Rate Ratios From Log-Link Models

Why Rate Ratios Need an Exposure Offset

Count data often come from observation periods or populations of different sizes. For example, one group of policies may be observed for twice as many policy-years as another. A larger expected number of claims could then reflect greater exposure, a higher claim rate, or both. A log-link model with an exposure offset separates these effects: the offset accounts for the amount observed, while predictor coefficients describe multiplicative changes in the rate.

The offset is the logarithm of exposure, entered with its coefficient fixed at 1. This is not the same as estimating an ordinary coefficient for log exposure. Fixing the coefficient at 1 encodes the assumption that, all else equal, doubling exposure doubles the expected count.

Definition: Let $N_i$ be an event count for observation $i$, and let $e_i>0$ be its exposure measured in a specified unit. In a Poisson log-link model with an exposure offset, $$ N_i\mid \mathbf{x}_i,e_i \sim \operatorname{Poisson}(\mu_i), \qquad \log(\mu_i)=\log(e_i)+\beta_0+\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta}. $$ The conditional expected count is $\mu_i$, and the modeled rate per exposure unit is $\rho_i=\mu_i/e_i$. This specification assumes the count has a Poisson distribution conditional on the predictors and exposure, and that expected count is proportional to exposure when the predictors are held fixed. The logarithm requires positive exposure.

The rate ratio compares two modeled rates while holding the other predictors fixed. In a model without interactions, increasing a predictor $x$ by $\Delta$ changes the log rate by $\beta\Delta$, so the rate is multiplied by $\exp(\beta\Delta)$. For a one-unit increase, the multiplier is $\exp(\beta)$. If $x$ is an indicator changing from 0 to 1, its rate ratio is also $\exp(\beta)$, with all other predictors held fixed.

\rho_i=\frac{\mu_i}{e_i} =\exp\!\left(\beta_0+\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta}\right), \qquad \frac{\rho_{\text{new}}}{\rho_{\text{old}}} =\exp(\beta\Delta)

A rate ratio is a relative comparison, not an absolute difference in rates. For example, a rate ratio of 1.25 means the modeled rate is 25% higher for the specified predictor change. It does not mean the rate rises by 0.25 events, nor does it by itself specify the difference in expected counts. The corresponding expected-count comparison also depends on exposure.

Deriving the Rate Ratio

The result follows directly from dividing the modeled mean count by exposure. Consider two otherwise identical observations whose predictor values differ by $\Delta$ for $x$. Their log expected counts differ by $\beta\Delta$; the offset terms cancel if exposure is the same. Exponentiating gives the ratio of their expected counts. Since equal exposure makes the ratio of expected counts equal to the ratio of rates, this is the rate ratio. More generally, even if exposures differ, dividing each expected count by its own exposure leaves the same rate comparison.

Write the two linear predictors.
For observations differing only by $\Delta$ in $x$, their linear predictors excluding the offset differ by $\beta\Delta$.

Express each rate.
Because $\rho_i=\mu_i/e_i$, the exposure term cancels from the expression for each modeled rate.

Divide the rates.
The ratio is $\exp(\beta\Delta)$, because $\exp(a+b)/\exp(a)=\exp(b)$.

\frac{\rho_{\text{new}}}{\rho_{\text{old}}} = \frac{\exp(\eta_{\text{old}}+\beta\Delta)} {\exp(\eta_{\text{old}})} =\exp(\beta\Delta)

Here $\eta_{\text{old}}$ denotes the part of the linear predictor excluding the offset and the changed contribution $\beta\Delta$. This derivation depends on a log link and on comparing observations with the other modeled predictors fixed. If the predictor change also changes another modeled term, that term must be included in the log-rate difference before exponentiating.

Worked Interpretations

Worked Example: From an Indicator Coefficient to a Rate Ratio

A hypothetical service program models the number of equipment incidents using a Poisson log-link model. Exposure $e$ is measured in machine-years, and $z=1$ indicates use of a specified maintenance schedule while $z=0$ indicates the comparison schedule. The fitted model is $\log(\mu)=\log(e)+\log(0.04)+\log(1.5)z$. Find the rate ratio, the rates for the two schedules, and the expected counts for $e=10$ machine-years. Use the model’s assumptions that counts are conditionally Poisson and expected counts are proportional to exposure.

Method. First remove exposure by dividing the expected count by $e$. Then exponentiate the coefficient difference to obtain the rate ratio.

For $z=0$, the rate is $\exp(\log(0.04))=0.04$ incidents per machine-year. Its expected count at $e=10$ is $10(0.04)=0.40$ incidents.

For $z=1$, the rate is $\exp(\log(0.04)+\log(1.5))=0.04(1.5)=0.06$ incidents per machine-year. Its expected count at $e=10$ is $10(0.06)=0.60$ incidents.

The rate ratio is $0.06/0.04=1.5$, which is also $\exp(\log(1.5))$. Thus the modeled rate is 50% higher under the specified schedule. The absolute rate difference is $0.06-0.04=0.02$ incidents per machine-year, and the expected-count difference for this exposure is $0.60-0.40=0.20$ incidents.

Check. The rate ratio from the two rates is 1.5, agreeing with the exponentiated coefficient. A common mistake is to describe $\log(1.5)$ itself as a 50% increase; the percentage increase comes from exponentiating the coefficient first.

Worked Example: Rate Ratio Versus Total Expected Count

A hypothetical portfolio model has the form $\log(\mu)=\log(e)+\log(0.08)+\log(1.25)z+\gamma w$, where $e$ is exposure in a common unit, $z$ is a binary rating characteristic, and $w$ is another predictor. Compare a risk with $z=1$, $e=400$, and $w=2$, with a risk having $z=0$, $e=250$, and $w=2$. Use the model assumptions of conditional Poisson counts, proportionality to exposure, and the same value of $w$ in both predictions.

