Distribution Parameter Calculation Workshop

From Supplied Parameters to Useful Quantities

In ratemaking work, a distribution may be specified by parameters that are not themselves the quantities of interest. A frequency model might give a Poisson mean, while a severity model might give a Gamma mean and dispersion. The calculation task is to translate that information into the requested mean, variance, or probability without silently changing parameterizations.

The main safeguard is to write down what each parameter means before substituting it. In particular, a negative binomial probability parameter can be defined in more than one way, and a Gamma distribution may be parameterized by shape and scale or by shape and rate. Here, each convention is stated explicitly. Loss amounts are nonnegative, and calculations that use continuous distribution functions refer to the model’s stated support.

Definition: For a Poisson random variable \(N\) with mean parameter \(\lambda>0\), \(E[N]=\operatorname{Var}(N)=\lambda\), and \(P(N=k)=e^{-\lambda}\lambda^k/k!\) for \(k=0,1,2,\ldots\). For a negative binomial random variable \(N\), this tutorial uses the convention that \(N\) counts failures before \(r\) successes, each trial having success probability \(p\). Thus \(P(N=k)=\binom{k+r-1}{k}p^r(1-p)^k\), \(E[N]=r(1-p)/p\), and \(\operatorname{Var}(N)=r(1-p)/p^2\).

For the continuous families, this tutorial uses a mean-dispersion parameterization. For a Gamma variable with mean \(\mu>0\) and dispersion \(\phi>0\), the shape is \(\alpha=1/\phi\), the scale is \(\theta=\mu\phi\), and the variance is \(\phi\mu^2\). For an inverse Gaussian variable with mean \(\mu>0\) and dispersion \(\phi>0\), the variance is \(\phi\mu^3\). A Tweedie variable with power \(1<p<2\) has mean \(\mu>0\) and variance \(\phi\mu^p\); in this range it has a compound-Poisson representation and can have probability mass at zero.

Frequency Calculations: Poisson and Negative Binomial

A Poisson parameter is already the mean, so the mean and variance require no conversion. A negative binomial parameter pair \((r,p)\), by contrast, describes a count distribution under the stated failures-before-successes convention. If a problem instead supplies the mean \(\mu\) and size \(r\), solve \(\mu=r(1-p)/p\) to obtain \(p=r/(r+\mu)\), then use the probability formula.

Worked Example: A Poisson Count and an Overdispersed Count

A hypothetical policy segment has a Poisson claim count with \(\lambda=3\). A second hypothetical segment is modeled with a negative binomial count having mean \(\mu=6\) and size \(r=4\). For each model, calculate the mean, variance, and probability of exactly two claims. For the negative binomial, use the failures-before-successes convention.

Poisson method. The mean and variance are both \(\lambda=3\). For exactly two claims,

P(N=2)=e^{-3}\frac{3^2}{2!} =e^{-3}\frac{9}{2} =4.5e^{-3} \approx 0.2240.

The result is a valid probability, and two is within the Poisson support of the nonnegative integers.

Negative binomial method. First convert the supplied mean and size to the success probability:

p=\frac{r}{r+\mu}=\frac{4}{4+6}=0.4, \qquad 1-p=0.6.

The mean is the supplied value, \(6\). The variance is

\operatorname{Var}(N)=\frac{r(1-p)}{p^2} =\frac{4(0.6)}{(0.4)^2} =\frac{2.4}{0.16}=15.

For exactly two claims,

P(N=2)=\binom{2+4-1}{2}(0.4)^4(0.6)^2 =10(0.0256)(0.36) =0.09216.

The check is that the negative binomial variance, \(15\), exceeds its mean, \(6\), consistent with this parameterization’s extra count variability. The method avoids the common mistake of using the Poisson rule \(\operatorname{Var}(N)=E[N]\) for the negative binomial.

The negative binomial probability calculation depends on the meaning assigned to \(p\). Under the convention used here, \(p\) is the probability of success and \(N\) counts failures. A formula that treats \(p\) as the probability of a failure would produce different substitutions. Always match the probability formula to the stated convention, not just to a remembered symbol.

