Variance Function

Study guide: Exam MAS-II

The variance function $V(\mu)$ specifies how the variance of a response depends on its mean. In a Generalized Linear Model it is fixed by the choice of exponential-family distribution. In a Linear Mixed Model the residual variance is constant by default, and a variance function is added when the data say it is not.

$\mathrm{Var}(Y_i) = \frac{\phi\,V(\mu_i)}{w_i}$

\phi is the dispersion parameter and w_i the prior weight (exposure, claim count). The standard choices of V(\mu):
- normal: $1$ (constant variance)
- Poisson: $\mu$
- binomial: $\mu(1 - \mu)$
- gamma: $\mu^2$ (constant coefficient of variation)
- inverse Gaussian: $\mu^3$
- Tweedie with $1 < p < 2$ : $\mu^p$
What it does in the fit. A GLM's estimating equations weight each residual $y_i - \mu_i$ by $w_i/V(\mu_i)$ , along with a factor from the link. Under a gamma model, a $\$3{,}000$ miss on a $\$50{,}000$ claim counts far less than the same miss on a $\$5{,}000$ claim. The Pearson residual $(y_i - \hat{\mu}_i)/\sqrt{V(\hat{\mu}_i)}$ puts all residuals on one scale. A fan shape in a residual plot of such residuals means the wrong $V$ was chosen.
Choosing it. Plot group variances against group means on log–log axes. The slope estimates $p$ in $V(\mu) = \mu^p$ : $0$ for normal, $1$ for Poisson, $2$ for gamma, and between $1$ and $2$ for Tweedie.
In a linear mixed model. The default residual covariance is \mathbf{R}_i = \sigma^2\mathbf{I}: constant variance. It can be relaxed in two ways:
- heterogeneous variances, a separate $\sigma_g^2$ for each level of a factor (a treatment group, a region)
- variance as a function of a covariate or of the fitted mean, e.g. the power form $\mathrm{Var}(\varepsilon_{ij}) = \sigma^2\lvert\mu_{ij}\rvert^{2\delta}$ (the varPower structure in R's nlme)
Testing it. Unlike in a GLM, the LMM's mean–variance link is a modelling choice with its own estimated parameters. The competing models share their fixed effects, so they are compared by a likelihood ratio test on REML fits, or by AIC. The choice feeds the covariance structure, and so the standard errors of every fixed effect.

Same Miss, Different Claims A gamma severity GLM has $\phi = 0.5$ . Claim 1 has $\hat{\mu} = \$5{,}000$ and settles for $\$8{,}000$ . Claim 2 has $\hat{\mu} = \$50{,}000$ and settles for $\$53{,}000$ . Compare the claims' standard deviations and their standardised Pearson residuals.
Answer With $V(\mu) = \mu^2$ , $\mathrm{SD} = \sqrt{\phi}\,\mu = 0.7071\mu$ : $\$3{,}536$ for claim 1 and $\$35{,}355$ for claim 2.
 $\begin{align*} r_1 &= \frac{8{,}000 - 5{,}000}{\sqrt{0.5}\,(5{,}000)} \\ &= 0.849 \\ r_2 &= \frac{53{,}000 - 50{,}000}{\sqrt{0.5}\,(50{,}000)} \\ &= 0.085 \end{align*}$ 
The same $\$3{,}000$ miss is ten times as surprising on the small claim. A normal model with constant variance would have treated the two identically.

Reading a Mean–Variance Plot Pure premium data are grouped into three cells with means $200$ , $400$ and $800$ , and variances $60{,}000$ , $180{,}000$ and $540{,}000$ . Which variance function is indicated?
Answer Each doubling of the mean triples the variance, so on log–log axes:
 $\begin{align*} p &= \frac{\ln 3}{\ln 2} \\ &= 1.585 \end{align*}$ 
$V(\mu) \approx \mu^{1.6}$ lies between Poisson ( $p = 1$ ) and gamma ( $p = 2$ ). That is a Tweedie with $p \approx 1.6$ , the compound Poisson–gamma shape expected of pure premium, with its point mass at zero.

Testing Heterogeneous Residual Variance in an LMM A random-intercept model of loss ratios by insured is fitted by REML twice, with identical fixed effects. With one common residual variance, $-2\ell_R = 1{,}254.6$ . With a separate residual variance for each of three regions, $-2\ell_R = 1{,}241.2$ . Test at $5\%$ , given $\chi^2_{0.05,\,2} = 5.991$ .
Answer The heterogeneous model has $2$ extra variance parameters.
 $\begin{align*} \text{LRT} &= 1{,}254.6 - 1{,}241.2 \\ &= 13.4 \end{align*}$ 
$13.4 > 5.991$ , so reject a common residual variance. The regions differ in volatility, and a constant-variance model would give wrong standard errors for the fixed effects. REML fits are the right basis here because only the variance structure differs between the models. Equal variances is not a boundary hypothesis, so the plain $\chi^2_2$ applies.