Predictors

Convert continuous-predictor changes into expected-count ratios, and re-express a Poisson model under new predictor units without changing its fitted means.

Intermediate 9 min read

What You'll Learn

  • Derive the expected-count ratio for an arbitrary finite change in a continuous predictor.
  • Interpret the coefficient on the log-mean scale and after exponentiation.
  • Calculate fitted means from a baseline mean and a predictor change.
  • Convert a coefficient when a predictor is expressed in different units.
  • Reparameterize an affine transformation, including a change in predictor origin.
  • Identify errors in interpreting unit-specific count effects.

Why Predictor Units Matter

A coefficient for a continuous predictor is tied to the units in which that predictor is recorded. In a Poisson model with a log link, the coefficient describes a change in the log expected count for a one-unit increase. If one unit means one year, one kilogram, or one millimetre, the numerical coefficient has a different scale. This does not mean the fitted means have changed: it means the model has been written using a different numerical representation of the same predictor.

The useful interpretation is not restricted to a one-unit increase. A finite change of any size produces an expected-count ratio obtained by multiplying the coefficient by the change and exponentiating. This lets us interpret a coefficient in the units that are meaningful for the application, even when those units differ from the units used to fit the model.

Finite Changes in a Continuous Predictor

Suppose \(Y\) is a count response and, at predictor value \(x\), its conditional mean is \(\mu(x)=E[Y\mid X=x]\). Consider the Poisson log-mean model

$$ Y\mid X=x\sim\operatorname{Poisson}(\mu(x)), \qquad \log\mu(x)=\beta_0+\beta_1x. $$

Here \(\beta_0\) and \(\beta_1\) are fixed model parameters. The mean is positive, so the logarithm is defined. If the predictor changes from \(x\) to \(x+\Delta\), then \(\Delta\) can be positive or negative; the resulting ratio compares the model means at those two predictor values.

Definition: In the model \(\log\mu(x)=\beta_0+\beta_1x\), the expected-count ratio for a change of \(\Delta\) predictor units is \(\mu(x+\Delta)/\mu(x)\). It is \(\exp(\beta_1\Delta)\), independent of the starting value \(x\). The corresponding percent change in the expected count is \(100\{\exp(\beta_1\Delta)-1\}\%\).

For \(\Delta=1\), the ratio is \(\exp(\beta_1)\), the expected-count ratio for a one-unit increase. For a change of \(k\) units, it is \(\exp(k\beta_1)\), not \(k\exp(\beta_1)\). A negative \(\Delta\) reverses the comparison: the ratio is then less than one when \(\beta_1>0\).

Worked Example: A Ten-Year Increase in Age

Suppose a model for an expected count has \(\log\mu(x)=\beta_0+0.04x\), where \(x\) is age in years. At age 40, the fitted expected count is 8. Find the expected-count ratio for a ten-year increase and the fitted mean at age 50.

The change is \(\Delta=10\), so the ratio of the age-50 mean to the age-40 mean is

$$ \frac{\mu(50)}{\mu(40)} =\exp(0.04\times10) =\exp(0.40) \approx1.4918. $$

Thus the model implies an expected count about 1.4918 times as large, or a percentage increase of \(100(1.4918247\ldots-1)\%\approx49.18\%\). The fitted mean at age 50 is calculated from the unrounded exponential:

$$ \mu(50)=8\exp(0.40)\approx11.9346. $$

The starting age is not needed to calculate the ratio, but it is needed to label the two means being compared. The ratio describes model-based expected counts, not a guarantee that any particular observed count will increase by that factor.

Why the Ratio Formula Holds

The formula follows directly from the log-mean specification. The proof also explains why the starting value cancels.

