Common Distribution Assumption Mistakes

Why Distribution Assumptions Matter

A distribution is more than a name attached to a data set. Its parameterization determines how to calculate expected values and variances, and its support determines which outcomes are possible. In ratemaking, a mistaken parameter convention can produce an implausible severity estimate; a variance assumption that is too restrictive can understate uncertainty; and excluding valid zero losses can change the quantity being estimated.

This tutorial focuses on four errors involving Gamma, Poisson, negative binomial, and Tweedie distributions. In each case, first identify what the random variable represents, then state the distribution and parameter convention, and only then calculate. The examples are hypothetical. Interest and mortality are not modeled; amounts and counts are considered over the stated insurance periods.

Definition: A parameterization specifies how a distribution's parameters determine its probability law. Two parameterizations can describe the same distribution but use different parameter values. A conditional variance is a variance calculated under stated conditions, such as a fixed exposure or fixed covariates; it need not equal the variance in a population that mixes different conditions.

Gamma Scale Is Not Gamma Rate

A Gamma random variable is often parameterized by shape and scale, or by shape and rate. The shape parameter is the same in both conventions, but scale and rate are reciprocals. Confusing them can change a calculated mean by orders of magnitude.

Definition: If $X$ has a Gamma distribution with shape $\alpha>0$ and scale $\theta>0$, write $X\sim\operatorname{Gamma}(\alpha,\theta)$ using the scale convention. Then $E[X]=\alpha\theta$ and $\operatorname{Var}(X)=\alpha\theta^2$. Under the rate convention, with rate $\beta=1/\theta$, the same distribution has $E[X]=\alpha/\beta$ and $\operatorname{Var}(X)=\alpha/\beta^2$.

The density makes the distinction visible. In scale form it is proportional to $x^{\alpha-1}e^{-x/\theta}$; in rate form the exponential term is $e^{-\beta x}$. Thus the parameter multiplying $x$ in the exponent is the rate, while the parameter dividing $x$ is the scale. These are not interchangeable numerical values.

Worked Example: Converting a Gamma Claim Severity

A hypothetical claim severity $X$, measured in dollars, is Gamma with shape $\alpha=4$ and scale $\theta=\$250$. Find its mean, variance, standard deviation, and the equivalent rate parameter. Avoid treating 250 as the rate.

Method and assumptions. Use the shape-scale convention, with $\alpha=4$ and $\theta=250$ dollars. The severity is positive and follows the stated Gamma model. No discounting or mortality assumption is involved.

Mean. The shape-scale formula gives $E[X]=\alpha\theta=4(250)=\$1{,}000$.

Variance. $\operatorname{Var}(X)=\alpha\theta^2=4(250^2)=4(62{,}500)=250{,}000$ dollars squared. Therefore the standard deviation is $\sqrt{250{,}000}=\$500$.

Equivalent rate. The rate is the reciprocal of the scale: $\beta=1/250=0.004$ per dollar. Checking in rate form, $E[X]=4/0.004=\$1{,}000$, and $\operatorname{Var}(X)=4/(0.004)^2=250{,}000$ dollars squared.

Common mistake and check. Treating 250 as a rate would give the incorrect mean $4/250=\$0.016$. The correct conversion is $\beta=1/\theta$, not $\beta=\theta$. Both valid parameterizations agree on the mean and variance, confirming the result.

The derivation of the scale formulas also shows why the reciprocal matters. If $Z$ has a Gamma distribution with shape $\alpha$ and unit scale, then $X=\theta Z$ has scale $\theta$. Multiplying a random variable by $\theta$ multiplies its mean by $\theta$ and its variance by $\theta^2$. Since $E[Z]=\alpha$ and $\operatorname{Var}(Z)=\alpha$, it follows that $E[X]=\alpha\theta$ and $\operatorname{Var}(X)=\alpha\theta^2$. Replacing scale by rate therefore requires the reciprocal substitution $\theta=1/\beta$.

Poisson Variance Depends on Its Mean

A Poisson model is commonly used for a count over a specified exposure period. Under a Poisson distribution with parameter $\lambda$, both the expected count and its variance equal $\lambda$. The equality is a defining feature of the model, not a rule that can be detached from the distributional assumption.

Definition: If $N\sim\operatorname{Poisson}(\lambda)$, where $\lambda>0$, then $E[N]=\lambda$ and $\operatorname{Var}(N)=\lambda$. If independent counts over nonoverlapping exposure periods are Poisson with parameters $\lambda_1,\ldots,\lambda_k$, their sum is Poisson with parameter $\sum_{j=1}^k\lambda_j$.

A frequent error is to hold the variance fixed while changing the Poisson mean. Another is to assume that observed counts must have variance equal to their overall mean even when they mix risks with different underlying rates. The Poisson relationship applies to a Poisson count under its stated conditions; heterogeneity across risks can create extra variation in a combined population.

