Identity And Reciprocal Links For Loss Means

Two Ways to Connect a Predictor to a Loss Mean

A generalized linear model separates the systematic part of a prediction from the rule that converts it to a mean. For a risk with covariate vector $x$, the linear predictor is $\eta=x^\mathsf{T}\beta$, where $\beta$ is the coefficient vector. The link function $g$ relates the conditional mean $\mu=E[X\mid x]$ to that predictor by $g(\mu)=\eta$. To obtain a fitted mean, apply the inverse link: $\mu=g^{-1}(\eta)$.

This tutorial compares the identity and reciprocal links when the target is a positive expected loss. Loss amounts below are sometimes stated in thousands of dollars; the units of the predictor and coefficients must then match the chosen link. A link specifies the mean relationship, not by itself the response distribution or its variance.

Definition: Under the identity link, $g(\mu)=\mu$, so $\mu=\eta$. Under the reciprocal link, $g(\mu)=1/\mu$, so $\mu=1/\eta$ wherever $\eta\ne 0$. For a model intended to predict positive loss means, the identity-link prediction must satisfy $\eta>0$, and the reciprocal-link predictor must also satisfy $\eta>0$. A zero or negative fitted mean is not a valid positive expected loss.

The reciprocal link is also called the inverse link in some contexts. Here “reciprocal link” means specifically $g(\mu)=1/\mu$; it does not mean taking the reciprocal of each coefficient or covariate separately. Its inverse transformation is applied to the entire linear predictor.

Identity Link: The Predictor Is the Mean

With the identity link, a model such as $\eta=\beta_0+\beta_1x_1+\beta_2x_2$ gives the fitted mean directly: $\mu=\beta_0+\beta_1x_1+\beta_2x_2$. Holding other covariates fixed, increasing $x_j$ by one unit changes the fitted mean by $\beta_j$ units. This direct interpretation can be convenient when the desired effects are additive.

The limitation is equally direct: a linear predictor is not automatically positive. A model may be algebraically well-defined for any real $\eta$, but if the response represents a positive expected loss, predictions with $\eta\le 0$ fail the target’s basic constraint. Positivity therefore has to be checked on the relevant range of covariates, not just at the reference risk.

Worked Example: Checking an Identity-Link Loss Model

A hypothetical insurer models expected claim cost in thousands of dollars using $\eta=18+6x-20z$, where $x$ is a vehicle-use score between 0 and 4 and $z=1$ indicates a specified vehicle feature (otherwise $z=0$). The model uses the identity link. Find the fitted means for two risks and determine where the model ceases to give a positive mean when $z=1$.

Method and units. For an identity link, the fitted mean is $\mu=\eta$. Since the response is in thousands of dollars, each value of $\mu$ is a number of thousands of dollars.

Risk A: $x=2,\ z=1$. Substitute into the predictor:

\eta=18+6(2)-20(1)=18+12-20=10.

Thus $\mu=10$ thousand dollars, or $10,000. For Risk B, with $x=2,\ z=0$, $\eta=18+12=30$, so $\mu=30$ thousand dollars, or $30,000. The fitted difference is $10-30=-20$ thousand dollars, consistent with the coefficient on $z$.

For $z=1$, the predictor as a function of $x$ is $\eta=18+6x-20=6x-2$. It is zero when $6x-2=0$, hence $x=1/3$. It is negative for $0\le x<1/3$, zero at $x=1/3$, and positive for $x>1/3$. For example, at $x=0$, $\mu=-2$ thousand dollars, which is not a valid positive expected loss.

Check. At $x=2,z=1$, $6(2)-2=10$, agreeing with the first fitted value. The common mistake is to assume that an identity link guarantees a valid mean; it only makes the mean equal to the predictor. The sign still needs to be checked.

Reciprocal Link: Invert the Whole Predictor

For a reciprocal link, $g(\mu)=1/\mu=\eta$. Solving this equation gives $\mu=1/\eta$, provided $\eta\ne0$. For a positive loss mean, the stricter practical condition is $\eta>0$: a positive predictor gives a positive fitted mean, a zero predictor makes the fitted mean undefined, and a negative predictor gives a negative fitted mean.