Theorem: If \(\mu(x)>0\) and \(\log\mu(x)=\beta_0+\beta_1x\) for the predictor values being compared, then for any finite change \(\Delta\), \[ \frac{\mu(x+\Delta)}{\mu(x)}=\exp(\beta_1\Delta). \]

Proof. Exponentiating the log-mean equation gives \(\mu(x)=\exp(\beta_0+\beta_1x)\). Apply the same model at \(x+\Delta\) and divide:

$$ \frac{\mu(x+\Delta)}{\mu(x)} = \frac{\exp\{\beta_0+\beta_1(x+\Delta)\}} {\exp(\beta_0+\beta_1x)} = \exp\{\beta_1\Delta\}. $$

The intercept and the term involving the starting value cancel, leaving a ratio that depends only on the coefficient and the size of the change. Percent change is the ratio minus one, multiplied by 100. \(\square\)

The same algebra applies when other predictors are held fixed. For example, if \(\log\mu=\beta_0+\beta_1x+\gamma z\), then changing \(x\) by \(\Delta\) while keeping \(z\) fixed gives the same ratio \(\exp(\beta_1\Delta)\). This statement relies on the specified additive linear predictor: the coefficient of \(x\) does not vary with the value of \(z\).

Worked Example: A Change Measured in Five-Kilogram Units

Suppose the log mean is \(\log\mu(x)=-1+0.08x\), where \(x\) is weight in kilograms. Find the fitted means at 25 kg and 35 kg, then rewrite the model using \(x^*=x/5\), so that the new predictor is measured in five-kilogram units.

At 25 kg, the linear predictor is \(-1+0.08(25)=1\), so \(\mu(25)=\exp(1)\approx2.7183\). The increase to 35 kg is \(\Delta=10\) kg, giving the ratio

$$ \frac{\mu(35)}{\mu(25)} =\exp(0.08\times10) =\exp(0.8) \approx2.2255. $$

Equivalently, the linear predictor at 35 kg is \(-1+0.08(35)=1.8\), so \(\mu(35)=\exp(1.8)\approx6.0496\). The ratio of these fitted means is \(\exp(1.8)/\exp(1)=\exp(0.8)\), as required.

Since \(x=5x^*\), substitution gives

$$ \log\mu=-1+0.08(5x^*)=-1+0.4x^*. $$

The coefficient is now 0.4 per five-kilogram unit rather than 0.08 per kilogram. The 10 kg increase is two units of \(x^*\), so its ratio is \(\exp(0.4\times2)=\exp(0.8)\), unchanged. At 25 kg, \(x^*=5\) and the new linear predictor is \(-1+0.4(5)=1\); at 35 kg, \(x^*=7\) and it is \(-1+0.4(7)=1.8\). Both fitted means are therefore unchanged.

Rescaling and Re-expressing the Model

It is important to state exactly how a predictor is transformed. If \(x^*=x/c\), then \(x=cx^*\), so the coefficient of \(x^*\) is \(c\beta_1\). If instead \(x^*=cx\), then \(x=x^*/c\), so the coefficient is \(\beta_1/c\). Both transformations can be described informally as “scaling,” but they produce different coefficient formulas. Substituting the relationship between the old and new variables avoids confusion.

Theorem: Consider \(\log\mu=\beta_0+\beta_1x+\sum_{j=2}^p\beta_jz_j\). For a nonzero constant \(b\) and a constant \(a\), define \(x^*=a+bx\). The same linear predictor, and hence the same fitted mean at every profile, is obtained by using \[ \beta_0^*=\beta_0-\frac{a\beta_1}{b}, \qquad \beta_1^*=\frac{\beta_1}{b}, \] so that \(\log\mu=\beta_0^*+\beta_1^*x^*+\sum_{j=2}^p\beta_jz_j\).

Proof. Because \(b\ne0\), the transformation can be inverted: \(x=(x^*-a)/b\). Substitute this expression into the original linear predictor:

$$ \beta_0+\beta_1x+\sum_{j=2}^p\beta_jz_j = \beta_0+\frac{\beta_1}{b}(x^*-a)+\sum_{j=2}^p\beta_jz_j = \left(\beta_0-\frac{a\beta_1}{b}\right) +\frac{\beta_1}{b}x^* +\sum_{j=2}^p\beta_jz_j. $$

These are precisely the stated transformed coefficients. The linear predictor is equal at every corresponding profile, so exponentiating it gives the same fitted mean. \(\square\)

For a pure change of scale \(x^*=x/c\), take \(a=0\) and \(b=1/c\). The intercept is unchanged and \(\beta_1^*=c\beta_1\). For a change of origin as well, the intercept generally changes too. This is why converting Celsius to Fahrenheit affects both the slope and intercept, not just the slope.

