Worked Tweedie Pure Premium Calculation

Why Exposure Matters in a Tweedie Calculation

A Tweedie pure premium model can describe an aggregate loss that is zero for many policy periods and positive otherwise. For a policy with less than a full unit of exposure, both the expected loss and the chance of at least one loss change. The variance and probability of zero loss change as well, but not in the same way as the mean.

This tutorial uses a constant pure premium rate across exposure: the policy’s risk characteristics do not change during the period, and exposure scales the amount of risk. The calculations concern aggregate loss for a policy period, not the average loss conditional on a claim or on a positive policy-period loss. Monetary results below are rounded to cents only at the final step, and probabilities are rounded to four decimals.

Definition: Let $Y_e$ be aggregate loss for exposure $e$, where $e>0$ is measured in the model’s exposure units. Let $\mu$ be expected loss per unit of exposure, $\phi>0$ the dispersion parameter per unit of exposure, and $p$, with $1<p<2$, the Tweedie power parameter. Under the constant-rate Tweedie model, $Y_e$ has mean $e\mu$, variance $\phi e\mu^p$, and a probability mass at zero. The positive-loss distribution is continuous.

For $1<p<2$, this Tweedie model has a compound Poisson–gamma representation: the number of claims is Poisson, and claim amounts, conditional on a claim, are gamma distributed. The parameters $\mu,\phi,p$ determine both the frequency and severity components. The calculations here do not require fitting those parameters; they take them as supplied. Exposure is assumed to scale the Poisson claim rate while leaving the conditional claim severity distribution unchanged.

Exposure-Scaled Results

Under the assumptions just stated, expected aggregate loss is proportional to exposure. Variance is also proportional to exposure when $\mu$, $\phi$, and $p$ are held fixed. In particular, it is not obtained by simply raising the exposure-scaled mean to the power $p$ while keeping the unit-exposure dispersion unchanged.

E[Y_e]=e\mu,\qquad \operatorname{Var}(Y_e)=\phi e\mu^p

The Poisson claim rate per unit of exposure is $\lambda=\mu^{2-p}/\{\phi(2-p)\}$. Thus the claim count over exposure $e$ has mean $e\lambda$. Aggregate loss is zero exactly when there are no claims, so the probability of zero loss follows from the Poisson probability of zero events.

P(Y_e=0)=\exp\left(-\frac{e\mu^{2-p}}{\phi(2-p)}\right)

Assumption: These exposure formulas apply when the same risk rate and Tweedie parameters apply throughout the exposure being aggregated, with independent increments over disjoint exposure periods. They do not automatically apply if coverage, risk characteristics, or the exposure definition changes in a way that changes the underlying mean rate or severity distribution.

Why the Zero-Loss Formula Holds

The zero-loss result follows directly from the compound Poisson representation. Let $N_e$ be the number of claims during exposure $e$, with $N_e\sim\operatorname{Poisson}(e\lambda)$. A gamma claim amount is positive with probability one, so aggregate loss is zero if and only if $N_e=0$. For a Poisson random variable with mean $e\lambda$, the probability of zero is $\exp(-e\lambda)$. Substituting the Tweedie claim rate gives the stated expression.

\begin{aligned} P(Y_e=0) &=P(N_e=0)\\ &=\exp(-e\lambda)\\ &=\exp\left(-\frac{e\mu^{2-p}}{\phi(2-p)}\right). \end{aligned}

This argument also explains why zero probability is not determined by expected loss alone. The zero probability depends on the implied claim rate, which depends on $\mu,\phi,$ and $p$. Different combinations of parameters can produce the same rate of zero-loss periods while producing different expected losses and variances.

Worked Calculations

Worked Example: Three Exposure Periods With One Tweedie Rate

A hypothetical book is modeled with unit-exposure mean $\mu=\$400$, dispersion $\phi=20$, and power $p=1.5$. Calculate the expected aggregate loss, variance, and zero-loss probability for exposures $e=0.25$, $e=1$, and $e=2.5$.

Method and setup. Use the exposure-scaled Tweedie mean, variance, and zero-probability formulas. First calculate the common quantities:

\mu^p=400^{1.5}=400\sqrt{400}=400(20)=8{,}000

The unit-exposure claim rate is

\lambda=\frac{400^{2-1.5}}{20(2-1.5)} =\frac{\sqrt{400}}{20(0.5)} =\frac{20}{10}=2.

Therefore, per unit of exposure the mean is $\$400$, the variance is $20(8{,}000)=\$160{,}000$ in squared-dollar units, and the zero-loss probability over exposure $e$ is $\exp(-2e)$.

At $e=0.25$. The expected loss is $0.25(\$400)=\$100.00$. The variance is $20(0.25)(8{,}000)=\$40{,}000.00$. The zero-loss probability is $\exp[-2(0.25)]=\exp(-0.5)=0.6065$.

At $e=1$. The expected loss is $1(\$400)=\$400.00$. The variance is $20(1)(8{,}000)=\$160{,}000.00$. The zero-loss probability is $\exp(-2)=0.1353$.

At $e=2.5$. The expected loss is $2.5(\$400)=\$1{,}000.00$. The variance is $20(2.5)(8{,}000)=\$400{,}000.00$. The zero-loss probability is $\exp(-5)=0.0067$.

Check. The mean and variance both increase in direct proportion to exposure, while the probability of zero loss decreases as the Poisson claim rate grows. The common mistake this method avoids is treating each exposure as if it had a fresh unit-exposure claim rate.

Worked Example: Equal Zero-Loss Probabilities, Different Pure Premiums

Two hypothetical classes have the same exposure $e=1.25$. Class A has $\mu=\$400$, $\phi=20$, and $p=1.5$. Class B has $\mu=\$900$, $\phi=30$, and $p=1.5$. For each class, calculate expected aggregate loss, variance, and zero-loss probability. Explain whether equal zero-loss probabilities imply equal expected losses.

