The Gamma Mean-Variance Relationship

Why the Mean-Variance Relationship Matters

A regression model describes more than how the expected response changes with predictors. It also specifies how the response varies around that expectation. This matters for positive, right-skewed outcomes such as hypothetical claim severities: two portfolios can have different average severities and different absolute variability, even when their relative variability is similar.

For a Gamma regression, the key pattern is that the conditional variance is proportional to the square of the conditional mean, provided the shape parameter is held fixed. This is different from a Poisson regression, where the conditional variance equals the conditional mean. The distinction concerns the model for variability, not just the response's units or its average.

Definition: Under the shape-scale parameterization, $Y\sim\operatorname{Gamma}(\alpha,\theta)$ has shape $\alpha>0$ and scale $\theta>0$, with density $$ f_Y(y)=\frac{y^{\alpha-1}e^{-y/\theta}}{\Gamma(\alpha)\theta^\alpha},\qquad y>0. $$ Its mean is $\alpha\theta$, and its variance is $\alpha\theta^2$. In a Gamma regression with a common shape $\alpha$, conditional mean $\mu=E[Y\mid X]$, and dispersion $\phi=1/\alpha$, the conditional variance is $$ \operatorname{Var}(Y\mid X)=\phi\mu^2. $$

Deriving the Gamma Variance Function

The parameterization is important. In the density above, $\theta$ is a scale, not a rate. If a Gamma distribution is instead written with rate $\beta=1/\theta$, its mean and variance are $\alpha/\beta$ and $\alpha/\beta^2$, respectively. Both parameterizations give the same variance relationship once the mean is used.

To derive the relationship, start with the Gamma density and calculate its moments. For a nonnegative integer $r$, integration gives the $r$-th raw moment. The key integral is the defining Gamma-function integral, with its argument shifted from $\alpha$ to $\alpha+r$.

\begin{aligned} E[Y^r] &=\int_0^\infty y^r \frac{y^{\alpha-1}e^{-y/\theta}}{\Gamma(\alpha)\theta^\alpha}\,dy\\ &=\frac{\theta^r\Gamma(\alpha+r)}{\Gamma(\alpha)}. \end{aligned}

Using $\Gamma(\alpha+1)=\alpha\Gamma(\alpha)$, the first two moments follow. Subtracting the squared mean from the second raw moment yields the variance.

\begin{aligned} E[Y]&=\theta\frac{\Gamma(\alpha+1)}{\Gamma(\alpha)} =\alpha\theta,\\ E[Y^2]&=\theta^2\frac{\Gamma(\alpha+2)}{\Gamma(\alpha)} =\alpha(\alpha+1)\theta^2,\\ \operatorname{Var}(Y) &=E[Y^2]-\{E[Y]\}^2\\ &=\alpha(\alpha+1)\theta^2-\alpha^2\theta^2 =\alpha\theta^2. \end{aligned}

Now suppose the distribution is conditional on predictors, with a common shape $\alpha$ and a scale $\theta(X)$ that may depend on the predictors. The conditional mean is $\mu=E[Y\mid X]=\alpha\theta(X)$, so $\theta(X)=\mu/\alpha$. Substituting into the conditional variance gives the variance function.

\operatorname{Var}(Y\mid X) =\alpha\left(\frac{\mu}{\alpha}\right)^2 =\frac{\mu^2}{\alpha} =\phi\mu^2,\qquad \phi=\frac{1}{\alpha}.

This derivation explains the qualification “with a common shape.” The proportionality constant $\phi$ is fixed only when $\alpha$ is fixed across the conditional distributions being modeled. If shape varies with predictors, the formula still holds separately at each predictor value, but there is no single common $\phi$ for the regression.

Result: For a Gamma response with common shape $\alpha$, the variance function is $V(\mu)=\mu^2$, and the dispersion-scaled conditional variance is $\phi V(\mu)=\phi\mu^2$, where $\phi=1/\alpha$. The conditional coefficient of variation is constant: $$ \frac{\sqrt{\operatorname{Var}(Y\mid X)}}{E[Y\mid X]} =\frac{\sqrt{\phi\mu^2}}{\mu} =\sqrt{\phi}, \qquad \mu>0. $$

Worked Examples

Worked Example: Translating Shape and Mean Into Variability

A hypothetical portfolio has positive claim severities modeled conditionally as Gamma with shape $\alpha=4$ and conditional mean $\mu=\$800$. Find the scale, conditional variance, standard deviation, and coefficient of variation. Compare its variance with that of a Poisson variable having the same mean. No interest or mortality assumption is involved; the assumed severity distribution is Gamma with shape 4 and a scale to be determined.

