Exposure Scaling With Negative Binomial Frequency

Why Exposure Scaling Needs an Assumption

If a hypothetical portfolio is observed for twice as long, its expected claim count will generally be twice as large, all else equal. But a count model has more than a mean: its variance also matters when assessing the range of plausible outcomes. For a negative binomial model, the variance adjustment depends on the assumption made about dispersion. This tutorial uses a fixed shape parameter as exposure changes.

The key calculation is not to multiply the entire baseline variance by the exposure ratio. The mean scales in direct proportion to exposure, while the squared-mean part of the negative binomial variance scales with the square of that ratio. The distinction is important in frequency calculations and probability questions.

Definition: In this tutorial, a negative binomial claim count $N$ has mean $m>0$ and shape parameter $r>0$. Its probability function, mean, and variance are $$ P(N=k)=\frac{\Gamma(k+r)}{\Gamma(r)k!} \left(\frac{r}{r+m}\right)^r \left(\frac{m}{r+m}\right)^k,\quad k=0,1,2,\ldots, $$ $$ E[N]=m,\qquad \operatorname{Var}(N)=m+\frac{m^2}{r}. $$ Here $r$ is held fixed in the exposure adjustment considered below. This is the mean-and-shape parameterization; other negative binomial parameterizations may use a different parameter to play the role of $r$.

The Fixed-Shape Exposure Adjustment

Let the baseline exposure be $e_0>0$, the new exposure be $e_1>0$, and define the exposure ratio

c=\frac{e_1}{e_0}.

Suppose the baseline count has a negative binomial distribution with mean $m_0$ and shape $r$. Under a proportional exposure change with the same claim conditions and risk mix, and with the shape $r$ held fixed, the adjusted count has mean $cm_0$ and shape $r$. Thus,

E[N_1]=cm_0,\qquad \operatorname{Var}(N_1)=cm_0+\frac{(cm_0)^2}{r}.

This is a model assumption, not a universal rule for every way of changing exposure. It is appropriate when the problem specifies fixed dispersion, or otherwise gives a reason to keep the same negative binomial shape. Use the exposure ratio consistently: $c$ is new exposure divided by baseline exposure, not the reverse.

Why the Mean and Variance Scale Differently

A gamma-Poisson mixture gives a direct justification. For the baseline exposure, let $\Lambda$ denote the random Poisson intensity at that baseline exposure. Assume

\Lambda\sim\operatorname{Gamma}\left(r,\text{ rate } \frac{r}{m_0}\right), \qquad E[\Lambda]=m_0,\qquad \operatorname{Var}(\Lambda)=\frac{m_0^2}{r},

and, conditional on $\Lambda$, let the baseline count be Poisson with mean $\Lambda$. The word “baseline” matters: $\Lambda$ here is not a rate per unit of exposure. At the new exposure, proportional scaling gives conditional Poisson mean $c\Lambda$. The new random intensity $c\Lambda$ has mean $cm_0$ and variance $c^2m_0^2/r$.

The law of total expectation and the law of total variance give the adjusted moments. Conditional on the intensity, the mean and variance of a Poisson count are both that intensity. Therefore,

E[N_1]=E[E[N_1\mid\Lambda]]=E[c\Lambda]=cm_0, $$ $$ \begin{aligned} \operatorname{Var}(N_1) &=E[\operatorname{Var}(N_1\mid\Lambda)] +\operatorname{Var}(E[N_1\mid\Lambda])\\ &=E[c\Lambda]+\operatorname{Var}(c\Lambda)\\ &=cm_0+\frac{c^2m_0^2}{r}. \end{aligned}

The first variance term is the Poisson variation conditional on the intensity; it scales by $c$. The second comes from variation in the random intensity; because the intensity is multiplied by $c$, its variance is multiplied by $c^2$. Mixing a Poisson distribution with this gamma intensity produces the negative binomial model with mean $cm_0$ and the same shape $r$.

