Deriving the Power Variance Function

From a Variance Function to a Cumulant Function

In an exponential-dispersion model, the mean is not merely a location parameter: it determines the variance through a variance function. Specifying that function as a power, $V(\mu)=\mu^p$, imposes a precise relationship between the mean $\mu$, the canonical parameter, and the cumulant function. Deriving that relationship also shows how dispersion scales the variance and the higher cumulants.

The derivation is conditional on having a valid exponential-dispersion model on the parameter range under consideration. The algebra below does not establish that a normalized distribution exists for every value of $p$; whether a proposed power and parameter range are admissible is a separate question.

Definition: An exponential-dispersion model with fixed dispersion $\phi>0$ has density or mass function $$ f(y;\theta,\phi)=\exp\left\{\frac{y\theta-b(\theta)}{\phi}+c(y,\phi)\right\}, $$ where $\theta$ lies in an open natural-parameter interval, $b$ is the cumulant function, and $c$ is the carrier term. Under the regularity conditions that permit differentiating the normalizing integral or sum with respect to $\theta$, the mean is $\mu=b'(\theta)$, the variance is $\phi b''(\theta)$, and the variance function is $V(\mu)=b''(\theta)$, expressed as a function of $\mu$.

Here $\phi$ is held fixed when differentiating with respect to $\theta$. The power variance specification is $V(\mu)=\mu^p$, considered on a mean range where $\mu>0$. Thus the model identities become

\frac{d\mu}{d\theta}=b''(\theta)=\mu^p, \qquad \operatorname{Var}(X)=\phi\mu^p.

The first equation determines how quickly the mean changes as the canonical parameter changes. Inverting it gives $d\theta/d\mu=\mu^{-p}$; integrating that equation determines the canonical parameter up to an additive constant. Since $db/d\mu=b'(\theta)\,d\theta/d\mu=\mu^{1-p}$, a second integration determines the cumulant function.

Deriving the Canonical Parameter and Cumulant Function

Integrate the parameter-mean relationship.
Use $d\theta/d\mu=\mu^{-p}$. The integral is logarithmic when $p=1$, and a power otherwise.

Integrate for the cumulant function.
Since $db/d\mu=\mu^{1-p}$, this integral is logarithmic when $p=2$, and a power otherwise.

Check by differentiation.
Differentiate the resulting $b$ with respect to $\theta$; the result must be $\mu$, and its second derivative must be $\mu^p$.

Choosing zero integration constants gives the following relationships. Additive constants in the parameter coordinate or in $b$ change the coordinate convention or normalizing carrier, not the prescribed variance function.

\theta(\mu)= \begin{cases} \dfrac{\mu^{1-p}}{1-p}, & p\ne 1,\\[6pt] \log\mu, & p=1, \end{cases} \qquad b(\mu)= \begin{cases} \dfrac{\mu^{2-p}}{2-p}, & p\ne 2,\\[6pt] \log\mu, & p=2. \end{cases}

These are parametric expressions: both $\theta$ and $b$ are specified through the same mean $\mu$. For $p\ne1$, the first relationship can be inverted wherever it is defined, giving $\mu=((1-p)\theta)^{1/(1-p)}$. Substitution into $b(\mu)$ then gives $b$ as a function of $\theta$. The parametric form avoids imposing a parameter domain beyond the mean range on which the model is being considered.

Derived result: If an exponential-dispersion model has $V(\mu)=\mu^p$ on a positive mean range, then $d\theta/d\mu=\mu^{-p}$ and $db/d\mu=\mu^{1-p}$. Consequently, the canonical parameter and cumulant function are the integrals displayed above, with the logarithmic forms at $p=1$ and $p=2$.

Worked Example: A Power Index of Three

Suppose the variance function is $V(\mu)=\mu^3$. Derive the canonical parameter and cumulant function in terms of $\mu$, and verify the variance identity.

