Normal Responses And Constant Variance Functions

Why the Normal Variance Function Is Constant

For a normal response, the variance does not change as the mean changes, provided the dispersion parameter is held fixed. The exponential-dispersion representation makes that fact especially transparent: the variance function is \(V(\mu)=1\), while the response variance is \(\phi V(\mu)=\phi\). The distinction matters because the variance function describes how variance depends on the mean, whereas dispersion supplies the overall scale.

We derive both quantities from the normal density. In the course’s convention, \(N(\mu,\sigma^2)\) uses its second argument for the variance. We write the normal dispersion parameter as \(\phi>0\), so that a response with mean \(\mu\) and dispersion \(\phi\) has distribution \(N(\mu,\phi)\).

The Normal Exponential-Dispersion Form

The density of \(Y\sim N(\mu,\phi)\), for \(y\in\mathbb R\), is

Expanding the square in the exponent separates the terms involving the response \(y\) and the parameter \(\mu\):

Thus the density can be written in the exponential-dispersion form \(f(y;\theta,\phi)=\exp\{[y\theta-b(\theta)]/\phi+c(y,\phi)\}\). For the normal response, \(\theta=\mu\), \(b(\theta)=\theta^2/2\), and \(c(y,\phi)=-y^2/(2\phi)-\tfrac12\log(2\pi\phi)\).

The support is the whole real line, independent of \(\mu\) and \(\phi\), and \(b(\theta)\) is defined for every real \(\theta\). Differentiating the log-partition function gives \(b'(\theta)=\theta\), so the mean parameter is indeed \(\mu=\theta\). The same simple quadratic form will let us obtain the variance directly.

Deriving the Mean and Variance

A direct route from the exponential-dispersion form to moments is the moment-generating function. For the normal representation, multiplying the density by \(e^{ty}\) adds \(ty\) to the exponent. This is equivalent to replacing \(\theta\) with \(\theta+\phi t\), so long as the resulting parameter is in the family’s domain. Here that domain is all of \(\mathbb R\), and therefore the calculation is valid for every real \(t\).

Derivation. Using the density’s exponential-dispersion form and the fact that it integrates to one at every real value of \(\theta\),

The integral in the third line is one: it is the density normalized at parameter value \(\theta+\phi t\). For the normal response, \(b(\theta)=\theta^2/2\) and \(\theta=\mu\). Expanding the difference in the last line gives

Hence \(M_Y(t)=\exp(\mu t+\phi t^2/2)\). This moment-generating function is finite in a neighborhood of zero, so it can be differentiated there to obtain the first two moments. Writing \(K_Y(t)=\log M_Y(t)\), we have \(K_Y'(0)=E[Y]\) and \(K_Y''(0)=\operatorname{Var}(Y)\). In this case,

This establishes the mean and variance directly from the exponential-family form. The quadratic term in \(b\) is what makes the variance independent of the mean.

Deriving the Constant Variance Function

The variance function records the part of the response variance that varies with the mean parameter; it does not include the dispersion multiplier. For the normal family, the mean-parameter relationship and its inverse are especially simple.

Derivation. From \(b(\theta)=\theta^2/2\), differentiation gives \(b'(\theta)=\theta\) and \(b''(\theta)=1\). The mean parameter is \(\mu=b'(\theta)=\theta\), and the inverse is \(\theta(\mu)=\mu\), which exists for every real mean. Substituting into the definition of the variance function gives

The exponential-dispersion mean-variance law then gives \(\operatorname{Var}(Y)=\phi V(\mu)=\phi\cdot1=\phi\). Thus the variance function is constant across all possible means, while the response variance equals the dispersion parameter. This derivation also agrees with the moment-generating-function calculation above. \(\square\)

Dispersion Sets the Variance Scale

Because \(V(\mu)=1\) regardless of \(\mu\), changing the mean alone does not change the variance in this family. Changing \(\phi\), on the other hand, changes the response variance directly. For instance, \(N(3,1)\) has variance \(1\), whereas \(N(3,16)\) has variance \(16\), despite having the same mean. Their standard deviations are \(1\) and \(4\), respectively.

The parameter \(\phi\) is a variance scale, not a standard-deviation scale: the standard deviation is \(\sqrt{\phi}\). It is also useful to keep the two layers of the variance law visible:

Common Mistakes and Why the Distinction Matters

• Calling the variance function \(\phi\). The normal variance function is \(V(\mu)=1\); the response variance is \(\phi V(\mu)=\phi\). These quantities have different roles.
• Using the standard deviation as the dispersion. If a normal response has standard deviation \(3\), its variance and dispersion are \(9\), not \(3\).
• Assuming that a constant variance function means every normal model has the same variance. The variance is constant with respect to \(\mu\) when \(\phi\) is held fixed. Different dispersions give different response variances.
• Thinking \(V(\mu)=1\) means \(\operatorname{Var}(Y)=1\). That conclusion omits the dispersion multiplier. The variance equals one only when \(\phi=1\).

This distinction is useful whenever a model’s mean and variance are described separately. The mean parameter specifies the response’s location; the variance function describes how the mean relates to variance within the family; and dispersion supplies the scale multiplying that function. For a normal response the relationship is particularly simple: the mean can vary freely, the variance function stays at one, and the dispersion is the response variance.