Changing Exposure Units Without Changing Rate Effects

Why Exposure Units Matter

A count model often describes how many events occur during a given amount of exposure: events per hour, per day, or per person-year, for example. Changing the exposure unit changes the numerical value of a rate. It should not, however, change the expected event count for the same observation or the relative rate associated with a covariate. Keeping these quantities distinct makes it easier to interpret fitted Poisson models and to check calculations.

Suppose \(Y_i\) is a count and \(e_i>0\) is its exposure, measured in some chosen unit. Let \(z_i\) denote the covariates for observation \(i\), and let \(\beta_0\) be the intercept. In a Poisson rate model,

Y_i\mid e_i,z_i\sim\operatorname{Poisson}(\mu_i), \qquad \log(\mu_i)=\log(e_i)+\beta_0+z_i^\mathsf{T}\beta.

Thus the fitted rate per exposure unit is \(\exp(\beta_0+z_i^\mathsf{T}\beta)\), and the fitted mean count is exposure multiplied by that rate. The offset \(\log(e_i)\) is determined by the recorded exposure; the conversion considered here changes its numerical value in a precisely matched way.

The Algebra of a Unit Conversion

Write \(e_i^*=a e_i\), where \(a>0\) is the number of new-unit exposure units corresponding to one old-unit exposure. For instance, if \(e_i\) is measured in hours and \(e_i^*\) in days, then \(a=1/24\). The model must use an intercept adjusted to the new rate unit.

Definition: A constant exposure-unit conversion replaces every exposure \(e_i\) by \(e_i^*=a e_i\) for the same positive constant \(a\). In the Poisson log-rate model, the corresponding intercept is \(\beta_0^*=\beta_0-\log(a)\); the covariate coefficient vector remains \(\beta\).

The direction of the adjustment follows directly from the model. Since \(\log(e_i^*)=\log(e_i)+\log(a)\), the intercept must decrease by \(\log(a)\) in the new parameterization to keep the sum of the offset and intercept unchanged. The new rate is per new exposure unit, so it equals the old rate divided by \(a\).

Theorem: Under a constant conversion \(e_i^*=a e_i\), \(a>0\), set \(\beta_0^*=\beta_0-\log(a)\) and leave \(\beta\) unchanged. Then every fitted mean is unchanged, the numerical fitted rate is divided by \(a\), and every covariate rate ratio is unchanged.

Proof. For every observation \(i\),

\begin{aligned} \log(\mu_i^*) &=\log(e_i^*)+\beta_0^*+z_i^\mathsf{T}\beta\\ &=\log(ae_i)+\beta_0-\log(a)+z_i^\mathsf{T}\beta\\ &=\log(e_i)+\beta_0+z_i^\mathsf{T}\beta\\ &=\log(\mu_i). \end{aligned}

Therefore \(\mu_i^*=\mu_i\). The rate in the old units is \(r_i=\exp(\beta_0+z_i^\mathsf{T}\beta)\), while the rate in the new units is

r_i^*=\exp(\beta_0^*+z_i^\mathsf{T}\beta) =\exp(\beta_0+z_i^\mathsf{T}\beta-\log(a)) =\frac{r_i}{a}.

For two covariate profiles \(z_A\) and \(z_B\), their rate ratio in either unit is

\frac{r_A^*}{r_B^*} =\frac{r_A/a}{r_B/a} =\frac{r_A}{r_B} =\exp\!\left((z_A-z_B)^\mathsf{T}\beta\right).

The common unit-conversion factor cancels. This proves the claims. \(\square\)

Worked Example: Converting an Hourly Rate to a Daily Rate

Suppose the fitted hourly rate for a reference group is \(\exp(-4.6)\) events per hour, and a binary covariate \(x\) has coefficient \(0.5\). The fitted rate is \(\exp(-4.6+0.5x)\) per hour. Convert the model to days and check the fitted means for \(x=0\) and \(x=1\) over 48 hours.

For hours converted to days, \(e_{\text{day}}=e_{\text{hour}}/24\), so \(a=1/24\). The new intercept is

\beta_0^*=-4.6-\log(1/24)=-4.6+\log(24)\approx-1.42195.

The coefficient of \(x\) stays \(0.5\). The reference rate is now \(\exp(-1.42195)\approx0.24124\) events per day, equal to \(24\exp(-4.6)\). For \(x=1\), the daily rate is \(\exp(-1.42195+0.5)\approx0.39770\) events per day. The rate ratio is \(\exp(0.5)\approx1.64872\) in either unit.

A 48-hour exposure is 2 days. For \(x=0\), the fitted mean is \(48\exp(-4.6)\approx0.48249\) using hours, and \(2\exp(-1.42195)\approx0.48249\) using days. For \(x=1\), the corresponding means are \(48\exp(-4.1)\approx0.79549\) and \(2\exp(-1.42195+0.5)\approx0.79549\). The counts’ fitted means agree, while the rate values have changed units.

Likelihood and Fitted-Model Equivalence

The equality of the means is not just a prediction-level observation. It implies that the entire Poisson likelihood is unchanged when the intercept is transformed along with the exposure. This gives a precise reason that fitting the same model after a consistent conversion does not alter the fitted counts.