The parameter-mean equation is $d\theta/d\mu=\mu^{-3}$. Integrating,

\theta(\mu)=\int \mu^{-3}\,d\mu=-\frac{1}{2}\mu^{-2}.

For the cumulant function, $db/d\mu=\mu^{1-3}=\mu^{-2}$, so

b(\mu)=\int \mu^{-2}\,d\mu=-\mu^{-1}.

To check, $d\mu/d\theta=\mu^3$ follows from $d\theta/d\mu=\mu^{-3}$. Also $db/d\mu=\mu^{-2}$, so the chain rule gives $db/d\theta=(db/d\mu)(d\mu/d\theta)=\mu^{-2}\mu^3=\mu$. Differentiating again gives $b''(\theta)=d\mu/d\theta=\mu^3$. The model variance, if the model exists on the specified range, is therefore $\phi\mu^3$.

Variance and Higher Cumulants

The variance follows immediately from the exponential-dispersion identity: $\operatorname{Var}(X)=\phi V(\mu)=\phi\mu^p$. Higher cumulants follow by differentiating the cumulant function further. The next result makes the dispersion scaling explicit.

Derived result: Suppose the moment-generating function of $X$ is finite in a neighborhood of zero. For every integer $r\ge1$ for which the required derivatives exist, the $r$th cumulant of $X$ is $$ \kappa_r(X)=\phi^{r-1}b^{(r)}(\theta). $$ For a power variance function, the third and fourth cumulants are $$ \kappa_3(X)=\phi^2p\mu^{2p-1}, \qquad \kappa_4(X)=\phi^3p(2p-1)\mu^{3p-2}. $$

Here $\kappa_1(X)=E[X]$, $\kappa_2(X)=\operatorname{Var}(X)$, and higher $\kappa_r$ are the ordinary cumulants, defined as derivatives of the cumulant-generating function at zero. The existence condition matters: the algebraic derivatives of $b$ alone do not guarantee that the moment-generating function exists around zero.

Obtain the cumulant-generating function.
Substitute $\theta+\phi t$ into the exponential-dispersion model and use normalization.

Differentiate at zero.
The $r$th derivative contributes $\phi^{-1}\phi^r b^{(r)}(\theta)$, giving the stated dispersion power.

Differentiate the power variance function.
Use $b'''=d(\mu^p)/d\theta$ and $d\mu/d\theta=\mu^p$, then repeat once for $b''''$.

For completeness, let $M_X(t)=E[\exp(tX)]$, and assume it is finite for $t$ in an open interval containing zero, with $\theta+\phi t$ in the natural-parameter interval there. Substitution into the model gives

M_X(t)=\exp\left\{\frac{b(\theta+\phi t)-b(\theta)}{\phi}\right\}, \qquad K_X(t)=\log M_X(t)=\frac{b(\theta+\phi t)-b(\theta)}{\phi}.

Differentiating $K_X$ $r$ times at zero yields $\kappa_r(X)=K_X^{(r)}(0)=\phi^{r-1}b^{(r)}(\theta)$. For the power case, the chain rule gives

b'''(\theta)=\frac{d}{d\theta}\mu^p =p\mu^{p-1}\frac{d\mu}{d\theta} =p\mu^{2p-1},

and one more differentiation gives

b''''(\theta) =p(2p-1)\mu^{2p-2}\frac{d\mu}{d\theta} =p(2p-1)\mu^{3p-2}.

Combining these derivatives with the cumulant formula proves the third- and fourth-cumulant expressions. In particular, the third cumulant is the third central moment when it exists.

Worked Example: Numerical Variance and Third Cumulant

Take $p=3/2$, mean $\mu=4$, and dispersion $\phi=0.2$. Calculate the canonical parameter, cumulant function, variance, and third cumulant.

Because $p\ne1$, $\theta=\mu^{1-p}/(1-p)=4^{-1/2}/(-1/2)=-1$. Because $p\ne2$, $b=\mu^{2-p}/(2-p)=4^{1/2}/(1/2)=4$. These are the parameter and cumulant values under the zero-constant convention.