Worked Example: Scaling a Poisson Count With Exposure

For a hypothetical portfolio segment, the number of claims on one policy-year is Poisson with $\lambda=3$. Assume two policy-years have independent counts with this same rate. Find the distribution, expected total count, variance, and probability of no claims across the two policy-years.

Method and assumptions. Add the two independent Poisson counts. Each count has parameter 3 per policy-year; the total covers exactly two policy-years. There is no discounting or mortality assumption.

Total-count distribution. The sum is Poisson with parameter $3+3=6$. Thus $N_{\mathrm{total}}\sim\operatorname{Poisson}(6)$.

Mean and variance. $E[N_{\mathrm{total}}]=6$ claims and $\operatorname{Var}(N_{\mathrm{total}})=6$ claims squared. The standard deviation is $\sqrt{6}$, approximately 2.4495 claims.

Probability of zero. For a Poisson count, $P(N_{\mathrm{total}}=0)=e^{-6}=0.0024788$, or 0.0025 to four decimals.

Common mistake and check. Keeping the one-policy-year variance of 3 after doubling the exposure would be wrong. The total parameter doubles, so both the mean and variance are 6. The calculated zero probability is small, as expected for a count with mean 6.

Negative Binomial Counts Allow Overdispersion

A negative binomial count model is often used when the variance exceeds the mean. One useful parameterization defines $N$ as the number of failures before $r$ successes, with success probability $p$, where $r>0$ and $0<p<1$. In this convention the mean is $r(1-p)/p$, and the variance is $r(1-p)/p^2$. Equivalently, writing the mean as $\mu$ and the size parameter as $r$, the variance is $\mu+\mu^2/r$.

The extra term $\mu^2/r$ is the overdispersion relative to a Poisson count. As $r$ increases without bound while $\mu$ remains fixed, that extra term approaches zero, and the variance approaches the Poisson value $\mu$. For finite $r$, using Poisson variance simply because the outcome is a count discards the model's additional variability.

Justification: A negative binomial count can be represented as Poisson conditional on a random rate $\Lambda$, where $E[\Lambda]=\mu$ and $\operatorname{Var}(\Lambda)=\mu^2/r$. The law of total variance gives $\operatorname{Var}(N)=E[\operatorname{Var}(N\mid\Lambda)]+\operatorname{Var}(E[N\mid\Lambda])=E[\Lambda]+\operatorname{Var}(\Lambda)=\mu+\mu^2/r$. The random rate represents rate variation across risks or periods.

Worked Example: Detecting Negative Binomial Overdispersion

A hypothetical class of insurance policies has claim counts modeled by a negative binomial distribution with size $r=5$ and success probability $p=0.5$, using the failures-before-successes convention. Calculate the mean, variance, and probability of zero claims. Compare its variance with a Poisson model having the same mean.

Method and assumptions. Apply the stated negative binomial convention to counts over one policy period. The model parameters are $r=5$ and $p=0.5$; interest and mortality are not modeled.

Mean. $E[N]=r(1-p)/p=5(0.5)/0.5=5$ claims.

Variance. $\operatorname{Var}(N)=r(1-p)/p^2=5(0.5)/(0.5^2)=2.5/0.25=10$ claims squared. Equivalently, with $\mu=5$, the mean-size form gives $5+5^2/5=5+5=10$.

Zero probability. Under this convention, $P(N=0)=p^r=(0.5)^5=1/32=0.03125$.

Comparison and check. A Poisson count with the same mean 5 would have variance 5, not 10. The negative binomial variance is twice as large because it includes the additional term $25/5=5$. The common mistake is to set the variance equal to 5 merely because the variable counts claims; that would replace the specified negative binomial model with a Poisson assumption.

Tweedie Models Include Zero Outcomes

For a compound Poisson-Gamma Tweedie model, the aggregate loss is the sum of a random number of positive claim amounts. A zero claim count therefore produces an aggregate loss of exactly zero. For the compound representation with Tweedie power $1<p<2$, mean $\mu>0$, and dispersion $\phi>0$, the count is Poisson with parameter $\lambda$, and positive severities are Gamma distributed.

Definition: In the compound Poisson-Gamma representation with $1<p<2$, the Tweedie aggregate $T$ has mean $\mu$, variance $\phi\mu^p$, and zero probability $P(T=0)=e^{-\lambda}$, where $\lambda=\mu^{2-p}/\{\phi(2-p)\}$. The positive claim amount has Gamma shape $a=(2-p)/(p-1)$ and scale $\theta=\phi(p-1)\mu^{p-1}$. These formulas use the Gamma scale convention.