Method. Calculate each class’s unit-exposure claim rate, then multiply that rate by $1.25$ to obtain the policy-period Poisson mean. Separately calculate the Tweedie mean and variance using the class’s own parameters.

Class A. Its unit-exposure claim rate is

\lambda_A=\frac{\sqrt{400}}{20(0.5)}=2.

Its policy-period expected loss is $1.25(400)=\$500.00$. Since $400^{1.5}=8{,}000$, its variance is $20(1.25)(8{,}000)=\$200{,}000.00$. Its zero-loss probability is $\exp[-1.25(2)]=\exp(-2.5)=0.0821$.

Class B. Its unit-exposure claim rate is

\lambda_B=\frac{\sqrt{900}}{30(0.5)}=\frac{30}{15}=2.

Its policy-period expected loss is $1.25(900)=\$1{,}125.00$. Since $900^{1.5}=900(30)=27{,}000$, its variance is $30(1.25)(27{,}000)=\$1{,}012{,}500.00$. Its zero-loss probability is also $\exp[-1.25(2)]=0.0821$.

Conclusion and check. The zero-loss probabilities agree because the two classes have the same implied claim rate per unit of exposure. Their expected losses differ by $\$625.00$, and their variances differ as well. Equal probabilities of no loss do not imply equal pure premiums. This calculation avoids the mistake of inferring the mean from the zero-loss probability alone.

Worked Example: Finding Exposure From a Target Zero-Loss Probability

A hypothetical risk has $\mu=\$625$, $\phi=25$, and $p=1.5$. Find the exposure for which the zero-loss probability is $e^{-1}$, then calculate the expected aggregate loss and variance at that exposure.

Method. First calculate the unit-exposure claim rate, then solve the zero-probability equation for exposure. Here $625^{0.5}=25$, so

\lambda=\frac{625^{0.5}}{25(0.5)} =\frac{25}{12.5}=2.

The target probability requires $\exp(-2e)=e^{-1}$. Equating the exponents gives $2e=1$, so the required exposure is $e=0.5$. The zero-loss probability is therefore $e^{-1}=0.3679$ to four decimals.

The expected aggregate loss is $0.5(\$625)=\$312.50$. For the variance, first calculate $625^{1.5}=625(25)=15{,}625$. Then

\operatorname{Var}(Y_{0.5}) =25(0.5)(15{,}625) =\$195{,}312.50.

Check. Substitution into the zero-probability formula gives $\exp[-0.5(2)]=e^{-1}$, as required. Solving for exposure using the unit claim rate avoids applying the target probability to the expected loss instead of to the Poisson rate.

A Common Exposure-Scaling Trap

A frequent error is to replace the unit-exposure mean $\mu$ with $e\mu$ in the variance-power formula, while leaving the unit-exposure dispersion $\phi$ unchanged. That calculation gives $\phi(e\mu)^p$, but the variance for the aggregate over $e$ exposure units is $\phi e\mu^p$. The incorrect expression scales as $e^p$, not $e$.

For example, using the first worked example’s parameters at $e=2$, the correct variance is $20(2)(400^{1.5})=20(2)(8{,}000)=\$320{,}000.00$. The incorrect calculation would be $20(800^{1.5})$, or approximately $\$452{,}548.34$. The discrepancy arises because the aggregate mean $e\mu$ cannot be inserted into the unit-exposure formula with the same dispersion parameter and interpreted as an exposure-scaled aggregate model.

The same consistency issue affects zero probabilities. At $e=2$ in that example, the correct probability is $\exp[-2(2)]=0.0183$. Plugging $800$ into the unit-exposure zero-probability formula instead would give $\exp[-\sqrt{800}/10]=0.0591$, which fails to scale the Poisson claim rate by exposure.

Key takeaway: With unit-exposure parameters $\mu,\phi,p$, calculate aggregate policy-period quantities as $e\mu$, $\phi e\mu^p$, and $\exp\{-e\mu^{2-p}/[\phi(2-p)]\}$. Keep the exposure scaling explicit in all three calculations.

Check Your Understanding

For each problem, use the stated Tweedie parameters and exposure. Round probabilities to four decimals and monetary amounts to cents.

A risk has $\mu=\$100$, $\phi=5$, and $p=1.5$. At exposure $e=0.4$, calculate expected aggregate loss, variance, and probability of zero loss. (Answer: mean $\$40.00$, variance $\$2{,}000.00$, zero probability $0.2019$.)
A policy has $\mu=\$225$, $\phi=15$, and $p=1.5$ over exposure $e=2$. Calculate the expected loss, variance, and probability of zero loss. (Answer: mean $\$450.00$, variance $\$101{,}250.00$, zero probability $0.0183$.)
A class has $\mu=\$400$, $\phi=20$, and $p=1.5$. For exposure $e=0.75$, calculate expected loss, variance, and zero-loss probability. (Answer: mean $\$300.00$, variance $\$120{,}000.00$, zero probability $0.2231$.)
For $\mu=\$256$, $\phi=16$, and $p=1.5$, find the exposure that gives zero-loss probability $e^{-2}$. Then calculate the expected loss and variance at that exposure. (Answer: exposure $1$, mean $\$256.00$, variance $\$65{,}536.00$.)
A risk has $\mu=\$64$, $\phi=4$, $p=1.5$, and exposure $e=0.5$. Calculate its policy-period mean, variance, and probability of zero loss. (Answer: mean $\$32.00$, variance $\$1{,}024.00$, zero probability $0.1353$.)

What You'll Learn