The reciprocal transformation also changes how coefficients act. A coefficient changes $\eta$ additively, but it does not change $\mu$ by a constant amount. If a covariate changes the predictor from $\eta$ to $\eta+b$, with both predictors positive, the change in mean is

\frac{1}{\eta+b}-\frac{1}{\eta} = \frac{\eta-(\eta+b)}{\eta(\eta+b)} = -\frac{b}{\eta(\eta+b)}.

The size of this change depends on the starting predictor as well as on $b$. In particular, as a positive $\eta$ approaches zero, $1/\eta$ grows without bound. Near-zero positive predictors can therefore produce extremely large fitted means and may signal an unstable or implausible prediction.

Worked Example: Fitting Means and Finding the Reciprocal-Link Boundary

For a hypothetical portfolio, expected claim cost is measured in thousands of dollars. A reciprocal-link model has $\eta=0.020+0.010x-0.025z$, where $x$ is a rating score and $z$ is an indicator equal to 1 for a selected coverage option. Calculate fitted means at $x=2,z=1$ and $x=0,z=0$, then find the boundary for positive predictions when $z=1$.

Method and units. The reciprocal-link equation is $\mu=1/\eta$. Since $\mu$ is in thousands of dollars, the predictor has units reciprocal to thousands of dollars. The inverse is applied after evaluating the complete predictor.

At $x=2,z=1$,

\eta=0.020+0.010(2)-0.025(1) =0.020+0.020-0.025 =0.015.

Therefore $\mu=1/0.015=66.6667$ thousand dollars, or $66,666.67 when rounded to cents. At $x=0,z=0$, $\eta=0.020$, so $\mu=1/0.020=50$ thousand dollars, or $50,000.

When $z=1$, $\eta=0.020+0.010x-0.025=0.010x-0.005$. The boundary is found by setting $\eta=0$: $0.010x-0.005=0$, so $x=0.5$. At $x=0.5$, the fitted mean is undefined. For $x<0.5$, $\eta<0$ and the resulting mean is negative; only $x>0.5$ gives a positive fitted mean under this specification.

Check and common mistake. The first calculation uses the full predictor $0.015$, whose reciprocal is $66.6667$. A wrong calculation would take reciprocals term by term and combine them, for example $1/0.020+1/0.020-1/0.025=50+50-40=60$. In general, $1/(a+b)\ne 1/a+1/b$; that calculation does not implement the reciprocal link.

Deriving Positivity Conditions Over a Risk Range

A coefficient set is acceptable for positive loss-mean predictions only if the model gives $\mu>0$ for every covariate combination in the intended prediction range. The condition can be stated precisely for both links.

Write the predictor.
For the relevant covariates, express $\eta=x^\mathsf{T}\beta$. Keep the stated ranges and any indicator restrictions explicit.

Impose the positive-mean condition.
For the identity link, $\mu=\eta$, so require $\eta>0$. For the reciprocal link, $\mu=1/\eta$, so again require $\eta>0$; $\eta=0$ is undefined and $\eta<0$ gives a negative mean.

Check the extremes of the specified range.
For a linear predictor in one covariate over a closed interval, its minimum occurs at an endpoint. For several covariates over a rectangular range, check the relevant corner combinations. If a predictor is not linear over the range or the range has other constraints, find its minimum under those constraints instead.

For example, let $\eta=\beta_0+\beta_1x$, with $0\le x\le M$ and $M>0$. If $\beta_1\ge0$, the minimum is $\beta_0$, so strict positivity throughout the range requires $\beta_0>0$. If $\beta_1<0$, the minimum is $\beta_0+\beta_1M$, so positivity requires $\beta_0+\beta_1M>0$. Equality gives a zero mean at an endpoint; a value below zero gives a negative mean there. These conditions apply to the linear predictor, and therefore enforce positive means under either link considered here.

The shared positivity condition does not make the two links interchangeable. With identity, a positive predictor is itself the mean. With reciprocal, a positive predictor is inverted, so larger predictors correspond to smaller means. The scale and size of coefficient effects consequently differ, even when the same predictor expression is used.

