Poisson Canonical Link Worked Derivation

Why the Poisson Link Is the Log Link

For claim frequency, the response is a nonnegative integer count. The Poisson model is useful when the expected count is nonnegative and the conditional variance is taken to equal the conditional mean. Its canonical link follows directly from the Poisson probability mass function; it is not merely a convenient way to keep fitted means positive.

Let \(N\) be a Poisson random variable with mean \(\mu>0\). Its probability mass function is \(P(N=n)=e^{-\mu}\mu^n/n!\), for \(n=0,1,2,\ldots\). A one-parameter exponential-family form is

P(N=n)=\exp\left\{n\theta-b(\theta)+c(n)\right\}, \qquad \theta=\log\mu,\quad b(\theta)=e^\theta,\quad c(n)=-\log(n!).

In an exponential family, the natural parameter is \(\theta\), and the canonical link is the function that maps the mean \(\mu\) to that natural parameter. Since \(\theta=\log\mu\), the Poisson canonical link is the log link. The exponential-family identities also give \(E[N]=b'(\theta)=e^\theta=\mu\) and \(\operatorname{Var}(N)=b''(\theta)=e^\theta=\mu\).

Definition: For a Poisson response with mean \(\mu>0\), the canonical link is \(g(\mu)=\log\mu\), with inverse link \(g^{-1}(\eta)=e^\eta\). A linear predictor \(\eta\) therefore gives a fitted mean \(\hat{\mu}=e^\eta\), which is always positive.

The distinction between the natural parameter and the mean matters: \(\theta\) can be any real number, while \(\mu=e^\theta\) must be positive. A linear predictor can range over all real values, so using it for \(\log\mu\) respects the required range of the Poisson mean.

Adding a Known Exposure Offset

In ratemaking, an observed count usually arises over an exposure amount, such as policy-years or vehicle-years. Let \(e>0\) be known exposure and let \(\lambda\) be the expected claim count per exposure unit. Under the Poisson frequency assumption, \(N\mid e,x\sim\operatorname{Poisson}(e\lambda(x))\), where \(x\) denotes rating characteristics. Thus the Poisson mean is the expected count over the observed exposure, \(e\lambda(x)\), while \(\lambda(x)\) is the expected count per exposure unit.

Suppose the rate has a log-linear rating structure, \(\lambda(x)=\exp(x^\mathsf{T}\beta)\), where \(x^\mathsf{T}\beta\) is the linear predictor for the rate. Then

\log E[N\mid e,x] =\log(e\lambda(x)) =\log e+x^\mathsf{T}\beta.

The known quantity \(\log e\) is an offset: its coefficient is fixed at one. It shifts the predictor for the expected count but does not change the modeled rate \(e^{x^\mathsf{T}\beta}\). This is the Poisson canonical link applied to the count mean, expressed so the rating model itself describes a rate.

Key result: With positive known exposure \(e\), a Poisson frequency model with rate \(\lambda(x)=\exp(x^\mathsf{T}\beta)\) has fitted count mean \(\hat{\mu}=e\exp(x^\mathsf{T}\beta)\). Divide that fitted count mean by \(e\) to recover the fitted rate.

The offset form requires \(e>0\), because \(\log e\) is undefined at zero. A zero-exposure record does not represent a period at risk for a Poisson count model with this offset; it should not be handled by inserting an arbitrary small exposure. The assumed Poisson model also implies equality of conditional mean and variance. If that assumption is unsuitable, the model’s variance structure may need review; the canonical-link derivation alone does not establish that Poisson variation fits every data set.

Deriving the Canonical Link and Offset Form

The following steps connect the probability model to a usable rating prediction. The derivation also clarifies why the offset belongs beside the rate predictor rather than inside the estimated rating factors.

Write the Poisson mass function.
For a count with mean \(\mu\), \(P(N=n)=e^{-\mu}\mu^n/n!\). Taking the logarithm of the mass gives \(n\log\mu-\mu-\log(n!)\).

Identify the natural parameter.
Set \(\theta=\log\mu\). Then \(\mu=e^\theta\), so the log mass becomes \(n\theta-e^\theta-\log(n!)\), the exponential-family form with \(b(\theta)=e^\theta\).