Method. Use the shape-scale mean $\mu=\alpha\theta$ to find $\theta$, then apply the Gamma variance formula. The Poisson comparison uses $\operatorname{Var}(N)=E[N]$ for $N\sim\operatorname{Poisson}(\lambda)$.

First solve for the Gamma scale: $$ \theta=\frac{\mu}{\alpha}=\frac{800}{4}=200. $$

The Gamma variance and standard deviation are $$ \operatorname{Var}(Y)=\alpha\theta^2=4(200)^2 =4(40{,}000)=160{,}000, \qquad \operatorname{SD}(Y)=\sqrt{160{,}000}=400. $$

The coefficient of variation is $400/800=0.5$, or 50%. For a Poisson variable with mean $\lambda=800$, the variance would be 800 and the standard deviation would be $\sqrt{800}\approx28.2843$. Thus, the Gamma variance is $160{,}000/800=200$ times as large as the Poisson variance at this numerical mean.

Check. The dispersion is $\phi=1/\alpha=0.25$, so $\phi\mu^2=0.25(800)^2=0.25(640{,}000)=160{,}000$, agreeing with the scale calculation. The comparison is mathematical; a Poisson model describes a count, whereas this Gamma model describes a positive continuous severity.

Common mistake avoided. The Poisson variance is not a default variance for every response with mean 800. It applies to a Poisson count. Under the stated Gamma severity model, the variance is determined by the shape and is 160,000, not 800.

Worked Example: Identifying a Common Gamma Dispersion

Two hypothetical risk segments have positive continuous outcomes. Their conditional means and variances are estimated as follows: Segment A has mean 250 and variance 12,500; Segment B has mean 600 and variance 72,000. Assess whether both are consistent with a common-shape Gamma variance relationship. If so, find the dispersion, shape, and scale in each segment. No interest or mortality assumption is involved; use a Gamma distribution for each segment's conditional outcome.

Method. Under a common-shape Gamma model, divide each variance by the square of its mean. The ratios should agree, and their common value is $\phi$.

For Segment A: $$ \frac{12{,}500}{250^2} =\frac{12{,}500}{62{,}500}=0.2. $$ For Segment B: $$ \frac{72{,}000}{600^2} =\frac{72{,}000}{360{,}000}=0.2. $$

The ratios agree, so the estimates are consistent with the same dispersion $\phi=0.2$. Therefore, the common shape is $$ \alpha=\frac{1}{\phi}=\frac{1}{0.2}=5. $$

For Segment A, the scale is $\theta_A=\mu_A/\alpha=250/5=50$. Its implied variance is $5(50)^2=5(2{,}500)=12{,}500$. For Segment B, the scale is $\theta_B=\mu_B/\alpha=600/5=120$. Its implied variance is $5(120)^2=5(14{,}400)=72{,}000$.

Check. Both calculated variances match the supplied values, and both coefficients of variation are $\sqrt{0.2}\approx0.4472$. The segment means differ, so their variances differ, but their relative variability is the same under this common-shape model.

Common mistake avoided. Equal dispersion does not mean equal variance. The variance is $\phi\mu^2$, so changing the mean changes the variance quadratically even when $\phi$ stays fixed.

Worked Example: Using Relative Variability to Recover the Gamma Parameters

A hypothetical severity study assumes a Gamma distribution with mean $\$360$ and coefficient of variation 0.5. Determine the shape, dispersion, scale, and variance. Then find the variance if a second risk group has mean $\$720$ but the same assumed shape. No interest or mortality assumption is involved; both groups are modeled with Gamma distributions of common shape.

Method. For a common-shape Gamma distribution, the coefficient of variation is $1/\sqrt{\alpha}=\sqrt{\phi}$. Square the given coefficient to find $\phi$, then recover the other parameters.

The dispersion and shape are $$ \phi=(0.5)^2=0.25, \qquad \alpha=\frac{1}{0.25}=4. $$

For the first group, the scale and variance are $$ \theta_1=\frac{360}{4}=90, \qquad \operatorname{Var}(Y_1)=\phi(360)^2 =0.25(129{,}600)=32{,}400. $$ Its standard deviation is $\sqrt{32{,}400}=180$, and $180/360=0.5$, as required.