Worked Example: Finding Coefficient Ranges for Both Links

A hypothetical model uses a single covariate $x$ between 0 and 3. Its predictor is $\eta=18+\gamma x$. The response is expected loss in thousands of dollars. For the identity and reciprocal links separately, find the condition on $\gamma$ that ensures positive fitted means throughout the full interval. Then assess $\gamma=-7$ at $x=3$ under each link.

Method. For either link, a positive loss mean requires $\eta>0$. Because the predictor is linear in $x$, its minimum over $0\le x\le3$ is at an endpoint.

If $\gamma\ge0$, the minimum is $\eta(0)=18$, which is positive. If $\gamma<0$, the minimum is $\eta(3)=18+3\gamma$. Thus the condition for all $x$ in the interval is $18+3\gamma>0$, or $\gamma>-6$. Combining the cases gives the same condition $\gamma>-6$: every nonnegative $\gamma$ already satisfies it, and the admissible negative values are greater than $-6$. At $\gamma=-6$, the predictor is zero at $x=3$; for $\gamma<-6$, it is negative there.

For $\gamma=-7$ at $x=3$, $\eta=18+(-7)(3)=18-21=-3$. Under the identity link, $\mu=-3$ thousand dollars, an invalid negative loss mean. Under the reciprocal link, $\mu=1/(-3)=-1/3$ thousand dollars, also invalid. The reciprocal link does not repair a negative predictor.

Check. At the boundary $\gamma=-6$, $\eta(3)=18-18=0$. The identity-link mean is zero, while the reciprocal-link mean is undefined. The strict inequality $\gamma>-6$, rather than $\gamma\ge-6$, is necessary for a strictly positive mean throughout the interval under either link.

What the Link Does—and Does Not—Guarantee

For an identity link, a model can be easy to interpret on the loss scale, but its predictions can cross zero as covariates change. For a reciprocal link, a positive predictor produces a positive mean, but the predictor must still be checked: a zero predictor is a singularity, and a negative predictor produces a negative result. Neither link automatically ensures valid predictions over every possible covariate combination.

A linear predictor may be positive on the observed data and nonpositive for a future risk outside the observed covariate range. Thus, validating a model’s prediction domain includes checking the range in which predictions will actually be used. The fitted mean alone also does not specify a probability distribution for individual losses; that requires additional model assumptions.

Key takeaway: First calculate the complete linear predictor $\eta$, then apply the specified inverse link. For a positive loss mean, require $\eta>0$ under both the identity and reciprocal links. At $\eta=0$, the identity prediction is zero and the reciprocal prediction is undefined; at $\eta<0$, both predictions are negative.

Check Your Understanding

For each problem, calculate the predictor before applying the link, and check the positive-mean condition.

A hypothetical identity-link model for loss in thousands of dollars is $\eta=22+5x-12z$. Find the fitted mean at $x=1,z=1$, and state whether it is positive. (Answer: $\eta=\mu=15$ thousand dollars, so the prediction is positive.)
A reciprocal-link model has $\eta=0.040+0.008x$, with loss measured in thousands of dollars. Find the fitted mean at $x=2$. (Answer: $\eta=0.056$, so $\mu=1/0.056=17.8571$ thousand dollars.)
An identity-link predictor is $\eta=12-3x$, where $0\le x\le5$. Find the value of $x$ where the mean reaches zero and state the positive-prediction range within the stated interval. (Answer: zero at $x=4$; positive for $0\le x<4$, zero at 4, and negative for $4<x\le5$.)
A reciprocal-link model has $\eta=0.030+0.006x-0.024z$, with $z=1$ and $x\ge0$. Find the boundary between negative and positive fitted means, and classify the prediction at the boundary. (Answer: $\eta=0.006x-0.006$, so the boundary is $x=1$; the reciprocal fitted mean is undefined there, negative for $x<1$, and positive for $x>1$.)
For $\eta=16+\gamma x$ on $0\le x\le4$, find the condition on $\gamma$ that guarantees strictly positive fitted means over the entire interval under either link. (Answer: $\gamma>-4$.)

What You'll Learn