Theorem: Suppose \(Y_1,\ldots,Y_n\) are conditionally independent Poisson variables with positive exposures \(e_i\), and the model has means \(\mu_i=e_i\exp(\beta_0+z_i^\mathsf{T}\beta)\). For a fixed \(a>0\), use exposures \(e_i^*=ae_i\) and parameter map \(T(\beta_0,\beta)=(\beta_0-\log(a),\beta)\). The log-likelihood at a parameter and at its image under \(T\) is equal. Consequently, if a maximum-likelihood estimate exists, the maximizers correspond under \(T\), and the fitted means are identical.

Proof. Conditional independence and the Poisson probability mass function give

L(\beta_0,\beta) =\prod_{i=1}^n \frac{\exp(-\mu_i)\mu_i^{y_i}}{y_i!}.

The first theorem shows that the transformed model has \(\mu_i^*=\mu_i\) for each \(i\). Substitution into the likelihood therefore gives

L^*(T(\beta_0,\beta)) =\prod_{i=1}^n\frac{\exp(-\mu_i^*) (\mu_i^*)^{y_i}}{y_i!} =\prod_{i=1}^n\frac{\exp(-\mu_i)\mu_i^{y_i}}{y_i!} =L(\beta_0,\beta).

Taking logarithms gives equality of the log-likelihoods as well. The map \(T\) is one-to-one and onto the corresponding parameter space: its inverse adds \(\log(a)\) to the intercept and leaves \(\beta\) unchanged. Thus each parameter value in one representation has exactly one corresponding value with the same likelihood in the other. If a maximum exists, its image is a maximum in the converted representation, and the fitted means coincide. \(\square\)

This is a reparameterization, not a change in the observations’ event counts or in the model’s covariate effects. The intercept has a different numerical value because it represents a rate in a different unit; the likelihood and fitted counts remain the same.

Worked Example: Converting Daily Rates to Hourly Rates

A fitted model has a reference rate of \(0.2\) events per day and a binary covariate coefficient \(\log(1.5)\). Calculate the two rates and expected counts for a reference observation with 3 days of exposure and a covariate-positive observation with 5 days of exposure. Then repeat the calculation in hours.

The daily intercept is \(\log(0.2)\), and the rate for \(x=1\) is \(0.2\exp(\log(1.5))=0.3\) events per day. The expected counts are \(3(0.2)=0.6\) for the reference observation and \(5(0.3)=1.5\) for the covariate-positive observation.

Converting days to hours gives \(e_{\text{hour}}=24e_{\text{day}}\), so \(a=24\). The hourly intercept is

\beta_0^*=\log(0.2)-\log(24)=\log(0.2/24).

Thus the reference hourly rate is \(0.2/24=0.008333\ldots\), and the \(x=1\) hourly rate is \(0.3/24=0.0125\). Three days is 72 hours and five days is 120 hours. The hourly calculations give \(72(0.008333\ldots)=0.6\) and \(120(0.0125)=1.5\), the same fitted means as before. The rate ratio remains \(0.3/0.2=0.0125/0.008333\ldots=1.5\).

Person-Days and Person-Years

The same algebra applies when exposure is measured in person-time. If exposure is recorded in person-days and the target unit is person-years, using 365.25 days per year gives \(e_{\text{year}}=e_{\text{day}}/365.25\). The numerical rate per person-year is therefore 365.25 times the rate per person-day. This conversion changes the rate scale, not the covariate rate ratios or fitted expected counts.

Worked Example: Converting a Rate to Person-Years

Suppose a model gives a rate of \(0.004\) events per person-day in the reference group and a rate ratio of 2 for a group indicator. Convert both rates to person-years and calculate the expected count for 500 person-days in the second group.

The second group’s rate is \(0.004(2)=0.008\) events per person-day. Since one person-year is 365.25 person-days, the corresponding rates are \(0.004(365.25)=1.461\) and \(0.008(365.25)=2.922\) events per person-year. The rate ratio remains \(2.922/1.461=2\).

The exposure of 500 person-days is \(500/365.25\approx1.368925\) person-years. The expected count in the second group is \(500(0.008)=4\) using person-days. Using person-years gives approximately \(1.368925(2.922)=4\), with any tiny discrepancy attributable only to rounding. Both calculations use the same exposure and rate on matching scales.

Common Mistake and Practical Checks

A common mistake is to convert the exposure but leave the intercept unchanged. For example, changing hours to days makes the numerical exposure 24 times smaller. If the rate is also left unchanged, the product of exposure and rate becomes 24 times smaller, so the fitted mean count changes. That is not a unit conversion of the same model; it is a different mean model.

The correction is to transform both parts together. Whenever \(e^*=ae\), change the intercept to \(\beta_0-\log(a)\), or equivalently divide the rate per old unit by \(a\) to obtain the rate per new unit. Then check that \(e^*r^*=er\). The covariate coefficients do not require conversion, and the rate ratio between covariate profiles remains the exponential of their coefficient contrast.

Key takeaway: A consistent exposure-unit conversion changes the offset and the numerical rate by reciprocal factors. The intercept adjusts, but the fitted means, likelihood, covariate coefficients, and covariate rate ratios are preserved.

Check Your Understanding

Use the conversion factor and check that the exposure-rate product is preserved.

A model uses exposure in hours. What value of \(a\) converts exposure to days, and how does the intercept change?
If a rate is 0.03 events per hour, what is the corresponding rate per day? Explain why the numerical rate increases.
A covariate has coefficient \(\log(1.8)\). What is its rate ratio after exposure is converted from days to hours?
A fitted mean is calculated using 4 days of exposure and a rate of 0.5 events per day. Show the equivalent calculation using hours.
Why does changing the exposure unit consistently leave the Poisson likelihood unchanged?