Read off the canonical link.
The natural parameter is \(\theta\), and the mean is \(\mu=b'(\theta)=e^\theta\). Therefore the function taking the mean to the natural parameter is \(g(\mu)=\log\mu\).

Separate count from rate.
For exposure \(e>0\), the count mean is \(\mu=e\lambda(x)\). Hence \(\log\mu=\log e+\log\lambda(x)\). If \(\log\lambda(x)=x^\mathsf{T}\beta\), the exposure term is the known offset and the fitted rate remains \(e^{x^\mathsf{T}\beta}\).

The result can also be checked by reversing the transformation: exponentiating \(\log e+x^\mathsf{T}\beta\) gives \(e\exp(x^\mathsf{T}\beta)\), exactly exposure multiplied by rate. This check catches a frequent error: the offset is added on the log scale, which means exposure multiplies the mean on the original scale.

Worked Example: Predicting a Count for One Rating Cell

A hypothetical insurer models claim counts using a Poisson rate per vehicle-year. The base rate is \(0.10\) claims per vehicle-year. A rating characteristic multiplies the rate by \(1.2\), and a second characteristic multiplies it by \(1.5\). A policy has \(1.5\) vehicle-years of exposure. Calculate the fitted rate and expected count. State the method and avoid confusing count with rate.

Method. Use the log-link rate model with multiplicative factors. The rate is the base rate times both relativities:

\hat{\lambda}=0.10(1.2)(1.5)=0.10(1.8)=0.18 \quad\text{claims per vehicle-year}.

Equivalently, the rate-scale predictor is \(\log(0.10)+\log(1.2)+\log(1.5)=\log(0.18)\). The count mean includes the exposure offset:

\log\hat{\mu}=\log(1.5)+\log(0.18)=\log(0.27), \qquad \hat{\mu}=1.5(0.18)=0.27.

The fitted rate is \(0.18\) claims per vehicle-year, and the fitted expected count over \(1.5\) vehicle-years is \(0.27\). Check: dividing \(0.27\) by \(1.5\) gives \(0.18\), the rate just calculated. The common mistake avoided is reporting \(0.27\) as the rate; it is the expected count for this policy’s stated exposure.

Worked Example: Using Coefficients and an Exposure Offset

A hypothetical frequency model has rate predictor \(\eta=\log(0.05)+\log(1.6)+\log(0.75)\) for a particular risk. The risk has \(4\) policy-years of exposure. Calculate the rate and fitted count mean using the canonical Poisson link. Explain why the offset is not another rating factor.

Method. The rate is the exponential of the rate predictor. Combine its factors:

\hat{\lambda} =\exp(\eta) =0.05(1.6)(0.75) =0.05(1.20) =0.06 \quad\text{claims per policy-year}.

The offset for the count mean is \(\log 4\), so

\log\hat{\mu}=\log 4+\eta, \qquad \hat{\mu}=\exp(\log4+\eta)=4\exp(\eta)=4(0.06)=0.24.

The answer is a fitted rate of \(0.06\) claims per policy-year and a fitted expected count of \(0.24\) claims over four policy-years. Check: \(0.24/4=0.06\). The common mistake avoided is treating exposure as an estimated multiplicative rating factor: it is known, and its log enters with coefficient one to scale the expected count.

Worked Example: Aggregating Fitted Counts Into a Rate

A hypothetical portfolio has three rating cells. Cell A has \(120\) vehicle-years and fitted rate \(0.03\) claims per vehicle-year. Cell B has \(80\) vehicle-years and fitted rate \(2.5\) times the base rate. Cell C has \(50\) vehicle-years and fitted rate \(4\) times the base rate. The base rate is \(0.03\). Calculate each fitted count mean, the total expected count, and the aggregate fitted rate. State why an unweighted average of cell rates is not appropriate.

Method. First calculate each rate, then multiply it by that cell’s exposure. For cell A, the rate is \(0.03\), so the count mean is \(120(0.03)=3.6\). For cell B, the rate is \(0.03(2.5)=0.075\), so the count mean is \(80(0.075)=6.0\). For cell C, the rate is \(0.03(4)=0.12\), so the count mean is \(50(0.12)=6.0\).

