The Gaussian Case at Power Zero

Why Power Zero Is a Gaussian Case

The power variance function $V(\mu)=\mu^p$ describes how an exponential-dispersion model’s variance changes with its mean. At $p=0$, the variance no longer depends on the mean: it is constant. This is exactly the mean–variance relationship of a Gaussian response. The parameter correspondence is direct: the Tweedie mean is the normal mean, and the Tweedie dispersion is the normal variance.

This case is worth handling explicitly because it makes the general exponential-dispersion notation concrete. It also clarifies an easy point of confusion: $\phi$ is a variance parameter, not a standard deviation, and $p=0$ means constant variance, not zero variance.

Power Zero and the Exponential-Dispersion Form

An exponential-dispersion model can be written, for response $Y$, as

f(y;\theta,\phi) = a(y,\phi)\exp\left\{\frac{y\theta-\kappa(\theta)}{\phi}\right\}, \qquad \phi>0.

Here $\theta$ is the canonical parameter, $\phi$ is the dispersion, and $\kappa$ is the cumulant function. In this parameterization, the mean is $\mu=\kappa'(\theta)$, and the variance is $\phi\kappa''(\theta)=\phi V(\mu)$. For the power variance function, $V(\mu)=\mu^p$.

Definition: At power $p=0$, take the power variance function to be the constant function $V_0(\mu)=1$. Thus the exponential-dispersion model has $E[Y]=\mu$ and $\operatorname{Var}(Y)=\phi$, where $\mu$ can be any real number and $\phi>0$.

The definition at power zero is the constant function $1$, including when $\mu=0$ or $\mu<0$. It is not necessary to evaluate an ambiguous expression $0^0$, nor to impose the positive-mean restriction that applies in some other power cases. The resulting Gaussian family has support on the whole real line and permits any real mean.

To identify the canonical parameter, use $d\mu/d\theta=\kappa''(\theta)=V_0(\mu)=1$. Choosing the conventional origin $\theta=0$ when $\mu=0$ gives $\theta=\mu$. Since $\kappa'(\theta)=\mu=\theta$, integration with $\kappa(0)=0$ gives $\kappa(\theta)=\theta^2/2$.

Key parameter correspondence: In the power-zero case, $\theta=\mu$ and $\kappa(\theta)=\theta^2/2$. The response has mean $\mu$ and variance $\phi$; equivalently, $Y\sim N(\mu,\phi)$, with the second normal parameter denoting variance.

Deriving the Gaussian Distribution

The power variance function already gives the first two moments, but identifying the entire distribution requires more. The following derivation obtains the Gaussian density and confirms the normal distribution from the exponential-dispersion form.

Theorem: Suppose $Y$ follows a normalized exponential-dispersion model with power-zero variance function, with $\phi>0$ and canonical parameter $\theta\in\mathbb{R}$. Under the conventional choice $\theta=\mu$ and $\kappa(\theta)=\theta^2/2$, its density is $$ f(y;\mu,\phi)=\frac{1}{\sqrt{2\pi\phi}}\exp\left\{-\frac{(y-\mu)^2}{2\phi}\right\},\qquad y\in\mathbb{R}. $$ Consequently, $Y\sim N(\mu,\phi)$.

Proof. For the normalized exponential-dispersion form, the cumulant-generating function of $Y$, for $t$ such that $\theta+\phi t$ belongs to the natural parameter space, is

K_Y(t)=\log E[\exp(tY)] =\frac{\kappa(\theta+\phi t)-\kappa(\theta)}{\phi}.

In this case the natural parameter space is all of $\mathbb{R}$, so this formula holds for every real $t$. Substituting $\kappa(\theta)=\theta^2/2$ and $\theta=\mu$,

K_Y(t) = \frac{(\mu+\phi t)^2-\mu^2}{2\phi} = \mu t+\frac{\phi t^2}{2}.

This is the cumulant-generating function of $N(\mu,\phi)$. Since a moment-generating function finite in a neighborhood of zero uniquely determines a probability distribution, $Y\sim N(\mu,\phi)$. The density can also be seen directly by completing the square in the exponential-dispersion kernel:

y\mu-\frac{\mu^2}{2}-\frac{y^2}{2} = -\frac{(y-\mu)^2}{2}.

With the Gaussian carrier $a(y,\phi)=(2\pi\phi)^{-1/2}\exp\{-y^2/(2\phi)\}$, the density becomes the displayed normal density. It integrates to one because it is a normal density with positive variance $\phi$. The cumulant-generating function also gives $E[Y]=K_Y'(0)=\mu$ and $\operatorname{Var}(Y)=K_Y''(0)=\phi$, as required. $\square$

Worked Parameter Translations

Worked Example: Reading the Mean and Variance

Suppose a response has a Tweedie power variance function with $p=0$, mean parameter $\mu=12$, and dispersion $\phi=4$. Identify its normal distribution and calculate the probability that the response lies between 10 and 14.

