A Continuous Predictor In Poisson Regression

Why Use a Continuous Predictor?

Count outcomes often vary with a measured quantity that can take many numerical values: a patient's age, a laboratory measurement, or the time since an event. A Poisson regression with one continuous predictor describes how the expected count changes across those values. Its defining feature is that the predictor enters the logarithm of the conditional mean, not the mean itself.

The model gives a smooth curve of expected counts as the predictor varies. It does not require the observed counts to lie on that curve: individual observations remain random. Here we focus on specifying the model and evaluating its fitted mean at selected covariate values.

Definition: For observations $i=1,\ldots,n$, let $Y_i$ be a count response and $x_i$ a measured continuous predictor. A one-predictor Poisson regression specifies $$ Y_i\mid x_i\sim \operatorname{Poisson}(\mu_i), \qquad \log(\mu_i)=\beta_0+\beta_1x_i, $$ where $\mu_i=E[Y_i\mid x_i]$, $\beta_0$ is the intercept, and $\beta_1$ is the slope. Since $\mu_i>0$, the model can equivalently be written $\mu_i=\exp(\beta_0+\beta_1x_i)$.

The conditional Poisson statement specifies the response distribution at each predictor value; the log-mean equation specifies how its mean depends on that value. In a fitted model, the coefficient estimates $\hat{\beta}_0$ and $\hat{\beta}_1$ give the fitted conditional mean $$ \hat{\mu}(x)=\exp(\hat{\beta}_0+\hat{\beta}_1x). $$ This fitted mean is an expected count. It need not be an integer, even though each observed response is a nonnegative integer.

Reading the Model on Its Two Scales

The linear expression $\beta_0+\beta_1x$ is the model for the log mean. Thus a one-unit increase in $x$ changes the modelled log mean by $\beta_1$. On the original count scale, the mean follows an exponential curve rather than a straight line. The direction of that curve is determined by the sign of the slope; its change in count units need not be constant across the predictor range.

The meaning of one unit depends on how the predictor is measured. A slope for age measured in years is expressed per year; if age is instead recorded in months, the numerical slope must be on the corresponding per-month scale. Always identify the units before calculating or describing fitted values.

Worked Example: Evaluating a Centered Predictor

Suppose a fitted model for a count response is $\log\hat{\mu}(x)=\log(3)+0.08(x-50)$, where $x$ is measured in years. Calculate the fitted means at $x=40$, $50$, and $65$.

At $x=40$, $x-50=-10$, so $$ \hat{\mu}(40)=\exp\{\log(3)+0.08(-10)\} =3\exp(-0.8) \approx 3(0.449329) \approx 1.348. $$

At $x=50$, the centered value is zero: $$ \hat{\mu}(50)=\exp\{\log(3)+0.08(0)\}=3. $$

At $x=65$, $x-50=15$, giving $$ \hat{\mu}(65)=\exp\{\log(3)+0.08(15)\} =3\exp(1.2) \approx 3(3.320117) \approx 9.960. $$

The fitted mean increases over these predictor values because the slope is positive. The three answers are conditional expected counts, not predictions that every individual response will equal those numbers.

How the Slope Determines the Direction of the Mean Curve

The model has a direct calculus consequence. Treating $x$ as a real-valued predictor, write $\mu(x)=\exp(\beta_0+\beta_1x)$. Differentiating shows how rapidly the conditional mean changes locally as $x$ changes.

Theorem: For the one-predictor log-mean model $\mu(x)=\exp(\beta_0+\beta_1x)$, the derivative is $$ \frac{d\mu(x)}{dx}=\beta_1\mu(x). $$ Consequently, the mean is strictly increasing in $x$ if $\beta_1>0$, strictly decreasing if $\beta_1<0$, and constant if $\beta_1=0$.

Proof. Apply the chain rule to the exponential function. The derivative of $\beta_0+\beta_1x$ with respect to $x$ is $\beta_1$, so $$ \frac{d}{dx}\exp(\beta_0+\beta_1x) =\exp(\beta_0+\beta_1x)\beta_1 =\beta_1\mu(x). $$ Since $\mu(x)>0$ for every real $x$, the derivative has the sign of $\beta_1$. A positive derivative throughout the predictor range implies strict increase; a negative derivative implies strict decrease; and if $\beta_1=0$, the derivative is zero everywhere and the mean is constant. $\square$

The derivative also clarifies why the change in expected counts is not generally constant per unit of $x$. It depends on $\mu(x)$, which itself varies with $x$. The slope gives a change on the log-mean scale; the derivative gives the corresponding local change on the mean scale.