Worked Example: Re-expressing Temperature in Celsius

Suppose \(\log\mu=-2+0.03F\), where \(F\) is temperature in degrees Fahrenheit. Write the same model using temperature \(C\) in degrees Celsius, given \(F=32+1.8C\). Then verify the fitted mean at \(20^\circ\)C in both forms.

Substitute the conversion into the original linear predictor:

$$ -2+0.03F =-2+0.03(32+1.8C) =-2+0.96+0.054C =-1.04+0.054C. $$

Thus the Celsius coefficient is 0.054 per degree Celsius, and the intercept is \(-1.04\). At \(20^\circ\)C, the equivalent Fahrenheit value is \(32+1.8(20)=68^\circ\)F. Using Fahrenheit gives the linear predictor \(-2+0.03(68)=0.04\). Using Celsius gives \(-1.04+0.054(20)=0.04\). Both forms therefore give the same fitted mean, \(\exp(0.04)\approx1.0408\).

A one-degree increase in Celsius corresponds to a 1.8-degree increase in Fahrenheit. The ratio from a one-degree Celsius increase is \(\exp(0.054)\), which is also \(\exp(0.03\times1.8)\) in the Fahrenheit formulation.

Common Mistakes and Practical Interpretation

A coefficient is not itself a percent change in the expected count. The coefficient is a change on the log-mean scale; exponentiating the coefficient multiplied by the predictor change gives the ratio on the mean scale. For a small coefficient, \(100\beta_1\%\) can be a useful rough approximation to the one-unit percent change, but the exact percent change is \(100\{\exp(\beta_1)-1\}\%\).

A second common mistake is to multiply the one-unit ratio by the number of units. If the one-unit ratio is \(\exp(\beta_1)\), the ratio for \(\Delta\) units is \(\{\exp(\beta_1)\}^{\Delta}=\exp(\beta_1\Delta)\), not \(\Delta\exp(\beta_1)\). For example, the ten-kilogram ratio in the preceding example is \(\exp(0.8)\), not \(10\exp(0.08)\).

Finally, changing units does not change the model's predictions when coefficients are transformed consistently, but it does change the numerical value and interpretation of the coefficient. Interpret the coefficient together with the predictor's units. These are model-based comparisons of conditional means; by themselves, they do not establish that changing the predictor would causally change an outcome.

1
Name the original units.
Identify what one unit of the predictor represents and state the change being interpreted.
2
Compute the mean ratio.
For a change \(\Delta\), calculate \(\exp(\beta_1\Delta)\), retaining full precision until rounding the final result.
3
Transform by substitution.
Write the old predictor in terms of the new one, substitute into the linear predictor, and collect the intercept and coefficient.
4
Check fitted means.
Evaluate both parameterizations at the same physical predictor value; their linear predictors and fitted means should agree.
Key takeaway: Under a linear log-mean model, a change of \(\Delta\) units multiplies the expected count by \(\exp(\beta_1\Delta)\). Rescaling or shifting the predictor changes the coefficient (and possibly the intercept), but a consistent reparameterization leaves every fitted mean unchanged.

Check Your Understanding

Use the log-mean model and unit-conversion rules to answer these questions.

  1. A coefficient is 0.025 per year. What is the expected-count ratio for a 12-year increase, and what is the corresponding percentage change?
  2. If \(x^*=x/100\) and the original coefficient of \(x\) is \(-0.6\), what is the coefficient of \(x^*\)? Explain the substitution used.
  3. Why does the expected-count ratio for a fixed change \(\Delta\) not depend on the starting value \(x\) in a linear log-mean model?
  4. A predictor is transformed by \(x^*=10+x\). Which coefficient changes, and which remains the same?
  5. Explain why \(\exp(\beta_1\Delta)\) is not generally equal to \(\Delta\exp(\beta_1)\).

Previous

Next

Back to Tutorials
Clinical Biostats