Summing the cell fitted means gives the portfolio expected count:

\hat{N}_{\mathrm{total}}=3.6+6.0+6.0=15.6.

Total exposure is \(120+80+50=250\) vehicle-years. Therefore the aggregate fitted rate is

\hat{\lambda}_{\mathrm{aggregate}} =\frac{15.6}{250} =0.0624 \quad\text{claims per vehicle-year}.

Check: the exposure-weighted rate calculation is \([120(0.03)+80(0.075)+50(0.12)]/250=15.6/250=0.0624\). The unweighted average of the three rates would be \((0.03+0.075+0.12)/3=0.075\), which gives equal influence to cells with different exposure. The common mistake avoided is averaging cell rates without weighting by exposure.

Interpretation and Common Exam Traps

A fitted Poisson count mean need not be a whole number. It is an expected count, not a prediction that the policy will produce a fractional observed claim. For example, a fitted mean of \(0.24\) is compatible with an integer-valued observed count; it summarizes the model’s conditional expectation over the specified exposure.

The canonical-link derivation applies to the Poisson response mean. In a frequency application with exposure, the response count has mean \(e\lambda(x)\), so the canonical link applies to that mean. The offset formulation then expresses the same model in terms of a rate. Keeping these quantities distinct avoids errors in both prediction and interpretation.

Using exposure as the rate. The rate is \(e^{x^\mathsf{T}\beta}\); the expected count is \(e\,e^{x^\mathsf{T}\beta}\).
Dropping the offset. Without \(\log e\), the predictor models a count mean that does not scale with the amount of time or risk exposure.
Multiplying on the wrong scale. The offset is added to the linear predictor, then exponentiation makes it multiplication by exposure on the mean scale.
Using the wrong aggregate. A portfolio rate is total fitted count divided by total exposure, not generally the unweighted mean of cell rates.
Assuming every response has Poisson variation. The canonical-link calculation identifies the mean structure; it does not verify that the Poisson mean-variance assumption is appropriate for a particular application.

Key takeaway: The Poisson probability mass function has natural parameter \(\theta=\log\mu\), so its canonical link is \(g(\mu)=\log\mu\). With known positive exposure \(e\), model the count mean as \(e\lambda(x)\): the log-exposure offset scales the count, while the log-linear rating predictor determines the rate.

Check Your Understanding

For each problem, identify whether the requested quantity is a rate or a fitted count mean before calculating.

A hypothetical Poisson frequency model has base rate \(0.08\) claims per vehicle-year and a rating factor of \(1.25\). A risk has \(2.5\) vehicle-years of exposure. Find the fitted rate and expected count. (Answer: rate \(0.10\) claims per vehicle-year; expected count \(0.25\))
A model has rate predictor \(\eta=-3+\log 2\), and a policy has exposure \(0.5\) policy-years. Calculate the rate and fitted count mean. (Answer: rate \(2e^{-3}\approx0.0996\) claims per policy-year; expected count \(e^{-3}\approx0.0498\))
Two hypothetical cells have exposures of \(40\) and \(60\) policy-years, with fitted rates \(0.02\) and \(0.05\). Calculate the total fitted count mean and the aggregate rate. (Answer: expected count \(3.8\); aggregate rate \(0.038\) claims per policy-year)
A base rate is \(0.04\) claims per policy-year. Cell A has factor \(1.5\) and exposure \(10\) policy-years; cell B has factor \(0.5\) and exposure \(20\) policy-years. Find both cell fitted counts and the aggregate rate. (Answer: cell counts \(0.6\) and \(0.4\); aggregate rate \(0.0333\) claims per policy-year)
For a Poisson response with mean \(\mu\), write its natural parameter and canonical link. Then explain how a known exposure \(e>0\) changes the linear predictor when the modeled rate is \(\lambda(x)=e^{x^\mathsf{T}\beta}\). (Answer: \(\theta=\log\mu\), canonical link \(g(\mu)=\log\mu\); the count-mean predictor is \(\log e+x^\mathsf{T}\beta\), and its fitted mean is \(e\exp(x^\mathsf{T}\beta)\))

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