Find the exposure ratio.
Calculate $c=e_1/e_0$ and check which exposure is in the numerator.

Keep the shape fixed.
Under this model, set the new negative binomial shape equal to the baseline $r$.

Adjust the mean and variance.
Use $m_1=cm_0$ and $\operatorname{Var}(N_1)=m_1+m_1^2/r$.

Calculate any requested probability.
Substitute the adjusted mean and unchanged shape into the negative binomial probability function.

Worked Example: Scaling to Two and a Half Times the Exposure

A hypothetical insurer models claims in a baseline block of exposure with a negative binomial distribution having mean $m_0=6$ claims and shape $r=3$. A proposed block has $2.5$ times the baseline exposure. Under a fixed-shape assumption, find its mean, variance, and probability of no claims.

Model and assumptions. The baseline count is negative binomial with mean 6 and shape 3. The new count is also negative binomial with shape 3, and exposure is proportional. No interest rate or mortality assumption is relevant to this claim-count calculation.

Exposure ratio. Here $c=2.5$. Therefore the new mean is $$ m_1=cm_0=2.5(6)=15. $$

Variance. Apply the negative binomial variance formula using the new mean and the unchanged shape: $$ \operatorname{Var}(N_1)=15+\frac{15^2}{3} =15+\frac{225}{3} =15+75=90. $$

Probability of no claims. Set $k=0$ in the probability function: $$ P(N_1=0)=\left(\frac{r}{r+m_1}\right)^r =\left(\frac{3}{3+15}\right)^3 =\left(\frac{1}{6}\right)^3 =\frac{1}{216} \approx 0.0046. $$ The probability is rounded to four decimal places. As a check, the adjusted mean is $15$, and $15+15^2/3=90$; the no-claim probability is between 0 and 1.

Method and trap avoided. The method is fixed-shape exposure scaling followed by the negative binomial probability formula. The variance is recalculated from the adjusted mean; it is not obtained by multiplying the baseline variance by $2.5$.

Worked Example: A Shorter Observation Period

A hypothetical fleet has $1{,}000$ vehicle-years of baseline exposure. Its claim count is modeled as negative binomial with mean 8 and shape $r=4$. Find the mean and variance for $500$ vehicle-years, then calculate the probability of at most one claim.

Model and assumptions. The baseline count has mean 8 and shape 4. The new count has the same shape, with unchanged claim conditions and risk mix. The exposure measure is vehicle-years, so no interest or mortality assumption is involved.

Exposure ratio and mean. $$ c=\frac{500}{1{,}000}=0.5,\qquad m_1=0.5(8)=4. $$

Variance. $$ \operatorname{Var}(N_1)=4+\frac{4^2}{4}=4+4=8. $$

Probability of at most one claim. For mean $4$ and shape $4$, the two terms are $$ P(N_1=0)=\left(\frac{4}{4+4}\right)^4 =\left(\frac{1}{2}\right)^4=\frac{1}{16}, $$ $$ P(N_1=1)= \frac{\Gamma(5)}{\Gamma(4)1!} \left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right) =4\left(\frac{1}{2}\right)^5 =\frac{1}{8}. $$ Thus, $$ P(N_1\leq 1)=\frac{1}{16}+\frac{1}{8} =\frac{3}{16}=0.1875. $$

Check. The adjusted mean is half the baseline mean, as expected for half the exposure. The variance is $8$. For comparison, the baseline variance was $8+8^2/4=24$; multiplying that whole variance by $0.5$ would give $12$, not the correct answer.

Method and trap avoided. The calculation uses the new mean $4$ in both the variance and probability formulas. It avoids scaling the baseline variance linearly, which would incorrectly scale its squared-mean component.

Worked Example: Inferring Shape Before Scaling

For a hypothetical baseline exposure, a negative binomial claim count has mean 2 and variance 4. The exposure is then increased to four times its baseline level. Find the adjusted mean and variance, and the probability of at least two claims.