Severity Calculations: Gamma and Inverse Gaussian

For a Gamma model specified by mean and dispersion, convert to shape and scale before applying a distribution function. The scale is \(\theta=\mu\phi\), not the rate; the rate would be \(1/\theta\). For an inverse Gaussian calculation, the dispersion form also determines the shape parameter in the density and cumulative distribution formula: \(\lambda=1/\phi\), so that \(\operatorname{Var}(X)=\mu^3/\lambda=\phi\mu^3\).

Worked Example: Gamma Severity Below a Threshold

A hypothetical claim severity \(X\), measured in thousands of dollars, has a Gamma distribution with mean \(\mu=10\) and dispersion \(\phi=0.5\). Calculate its variance and \(P(X\leq 10)\).

Method. Convert the mean-dispersion parameters into shape and scale:

\alpha=\frac{1}{\phi}=\frac{1}{0.5}=2, \qquad \theta=\mu\phi=10(0.5)=5.

The mean check is \(\alpha\theta=2(5)=10\), matching the supplied mean. The variance is

\operatorname{Var}(X)=\alpha\theta^2 =2(5^2)=50 \quad\text{(thousand dollars)}^2.

For shape \(2\) and scale \(5\), the Gamma cumulative distribution function is \(F_X(x)=1-e^{-x/\theta}(1+x/\theta)\) for \(x\geq0\). Therefore,

P(X\leq10)=1-e^{-10/5}\left(1+\frac{10}{5}\right) =1-3e^{-2} \approx 1-3(0.1353353) =0.5940.

The answer is between zero and one, and the event is evaluated at the mean, not at a threshold rescaled by the dispersion. The common mistake avoided here is treating \(\phi\) as the Gamma scale: the scale is \(\mu\phi=5\), not \(0.5\).

Worked Example: Inverse Gaussian Severity at Its Mean

A hypothetical positive claim severity \(X\), in thousands of dollars, follows an inverse Gaussian distribution with mean \(\mu=1\) and dispersion \(\phi=1\). Calculate its variance and \(P(X\leq1)\).

Method. The variance is \(\phi\mu^3\), so

\operatorname{Var}(X)=1(1^3)=1 \quad\text{(thousand dollars)}^2.

For the cumulative probability, use the inverse Gaussian shape parameter \(\lambda=1/\phi=1\). For \(x>0\), its cumulative distribution function is

F_X(x)= \Phi\left(\sqrt{\frac{\lambda}{x}}\left(\frac{x}{\mu}-1\right)\right) + \exp\left(\frac{2\lambda}{\mu}\right) \Phi\left(-\sqrt{\frac{\lambda}{x}}\left(\frac{x}{\mu}+1\right)\right),

where \(\Phi\) is the standard normal cumulative distribution function. At \(x=\mu=1\), the first argument is \(1(1-1)=0\), and the second argument is \(-1(1+1)=-2\). Thus,

P(X\leq1)=\Phi(0)+e^2\Phi(-2) =0.5+(7.389056)(0.0227501) \approx0.6681.

This probability is greater than one-half even though the threshold equals the mean; the inverse Gaussian distribution is right-skewed. The calculation checks the distinction between the mean and the median. It also avoids replacing the inverse Gaussian cumulative distribution with a normal approximation.

Tweedie Moments and the Probability of Zero

For a Tweedie model with \(1<p<2\), \(\mu\) is the mean, \(\phi\) is the dispersion, and \(p\) is the variance power. The variance follows directly from the stated parameterization, but the probability of zero uses the compound-Poisson representation. In that representation the number \(N\) of positive jumps is Poisson with mean

\lambda=\frac{\mu^{2-p}}{\phi(2-p)}.

The positive jump amounts have a continuous Gamma distribution, so a sum of one or more positive jumps is positive with probability one. Consequently, the aggregate equals zero exactly when the jump count is zero.

Derivation: If \(N\sim\operatorname{Poisson}(\lambda)\) is the number of positive jumps, the compound total is zero if and only if \(N=0\), because each jump is strictly positive. Therefore \(P(X=0)=P(N=0)=e^{-\lambda}\). Substituting the Tweedie compound-Poisson mean gives \(P(X=0)=\exp\left(-\mu^{2-p}/[\phi(2-p)]\right)\). This result applies to the compound-Poisson Tweedie range \(1<p<2\); it is not a general zero-probability formula for every Tweedie power.