The variance is

\operatorname{Var}(X)=\phi\mu^p =0.2(4^{3/2}) =0.2(8) =1.6.

The third cumulant is

\kappa_3(X)=\phi^2p\mu^{2p-1} =(0.2)^2\left(\frac{3}{2}\right)4^2 =0.04(1.5)(16) =0.96.

The powers of $\phi$ differ: variance scales with $\phi$, while the third cumulant scales with $\phi^2$. The calculation establishes the implications of the assumed power variance function; it does not by itself establish that a valid distribution with these parameters exists.

What the Mean and Power Control

The mean $\mu$ sets the scale at which the variance and higher cumulants are evaluated. At a fixed power $p$ and dispersion $\phi$, doubling the mean multiplies the variance by $2^p$, since $\phi(2\mu)^p=2^p\phi\mu^p$. The power $p$ controls this scaling. Equivalently, wherever $\mu>0$,

\frac{d\log V(\mu)}{d\log\mu}=p.

Thus $p$ is the elasticity of the variance function with respect to the mean: a one-percent increase in $\mu$ corresponds locally to a $p$-percent increase in $V(\mu)$. Dispersion then multiplies the variance by $\phi$, without changing this elasticity. It also enters the $r$th cumulant as $\phi^{r-1}$, under the stated existence conditions.

Worked Example: The Logarithmic Parameter at Power One

For the algebraic case $p=1$, derive the canonical parameter and cumulant function, then calculate the variance and third cumulant at $\mu=4$ and $\phi=0.25$.

The parameter equation is $d\theta/d\mu=\mu^{-1}$, hence $\theta=\log\mu$. The cumulant equation is $db/d\mu=\mu^{0}=1$, hence $b=\mu$, with integration constants set to zero. At $\mu=4$, this gives $\theta=\log 4$ and $b=4$.

The variance and third cumulant are

\operatorname{Var}(X)=0.25(4^1)=1, \qquad \kappa_3(X)=(0.25)^2(1)(4^1)=0.25.

These values use only the general derivative identities and the $p=1$ integrals. They identify neither a particular response distribution nor its support.

Common Mistake and Practical Interpretation

A common mistake is to treat $V(\mu)=\mu^p$ as if it were itself the model variance, or to assume that integrating the variance function automatically proves that a distribution exists. The correction is to keep the two roles separate: $V(\mu)$ is the variance function, while the model variance is $\phi V(\mu)$. The integrations determine the cumulant structure required if the exponential-dispersion model exists; they do not establish its normalization, support, or admissible parameter range.

A second potential error is to use the generic power formulas at $p=1$ or $p=2$ without taking account of their zero denominators. The underlying integrals remain well-defined, but their antiderivatives are logarithms: $\theta=\log\mu$ at $p=1$, and $b=\log\mu$ at $p=2$. These are not singularities in the variance function; they are the corresponding logarithmic integration cases.

Key takeaway: In an exponential-dispersion model, $d\mu/d\theta=V(\mu)$. Setting $V(\mu)=\mu^p$ determines the canonical parameter and cumulant function by integration, gives $\operatorname{Var}(X)=\phi\mu^p$, and implies higher cumulants through further derivatives. The mean sets the scale; $p$ sets how that scale changes.

Check Your Understanding

Use the exponential-dispersion identities and state any condition needed for your answer.

Starting from $V(\mu)=\mu^p$, derive $d\theta/d\mu$ and $db/d\mu$.
Why do the integrations for $\theta$ at $p=1$ and for $b$ at $p=2$ require separate logarithmic expressions?
For general $p$, write the model variance and the third cumulant in terms of $\mu$, $p$, and $\phi$.
At fixed $p$ and $\phi$, by what factor does the variance change when the mean is multiplied by three?
Why does deriving $b(\theta)$ from a proposed power variance function not, on its own, prove that a normalized response distribution exists?