At power zero, the normal mean is $\mu=12$, and the normal variance is $\phi=4$. Therefore the standard deviation is $\sqrt{4}=2$, and $Y\sim N(12,4)$.

Standardizing both endpoints gives $P(10\leq Y\leq14)=P((10-12)/2\leq Z\leq(14-12)/2)$, where $Z\sim N(0,1)$. Hence $P(10\leq Y\leq14)=\Phi(1)-\Phi(-1)\approx0.6827$. Thus the probability is about $68.27\%$. The interval is one standard deviation on either side of the mean.

Worked Example: A Negative Mean Is Permitted

Let $p=0$, $\mu=-1$, and $\phi=4$. Find the distribution and calculate $P(Y>1)$.

The power-zero variance is $V_0(\mu)=1$, regardless of the sign of $\mu$. Thus $\operatorname{Var}(Y)=\phi V_0(\mu)=4$, and $Y\sim N(-1,4)$. Its standard deviation is $2$.

For the probability, standardize the threshold: $P(Y>1)=P((Y-(-1))/2>(1-(-1))/2)=P(Z>1)$. Therefore $P(Y>1)=1-\Phi(1)\approx1-0.8413=0.1587$. A negative mean is valid here because the Gaussian response has real-valued support; power zero does not impose a positive-mean restriction.

Worked Example: Evaluating a Gaussian Likelihood

For three independent responses $y_1=1$, $y_2=2$, and $y_3=3$, evaluate their joint density at the power-zero parameters $\mu=2$, $\phi=1$.

Each response has density $f(y;2,1)=(2\pi)^{-1/2}\exp\{-(y-2)^2/2\}$. By independence, the joint likelihood at these parameters is

L(2) =\prod_{i=1}^3 f(y_i;2,1) =(2\pi)^{-3/2} \exp\left\{-\frac{(1-2)^2+(2-2)^2+(3-2)^2}{2}\right\}.

The squared deviations are $1$, $0$, and $1$, so their sum is $2$. The exponent is therefore $-2/2=-1$, giving $L(2)=(2\pi)^{-3/2}e^{-1}\approx0.02336$. This is a joint density value, not the probability of observing these three exact values. The calculation illustrates that the power-zero likelihood is the ordinary Gaussian likelihood with the Tweedie parameters inserted as the normal mean and variance.

What the Constant Variance Means in a GLM

For a response $Y_i$ with fitted mean $\mu_i$, the power-zero variance relation is $\operatorname{Var}(Y_i)=\phi V_0(\mu_i)=\phi$. Thus changing the fitted mean does not change the response variance under this model. This is the constant-variance feature of the Gaussian response, in contrast to power cases where the variance depends on the mean.

In the canonical parameterization, $\theta_i=\mu_i$. Therefore a model that specifies the canonical parameter through a linear predictor uses the identity relationship between that predictor and the mean. This observation is about the canonical parameter correspondence; it does not require every Gaussian regression model to use only that link.

The distinction between variance and standard deviation matters in interpretation and calculation. If $\phi=4$, the model variance is $4$, while the standard deviation used to standardize a response is $\sqrt{\phi}=2$. Treating $\phi$ as the standard deviation would incorrectly give variance $16$.

Common Mistakes

Reading $p=0$ as zero variance. The exponent is zero; the variance function is $V_0(\mu)=1$, so the response variance is $\phi$, not zero.
Assuming the mean must be positive. At power zero the variance function is defined as the constant one for every real $\mu$. The Gaussian case allows negative, zero, and positive means.
Calling $\phi$ the standard deviation. The normal parameterization here is $N(\mu,\phi)$, where the second argument is variance. Standard deviation is $\sqrt{\phi}$.
Checking only the first two moments to identify a distribution. Many distributions can share a mean and variance. The cumulant-generating-function derivation, or equivalently the Gaussian density, establishes the full distributional result.
Rounding a probability by subtracting rounded CDF values. For example, $\Phi(1)-\Phi(-1)$ is approximately $0.682689$, which rounds to $0.6827$. Subtracting the separately rounded values $0.8413-0.1587$ gives $0.6826$, an inaccurate four-decimal rounding of the probability.

Key takeaway: At power zero, $V_0(\mu)=1$, $\theta=\mu$, and $\kappa(\theta)=\theta^2/2$. The resulting response is $N(\mu,\phi)$: its mean is the Tweedie mean and its variance is the Tweedie dispersion.

Check Your Understanding

Use the power-zero parameterization and show the steps in any calculations.

If $\mu=6$ and $\phi=9$, state the normal distribution, its standard deviation, and its variance.
Why is a mean of $-2$ admissible at power zero?
Starting from $\kappa(\theta)=\theta^2/2$, derive the cumulant-generating function of the response.
If $Y\sim N(3,4)$, calculate $P(Y\leq5)$ by standardizing.
Explain why the first two moments alone do not establish that a response is Gaussian.