The zero probability follows directly from the count component: $T=0$ exactly when the Poisson claim count is zero, since each Gamma severity is strictly positive. Thus $P(T=0)=P(N=0)=e^{-\lambda}$. Removing zero observations is not a harmless data-cleaning step when the intended response is aggregate loss per exposure. It changes the data from the unconditional aggregate distribution to the distribution conditional on a positive aggregate.

Worked Example: Keeping Tweedie Zero Outcomes

A hypothetical portfolio's aggregate loss $T$, measured in thousands of dollars per exposure period, follows a compound Poisson-Gamma Tweedie model with $\mu=4$, $\phi=1$, and $p=1.5$. Calculate its Poisson frequency parameter, Gamma severity parameters, variance, and zero probability. Then find the expected aggregate conditional on a positive aggregate.

Method and assumptions. Use the compound representation for $1<p<2$. The aggregate is a sum of a Poisson number of independent positive Gamma amounts. All loss amounts are in thousands of dollars; interest and mortality are not modeled.

Frequency parameter. $\lambda=4^{2-1.5}/\{1(2-1.5)\}=4^{0.5}/0.5=2/0.5=4$ claims per exposure period.

Severity parameters. The Gamma shape is $a=(2-1.5)/(1.5-1)=0.5/0.5=1$. The Gamma scale is $\theta=1(1.5-1)4^{1.5-1}=0.5(4^{0.5})=0.5(2)=1$ thousand dollars. Its mean is $a\theta=1$ thousand dollars and its variance is $a\theta^2=1$ (thousands of dollars)$^2$.

Aggregate variance. The Tweedie variance is $\phi\mu^p=1(4^{1.5})=1(8)=8$ (thousands of dollars)$^2$. As a check, the compound sum has variance $\lambda E[Y^2]=4(1+1)=8$, because the Gamma severity's second moment is its variance plus its mean squared, $1+1=2$.

Zero probability. $P(T=0)=e^{-4}=0.0183156$, or 0.0183 to four decimals.

Conditional positive aggregate. Since $E[T]=P(T>0)E[T\mid T>0]$ and $T=0$ when it is not positive, $E[T\mid T>0]=4/(1-e^{-4})=4/0.9816844=4.0746$ thousand dollars, rounded to four decimals.

Common mistake and check. Deleting the zero outcomes and reporting 4 as the mean of positive aggregates confuses the unconditional mean with the conditional mean. The conditional positive mean, 4.0746, is larger than the unconditional mean, 4, as it must be when zero outcomes are excluded.

Practical Checks Before Using a Distribution

A calculation can be internally consistent and still answer the wrong question if the distribution or conditioning has been misread. Before using a formula, make a short audit of the assumptions:

For Gamma severity, identify whether the second parameter is scale or rate, and convert by taking a reciprocal before calculating moments.
For Poisson frequency, match the parameter to the exposure period. Under the Poisson model, variance equals the parameter and therefore equals the mean.
For negative binomial frequency, state the parameter convention. With finite size $r$, variance exceeds the mean by $\mu^2/r$.
For compound Poisson-Gamma Tweedie loss, retain the zero outcome whenever the target is the unconditional aggregate per exposure. A positive-loss-only analysis is conditional on $T>0$.

These distinctions matter because expected loss and variability support different decisions. A severity parameterization error distorts the scale of expected claims; a variance error changes measures of uncertainty; and excluding zeros changes the population represented by the estimate. Distribution names alone do not protect against these errors: the parameter convention, exposure basis, and conditioning event must all be explicit.

Check Your Understanding

For each problem, identify the stated convention before calculating. Amounts and counts are hypothetical; no interest or mortality assumptions apply.

A claim severity is Gamma with shape 3 and scale 200 dollars. Find its mean, variance, and equivalent rate. (Answer: Mean = $600; variance = 120,000 dollars squared; rate = 0.005 per dollar.)
Two independent policy-period counts are each Poisson with parameter 2. Find the total count's mean and variance. (Answer: Mean = 4 claims; variance = 4 claims squared.)
A negative binomial count has size $r=4$ and success probability $p=0.5$, under the failures-before-successes convention. Find its mean and variance. (Answer: Mean = 4 claims; variance = 8 claims squared.)
A compound Poisson-Gamma Tweedie aggregate has $\mu=9$, $\phi=1$, and $p=1.5$. Find $\lambda$ and $P(T=0)$. (Answer: $\lambda=6$; $P(T=0)=e^{-6}=0.0025$ to four decimals.)
For the model in Question 4, calculate the aggregate variance and the expected aggregate conditional on being positive. (Answer: Variance = 27; conditional positive mean $=9/(1-e^{-6})=9.0224$, in the same loss units as $\mu$.)

What You'll Learn