Method. Compare rates first, holding $w$ fixed, and then calculate the expected counts using each risk’s own exposure.

The rate for $z=0$ is $\exp(\log(0.08)+2\gamma)=0.08\exp(2\gamma)$. The rate for $z=1$ is $\exp(\log(0.08)+\log(1.25)+2\gamma)=0.10\exp(2\gamma)$. The common factor $\exp(2\gamma)$ cancels in the ratio, giving a rate ratio of $0.10/0.08=1.25$.

The expected counts are $\mu_1=400(0.10\exp(2\gamma))=40\exp(2\gamma)$ and $\mu_0=250(0.08\exp(2\gamma))=20\exp(2\gamma)$. Their ratio is $40/20=2$. The expected count is therefore twice as large for the first risk, despite its rate being only 1.25 times as large. The additional multiplier is the exposure ratio $400/250=1.6$, and $1.25(1.6)=2$.

Check. The common $w$ contribution cancels in both comparisons. The total-count ratio equals the rate ratio multiplied by the exposure ratio. A common mistake is to report 1.25 as the expected-count ratio while ignoring the unequal exposures.

Worked Example: Exponentiating a Nonzero Coefficient

A hypothetical monitoring model uses exposure measured in thousands of insured-unit-years. Its fitted form is $\log(\mu)=\log(e)+\log(25)+0.4x$, where $x$ is a one-unit rating measure. Assume conditional Poisson counts and the stated exposure-proportional mean. Compare $x=0$ with $x=1$ for exposures of $0.8$ and $1.2$, respectively. Calculate the rate ratio and expected-count ratio. Round the rate ratio to four decimal places and expected counts to two decimal places.

Method. Exponentiate the coefficient for the rate ratio. Calculate each expected count as exposure multiplied by its rate; do not substitute the rate ratio for the count ratio.

At $x=0$, the rate is $25$ events per thousand insured-unit-years, so $\mu_0=0.8(25)=20$. At $x=1$, the rate is $25\exp(0.4)$. Using $\exp(0.4)=1.4918247$, this rate is $25(1.4918247)=37.2956$ events per thousand insured-unit-years. Its expected count is $\mu_1=1.2(37.2956)=44.7547$.

The rate ratio is $\exp(0.4)=1.4918$ to four decimal places. The expected-count ratio is $44.7547/20=2.2377$, which also equals the exposure ratio $1.2/0.8=1.5$ multiplied by the rate ratio $1.4918247$. Rounded as requested, the expected counts are 20.00 and 44.75.

Check. The first rate is recovered as $20/0.8=25$; the second is recovered as $44.7547/1.2=37.2956$. A common error is to interpret $0.4$ as a 40% rate increase. The correct increase is $(\exp(0.4)-1)100\%=49.1825\%$, or 49.18% to two decimal places.

Reading Rate Ratios Carefully

The rate ratio is tied to the predictor’s scale. A coefficient applies to a one-unit increase as the variable is recorded in the model. If the model uses a predictor measured in tens of units, a one-unit increase in the recorded variable represents ten units on the original scale. For a change of $\Delta$ recorded units, use $\exp(\beta\Delta)$, not automatically $\exp(\beta)$.

The exposure unit must also be stated. If exposure is measured in machine-years, the rate is events per machine-year; if it is measured in thousands of machine-years, the numerical rate is events per thousand machine-years. Changing the exposure unit changes the numerical rate and the corresponding intercept, but does not change a rate ratio for a predictor contrast when the model is otherwise equivalent.

An estimated rate ratio describes a model-based association conditional on the included predictors. It does not establish that changing the predictor would cause the rate to change. Interpretation also depends on the model’s specification: for example, an omitted predictor or an unsuitable assumption of proportionality to exposure can make the fitted comparison misleading. A log-link rate-ratio interpretation is aligned with the SOA Exam PA (Predictive Analytics) syllabus.

Key takeaway: In a log-link count model with $\log(e)$ as an offset, a predictor change of $\Delta$ multiplies the modeled rate by $\exp(\beta\Delta)$. Expected counts also depend on exposure: $\mu=e\rho$. Keep those two comparisons separate.

Check Your Understanding

For each problem, state the rate comparison separately from any expected-count comparison. Assume the stated log-link models and exposure-proportional means apply.

A hypothetical count model has $\log(\mu)=\log(e)+\log(0.03)+\log(2)z$, where $e$ is exposure and $z$ changes from 0 to 1. Find the rate ratio and the two rates. (Answer: rate ratio $2$; rates $0.03$ and $0.06$ per exposure unit.)
A model has coefficient $\beta=0.2$ for predictor $x$. Find the rate ratio for a 3-unit increase in $x$, rounded to four decimal places. (Answer: $\exp(0.6)=1.8221$.)
In a hypothetical model, the rate is 12 events per exposure unit at $z=0$, and the coefficient for binary $z$ is $\log(1.4)$. Compare expected counts for $z=0,e=50$ and $z=1,e=30$. (Answer: rates $12$ and $16.8$; expected counts $600$ and $504$; rate ratio $1.4$, expected-count ratio $0.84$.)
A fitted coefficient is $-0.3$ for a one-unit increase in $x$. Calculate the rate ratio and percentage change in rate, rounding the ratio to four decimals and the percentage to two decimals. (Answer: rate ratio $\exp(-0.3)=0.7408$; rate decreases by $25.92\%$.)
A model estimates a rate ratio of $1.2$ for a predictor contrast. The comparison observations have exposures $75$ and $125$, respectively. What is the expected-count ratio for the first observation relative to the second? (Answer: $1.2(75/125)=0.72$.)

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