For the second group, the scale is $\theta_2=720/4=180$. Its variance is $$ \operatorname{Var}(Y_2)=0.25(720)^2 =0.25(518{,}400)=129{,}600. $$ This is four times the first group's variance because the mean doubled and the variance is proportional to the squared mean: $(720/360)^2=4$. Its standard deviation is 360, again giving a coefficient of variation $360/720=0.5$.

Check. The scale-based variances are $4(90)^2=32{,}400$ and $4(180)^2=129{,}600$, which agree with the mean-dispersion calculations.

Common mistake avoided. Doubling the mean does not merely double the Gamma variance when dispersion is fixed. It multiplies the variance by four.

Gamma and Poisson Variance Patterns

For a Poisson response $N$ with conditional mean $\lambda=E[N\mid X]$, the conditional variance equals $\lambda$. For a Gamma response $Y$ with conditional mean $\mu=E[Y\mid X]$ and common dispersion $\phi$, the conditional variance is $\phi\mu^2$. The Poisson variance grows linearly with the mean; the Gamma variance grows quadratically, up to the constant $\phi$.

Model	Conditional mean	Conditional variance	Relative variability
Poisson	$\lambda$	$\lambda$	Standard deviation divided by mean is $1/\sqrt{\lambda}$
Gamma, common shape	$\mu$	$\phi\mu^2$	Standard deviation divided by mean is $\sqrt{\phi}$

The different scaling has a direct consequence. As a Poisson mean increases, its coefficient of variation decreases. With a common-shape Gamma model, the coefficient of variation remains constant while the absolute standard deviation increases in proportion to the mean. These are distinct assumptions about conditional variability; they are not interchangeable simply because two models can be assigned the same numerical mean.

The variance function also helps identify whether a model's assumptions suit the response being analyzed. A positive continuous outcome with variability that grows roughly in proportion to the square of its mean may be consistent with the Gamma pattern. A count modeled as Poisson has the linear mean-variance relationship. This comparison alone does not establish that either model is appropriate: the response type, distributional shape, and other assumptions still matter.

Common Exam Traps

Confusing scale and rate. In the shape-scale form, the mean is $\alpha\theta$. In the shape-rate form, it is $\alpha/\beta$. Substituting a rate as though it were a scale reverses the parameter's role.
Forgetting to square the mean. The Gamma variance is $\phi\mu^2$, not $\phi\mu$. The latter is a linear variance pattern.
Calling the variance proportional to the mean. With fixed $\phi$, it is proportional to the squared mean. Doubling the mean multiplies the variance by four.
Ignoring the common-shape condition. A single dispersion parameter describes the variance pattern only when shape is common across conditional distributions.
Comparing unlike responses as if their distributions were identical. A Poisson variable represents counts; a Gamma variable represents positive continuous values. Equal means do not make their variance assumptions equal.

Key takeaway: A Gamma response with common shape has $\operatorname{Var}(Y\mid X)=\phi\{E[Y\mid X]\}^2$, where $\phi=1/\alpha$. A Poisson response instead has variance equal to its conditional mean. The Gamma relationship implies constant conditional relative variability; the Poisson relationship does not.

Check Your Understanding

For each problem, use the stated distribution and parameterization. Answers are provided at the end of each question.

A hypothetical Gamma severity has shape 3 and mean 300. Find its scale and variance. (Answer: scale $=100$; variance $=30{,}000$.)
Two positive outcomes have means 100 and 400. Under a common-shape Gamma model with shape 2, find both variances and state the factor by which the variance changes. (Answer: variances $=5{,}000$ and $80{,}000$; the second is 16 times the first.)
A Gamma outcome has mean 250 and variance 10,000. Find its dispersion and shape. (Answer: $\phi=0.16$; $\alpha=6.25$.)
Compare a Poisson response with mean 40 to a Gamma response with mean 40 and dispersion $\phi=0.25$. Find each variance and the ratio of the Gamma variance to the Poisson variance. (Answer: variances $=40$ and $400$; ratio $=10$.)
Two hypothetical Gamma groups have means 200 and 500, with variances 12,000 and 75,000, respectively. Are they consistent with a common-shape model? If so, find $\phi$ and $\alpha$. (Answer: yes; both have $\phi=0.3$, so $\alpha=10/3$.)

Model	Conditional mean	Conditional variance	Relative variability
Poisson	\(\lambda\)	\(\lambda\)	Standard deviation divided by mean is \(1/\sqrt{\lambda}\)
Gamma, common shape	\(\mu\)	\(\phi\mu^2\)	Standard deviation divided by mean is \(\sqrt{\phi}\)

What You'll Learn