Worked Example: A Predictor on Its Original Scale

Suppose the fitted equation is $\log\hat{\mu}(x)=0.4+0.03x$, where $x$ is a laboratory measurement recorded in its original units. Find the fitted means at $x=0$, $10$, and $20$.

For $x=0$, the fitted log mean is $0.4+0.03(0)=0.4$, hence $$ \hat{\mu}(0)=\exp(0.4)\approx 1.491825\approx 1.492. $$

For $x=10$, the fitted log mean is $0.4+0.03(10)=0.7$, hence $$ \hat{\mu}(10)=\exp(0.7)\approx 2.013753\approx 2.014. $$

For $x=20$, the fitted log mean is $0.4+0.03(20)=1.0$, hence $$ \hat{\mu}(20)=\exp(1.0)\approx 2.718282\approx 2.718. $$

The positive slope is consistent with the increasing sequence of fitted means. These calculations use the stated coefficient estimates; they do not show how those estimates were obtained.

Centering the Predictor

A continuous predictor may be centered by subtracting a chosen constant $c$, often a value near the middle of the observed range. Centering changes the numerical intercept but need not change the fitted curve. It can also make hand calculations more convenient, as in the first example.

Result: Replacing $x$ with $x-c$ preserves the one-predictor log-mean model if its intercept is re-expressed as $\gamma_0=\beta_0+c\beta_1$, while its slope remains $\gamma_1=\beta_1$.

Derivation. Substitute $x=(x-c)+c$ into the original linear predictor: $$ \beta_0+\beta_1x =\beta_0+\beta_1\{(x-c)+c\} =(\beta_0+c\beta_1)+\beta_1(x-c). $$ Set $\gamma_0=\beta_0+c\beta_1$ and $\gamma_1=\beta_1$. The expression for the log mean is then $\gamma_0+\gamma_1(x-c)$, exactly equal to the original expression for every $x$. Exponentiating equal log means gives equal conditional means at every predictor value. The re-expression changes the coordinate description of the same curve, not the curve itself. $\square$

Worked Example: A Decreasing Fitted Mean

Suppose $\log\hat{\mu}(x)=\log(2)-0.05(x-20)$, where $x$ is a continuous measurement. Calculate fitted means at $x=10$, $20$, $30$, and $40$.

At $x=10$, $x-20=-10$, so $$ \hat{\mu}(10)=\exp\{\log(2)-0.05(-10)\} =2\exp(0.5) \approx 2(1.648721) \approx 3.297. $$

At $x=20$, the centered value is zero, giving $\hat{\mu}(20)=2$. At $x=30$, $$ \hat{\mu}(30)=\exp\{\log(2)-0.05(10)\} =2\exp(-0.5) \approx 2(0.606531) \approx 1.213. $$

At $x=40$, $$ \hat{\mu}(40)=\exp\{\log(2)-0.05(20)\} =2\exp(-1) \approx 2(0.367879) \approx 0.736. $$

The negative slope gives a decreasing mean curve, as the calculated values confirm. The fitted mean can be below one; it remains positive, as required by the log link.

Practical Checks and Common Mistakes

A fitted mean is useful only when its predictor value and units are clear. Evaluate the model at values that make sense for the study, and be cautious about values far outside the range used to fit it: the equation returns a number there, but that numerical calculation alone does not establish that the model is a reliable description in that region.

Using the linear predictor as the expected count. The expression $\hat{\beta}_0+\hat{\beta}_1x$ is the fitted log mean. Exponentiate the entire expression to obtain $\hat{\mu}(x)$.
Assuming the fitted mean must be an integer. Counts are integer-valued; their conditional expectation can be any positive real number.
Reversing the slope’s direction. A positive slope increases the log mean and therefore increases the mean; a negative slope decreases it.
Changing predictor units without changing the coefficient. If the numerical scale changes, the numerical slope must be expressed on that new scale before making fitted calculations.
Rounding too early. Keep the unrounded log mean and exponential calculation until the final displayed value; otherwise small rounding differences can accumulate.

A practical calculation follows a consistent sequence: substitute the selected $x$, evaluate the complete linear predictor, exponentiate, and round the resulting mean. Keeping these steps separate makes errors easier to detect.

Check Your Understanding

Use the model definitions and calculation sequence above to answer these questions.

For $\log\hat{\mu}(x)=\log(5)+0.04(x-25)$, calculate the fitted means at $x=25$ and $x=35$.
In a one-predictor Poisson regression, what does the linear expression $\hat{\beta}_0+\hat{\beta}_1x$ represent, and what step gives the fitted conditional mean?
What is the direction of change in the fitted mean when the slope is negative? Justify your answer using the derivative.
Explain why a fitted conditional mean of $2.6$ is possible even though the response is a count.
Show algebraically how centering $x$ at $c$ can preserve the model’s fitted means.