Tweedie Power And Variance Function

Why the Tweedie Power Matters

A mean alone does not describe how variable an insurance response is. Two groups can have different expected losses and different degrees of dispersion. The Tweedie variance function expresses both effects in a compact way: dispersion sets the scale of variance, while the power determines how variance changes as the mean changes. This relationship is useful when comparing rating cells or checking whether a proposed variance calculation is consistent with a model.

The earlier material on exponential-family distributions established the role of a mean and dispersion parameter. Here the focus is the Tweedie variance function itself. It is a statement about the variance associated with a given mean; by itself, it does not specify every feature of a probability distribution.

Definition: In a Tweedie mean-variance model, for a response with mean $\mu>0$, dispersion $\phi>0$, and power $p$, the variance is $$ \operatorname{Var}(X)=\phi\mu^p. $$ Here $\mu=E[X]$, $\phi$ is the dispersion parameter, and $p$ is the variance power. The values of $\mu$, $\phi$, and $p$ must use a consistent parameterization and measurement scale.

The power $p$ describes how strongly variance responds to changes in the mean. If the mean is multiplied by a positive factor $c$, while $\phi$ and $p$ stay fixed, then $$ \frac{\operatorname{Var}(X\text{ at mean }c\mu)} {\operatorname{Var}(X\text{ at mean }\mu)} = \frac{\phi(c\mu)^p}{\phi\mu^p} =c^p. $$ Thus, doubling the mean multiplies variance by $2^p$, not generally by 2. Dispersion has a different role: multiplying $\phi$ by a factor multiplies variance by that same factor, with the mean and power unchanged.

Interpreting the Power and Dispersion

A useful interpretation follows by taking logarithms of the variance function: $$ \log\{\operatorname{Var}(X)\}=\log(\phi)+p\log(\mu). $$ When $\phi$ is fixed, a one-unit increase in $\log(\mu)$ changes $\log\{\operatorname{Var}(X)\}$ by $p$. Equivalently, the elasticity of variance with respect to the mean is $p$: a small percentage change in the mean corresponds approximately to $p$ times that percentage change in variance. This is a relative-change interpretation, not a claim that the variance increases by a fixed amount for each additional dollar of mean.

The coefficient of variation is the standard deviation divided by the mean. From the Tweedie variance function, for $\mu>0$, $$ \operatorname{CV}(X) = \frac{\sqrt{\operatorname{Var}(X)}}{\mu} = \sqrt{\phi}\,\mu^{(p-2)/2}. $$ This shows why the power affects relative variability as well as absolute variance. For $p=2$, the coefficient of variation is $\sqrt{\phi}$, independent of the mean. For other powers, it changes as the mean changes. This calculation describes the implication of the variance function; it does not determine the response distribution.

The numerical value and units of $\phi$ depend on the units used for $X$. If the response is expressed in dollars rather than thousands of dollars, the numerical mean changes, and the corresponding dispersion must be expressed consistently for the same variance model. Therefore, comparisons of dispersion values require compatible response scales and parameterizations.

Worked Calculations

Worked Example: Calculate Variance at Two Means

A hypothetical rating analysis models loss $X$ in dollars with a Tweedie variance function. At one rating-cell mean, $\mu=\$250$, the fitted dispersion is $\phi=0.04$ in the corresponding units, and the power is $p=1.5$. Calculate the variance and standard deviation. Then calculate the variance if the mean doubles to $\$500$, with $\phi$ and $p$ unchanged.

Method and assumptions. Use $\operatorname{Var}(X)=\phi\mu^p$, with the same dispersion parameterization and response units for both means. First evaluate the original variance: $$ \operatorname{Var}(X)=0.04(250)^{1.5}. $$ Since $250^{1.5}=250\sqrt{250}=250(15.811388\ldots)=3952.847\ldots$, $$ \operatorname{Var}(X)=0.04(3952.847\ldots)=158.1139\ldots. $$ The standard deviation is the square root of the variance: $$ \sqrt{158.1139\ldots}=12.5743\ldots. $$ Rounded only at the final step, the variance is $\$158.11^2$ and the standard deviation is $\$12.57$.

For the doubled mean, compute directly: $$ \operatorname{Var}(X)=0.04(500)^{1.5} =0.04(11180.3399\ldots) =447.2136\ldots. $$ The variance is therefore $\$447.21^2$, rounded to two decimals. As a check, the mean doubled, so the variance should be multiplied by $2^{1.5}=2\sqrt{2}=2.828427\ldots$. Indeed, $158.1139\ldots(2.828427\ldots)=447.2136\ldots$. The standard deviation is not needed for the second mean, but is approximately $\$21.15$.

Common mistake avoided: Doubling the mean does not simply double the variance here. That would use a scaling factor of $2$, but the correct factor is $2^{1.5}$.

Worked Example: Interpret a Quadratic Variance Function

For a hypothetical severity response measured in dollars, suppose $\phi=0.0025$ and $p=2$. Calculate the variance, standard deviation, and coefficient of variation when the mean is $\$400$. Then repeat the variance and coefficient-of-variation calculations at a mean of $\$200$, keeping $\phi$ and $p$ fixed.

Method and assumptions. The variance function applies with $\mu>0$, fixed positive $\phi$, and the same response units at both means. At $\mu=400$, $$ \operatorname{Var}(X)=0.0025(400)^2 =0.0025(160000)=400. $$ Thus the standard deviation is $\sqrt{400}=\$20$, and the coefficient of variation is $20/400=0.05$, or $5\%$.