Model and assumptions. The baseline and adjusted counts are negative binomial. Shape is fixed across the exposure change. There is no interest or mortality component in this frequency model.

Infer the baseline shape. From $\operatorname{Var}(N)=m+m^2/r$, $$ 4=2+\frac{2^2}{r},\qquad 2=\frac{4}{r},\qquad r=2. $$

Adjust the mean and variance. With $c=4$, $$ m_1=4(2)=8,\qquad \operatorname{Var}(N_1)=8+\frac{8^2}{2}=8+32=40. $$

Calculate the probability. For mean 8 and shape 2, $r/(r+m_1)=2/10=1/5$, and $m_1/(r+m_1)=8/10=4/5$. Hence, $$ P(N_1=0)=\left(\frac{1}{5}\right)^2=\frac{5}{125}, $$ $$ P(N_1=1)= \frac{\Gamma(3)}{\Gamma(2)1!} \left(\frac{1}{5}\right)^2\left(\frac{4}{5}\right) =2\left(\frac{1}{25}\right)\left(\frac{4}{5}\right) =\frac{8}{125}. $$ Therefore, $$ P(N_1\geq 2)=1-P(N_1=0)-P(N_1=1) =1-\frac{5}{125}-\frac{8}{125} =\frac{112}{125}=0.8960. $$

Check. The inferred shape is positive, and $8+8^2/2=40$. The probabilities for zero and one claim sum to $13/125$, so the complementary probability is $112/125$.

Method and trap avoided. First recover the shape from the two baseline moments; then hold that shape fixed while scaling. A common error is to treat the baseline variance as if it were enough to scale directly, without identifying how the negative binomial shape enters the model.

Common Mistakes and When to Use the Formula

For fixed shape, the adjusted variance can also be written directly in terms of the baseline mean and exposure ratio:

\operatorname{Var}(N_1)=cm_0+\frac{c^2m_0^2}{r}.

This expression highlights the two different powers of $c$. It also helps check a calculation: if $c=1$, the baseline moments are recovered; if $c>1$, the mean increases, and the variance is calculated using both the larger mean and the same $r$.

A specific alternative assumption about how the underlying risks change could lead to a different dispersion adjustment. Therefore, do not silently change $r$, or assume that it scales with exposure, when a problem says to keep dispersion fixed. Use the model stated in the question. For the fixed-shape model, the operational rule is simple: scale the mean by $c$, retain $r$, and recompute the variance and probabilities.

Key takeaway: For negative binomial frequency with fixed shape $r$, exposure ratio $c=e_1/e_0$ changes the mean from $m_0$ to $cm_0$, and the variance to $cm_0+(cm_0)^2/r$. The mean scales linearly; the squared-mean variance component scales quadratically.

Check Your Understanding

For each problem, use the fixed-shape assumption unless the question asks you to infer the shape. Give probabilities to four decimal places.

A hypothetical baseline count has negative binomial mean 5 and shape 5. Exposure doubles. Find the adjusted mean and variance. (Answer: mean $10$; variance $30$)
A hypothetical portfolio has baseline exposure of $1{,}200$ policy-years, mean 9 claims, and shape 3. At $400$ policy-years, find the mean, variance, and probability of zero claims. (Answer: mean $3$; variance $6$; $P(N=0)=0.1250$)
A baseline negative binomial count has mean 4 and variance 12. Exposure is reduced to half its baseline level. Find the shape, adjusted mean, and adjusted variance. (Answer: shape $2$; mean $2$; variance $4$)
A hypothetical count has baseline mean 3 and shape 2. Exposure triples. Find the adjusted mean, variance, and probability of zero claims. (Answer: mean $9$; variance $49.5$; $P(N=0)=(2/11)^2=0.0331$)
A hypothetical count has mean 10 and variance 35 at baseline exposure of $2$ units. Exposure increases to $5$ units. Infer the shape and find the adjusted mean and variance. (Answer: shape $4$; mean $25$; variance $181.25$)

What You'll Learn