Worked Example: Tweedie Mean, Variance, and Zero Mass

A hypothetical aggregate loss \(X\) follows a Tweedie model with \(\mu=1\), \(\phi=2\), and power \(p=1.5\). Calculate its variance and probability of zero.

Variance. Apply the variance function using the supplied power:

\operatorname{Var}(X)=\phi\mu^p =2(1^{1.5})=2.

Zero probability. First calculate the compound-Poisson mean:

\lambda=\frac{\mu^{2-p}}{\phi(2-p)} =\frac{1^{0.5}}{2(0.5)} =\frac{1}{1}=1.

Thus,

P(X=0)=e^{-1}\approx0.3679.

The result is consistent with the compound representation: a Poisson jump count with mean one has probability \(e^{-1}\) of no jumps. The method avoids using \(e^{-\mu}\) as a universal Tweedie zero-mass formula; the exponent depends on \(\mu\), \(\phi\), and \(p\).

A Reliable Calculation Sequence

A short sequence of checks reduces parameterization errors and makes a numerical answer auditable.

Identify the family and support.
Determine whether the variable is a count, a continuous positive amount, or a Tweedie aggregate that can include zero.

Write the parameter meanings.
Record the negative binomial convention, Gamma shape and scale, or the mean-dispersion form before using a formula.

Convert if necessary.
For example, turn a Gamma mean and dispersion into \(\alpha=1/\phi\) and \(\theta=\mu\phi\), or find \(p=r/(r+\mu)\) for a negative binomial supplied by mean and size.

Substitute, then check.
Verify a mean or variance against its parameterization, confirm probabilities lie between zero and one, and confirm the requested event is within the support.

A frequent exam trap is confusing a parameter with a moment. For example, the Poisson parameter is both mean and variance, but the negative binomial size is not its mean; Gamma dispersion is not its scale; and Tweedie dispersion alone does not determine the zero probability. A wrong calculation might set a negative binomial variance equal to its mean, giving \(6\) in the first worked example. The correction is to use \(\operatorname{Var}(N)=r(1-p)/p^2=15\). The contrast is not a matter of rounding: it follows from the distribution family and parameterization.

Key takeaway: Translate the parameters using the exact family-specific convention first. Then calculate the requested quantity and check it against the model’s support and moment relationships.

Check Your Understanding

For each problem, state the parameterization you use and keep intermediate calculations until the final answer.

A Poisson claim count has \(\lambda=2.5\). Find its mean, variance, and probability of zero claims. (Answer: mean \(=2.5\), variance \(=2.5\), \(P(N=0)=e^{-2.5}\approx0.0821\).)
A negative binomial count has mean \(\mu=8\) and size \(r=4\), counting failures before \(r\) successes. Find its variance and probability of exactly one claim. (Answer: \(p=4/(4+8)=1/3\), variance \(=24\), and \(P(N=1)=4(1/3)^4(2/3)=8/243\approx0.0329\).)
A Gamma severity has mean \(\mu=12\) and dispersion \(\phi=0.5\). Find its shape, scale, variance, and probability of a severity no greater than 6. (Answer: shape \(=2\), scale \(=6\), variance \(=72\), and \(P(X\leq6)=1-2e^{-1}\approx0.2642\).)
An inverse Gaussian severity has mean \(\mu=2\) and dispersion \(\phi=0.25\). Find its variance and \(P(X\leq2)\). Use \(\Phi(-2\sqrt{2})\approx0.0023389\). (Answer: variance \(=0.25(2^3)=2\); with \(\lambda=4\), \(P(X\leq2)=0.5+e^4\Phi(-2\sqrt2)\approx0.6277\).)
A Tweedie aggregate has \(\mu=4\), \(\phi=0.5\), and \(p=1.5\). Find its variance and probability of zero. (Answer: variance \(=0.5(4^{1.5})=4\), \(\lambda=2/(0.5(0.5))=8\), and \(P(X=0)=e^{-8}\approx0.0003355\).)

What You'll Learn