At $\mu=200$, $$ \operatorname{Var}(X)=0.0025(200)^2 =0.0025(40000)=100. $$ The coefficient of variation is $\sqrt{100}/200=10/200=0.05$, again $5\%$. The mean was halved, so the variance was multiplied by $(1/2)^2=1/4$, decreasing from $400$ to $100$. The unchanged coefficient of variation also follows directly from $\operatorname{CV}(X)=\sqrt{\phi}\mu^{(2-2)/2}=\sqrt{0.0025}=0.05$. All requested amounts are exact, so no intermediate rounding was needed.

Common mistake avoided: A quadratic variance function does not imply that standard deviation is proportional to the square of the mean. Standard deviation is the square root of variance; in this example it is proportional to the mean.

Worked Example: Infer the Power and Dispersion

A hypothetical analysis uses the same Tweedie variance function for two rating cells on the same response scale. The first cell has mean $\mu_1=100$ and variance $20$; the second has mean $\mu_2=400$ and variance $160$. Assume both cells share $\phi$ and $p$. Find $p$ and $\phi$, then verify both variances.

Method and assumptions. Write the model for each cell: $$ 20=\phi(100)^p,\qquad 160=\phi(400)^p. $$ Divide the second equation by the first to eliminate the common dispersion: $$ \frac{160}{20} = \left(\frac{400}{100}\right)^p, \qquad 8=4^p. $$ Since $4^{1.5}=4\sqrt{4}=4(2)=8$, the power is $p=1.5$.

Substitute this value into the first cell's equation. Since $100^{1.5}=100\sqrt{100}=1000$, $$ 20=\phi(1000), \qquad \phi=\frac{20}{1000}=0.02. $$ Check the first variance: $0.02(100)^{1.5}=0.02(1000)=20$. Check the second: $400^{1.5}=400\sqrt{400}=400(20)=8000$, so $0.02(8000)=160$. Both given variances are reproduced exactly; no rounding is needed.

Common mistake avoided: The variance ratio is not the mean ratio. The means have ratio $4$, whereas variances have ratio $8$; the power must satisfy $4^p=8$, giving $p=1.5$.

Why the Scaling Rule Holds

The scaling rule is a direct consequence of the variance function, not a separate distributional assumption. Suppose two modeled responses have means $\mu_1>0$ and $\mu_2>0$, and share the same $\phi$ and $p$. Their variances are $\phi\mu_1^p$ and $\phi\mu_2^p$. Taking the ratio cancels the shared dispersion:

\frac{\operatorname{Var}_2}{\operatorname{Var}_1} = \frac{\phi\mu_2^p}{\phi\mu_1^p} = \left(\frac{\mu_2}{\mu_1}\right)^p.

This justification also identifies a necessary condition: the same $\phi$ and $p$ must apply to both comparisons. If dispersion differs by rating cell, the variance ratio includes that difference: $$ \frac{\operatorname{Var}_2}{\operatorname{Var}_1} = \frac{\phi_2}{\phi_1} \left(\frac{\mu_2}{\mu_1}\right)^p. $$ It is not valid to infer the power from the mean and variance ratios alone when the dispersions are unequal or unknown.

Calculation Checks and Exam Traps

A reliable calculation separates the three roles in $\phi\mu^p$: identify the power, raise the mean to that power, and then multiply by dispersion. Keep the following checks in mind:

Variance versus standard deviation: $\phi\mu^p$ is variance. Take its square root only when standard deviation is requested.
Apply the power to the mean: calculate $\mu^p$, then multiply by $\phi$. Do not calculate $(\phi\mu)^p$; that incorrectly raises dispersion to the power too.
Use the correct scaling exponent: multiplying the mean by $c$ multiplies variance by $c^p$, provided the dispersion and power are unchanged.
Check what is shared: comparing two cells by the scaling rule requires a common dispersion and power. If those differ, include both parameters in the comparison.
Do not infer a full distribution from variance alone: the mean-variance relationship gives a moment calculation, not every distributional feature or model assumption.

For example, with $\phi=0.04$, $\mu=250$, and $p=1.5$, a common incorrect calculation is $(0.04\times250)^{1.5}=10^{1.5}$, approximately $31.62$. The correct calculation is $0.04\times250^{1.5}=158.11$, rounded to two decimals. The error is raising $\phi$ to the power along with the mean.

Key takeaway: A Tweedie variance function has the form $\operatorname{Var}(X)=\phi\mu^p$. Dispersion scales variance, while the power controls how variance changes with the mean. Always check the units, parameter sharing, and whether the requested quantity is variance or standard deviation.

Check Your Understanding

Use the Tweedie variance function and keep dispersion and power fixed within each comparison.

A hypothetical response has $\mu=80$, $\phi=0.05$, and $p=1$. Calculate its variance. (Answer: $4$)
For $\mu=144$, $\phi=0.01$, and $p=1.5$, calculate the variance. (Answer: $17.28$)
With $p=1.25$ and unchanged dispersion, the mean doubles. By what factor does variance change? (Answer: $2^{1.25}\approx2.3784$)
A response has mean $50$, variance $12.5$, and $p=1.5$. Find $\phi$. (Answer: $\phi=12.5/50^{1.5}\approx0.03536$)
Two cells share $\phi$ and $p$. Their means are $25$ and $100$, and their variances are $6$ and $48$, respectively. Find $p$ and $\phi$, and verify the second variance. (Answer: $p=1.5$, $\phi=0.048$; $0.048(100^{1.5})=48$)

What You'll Learn