Normal distribution

For normal distribution, \(Var[y]=\sigma^2\), which has nothing to do with the mean \(\mu\) so

• Dispersion parameter \(\phi = \sigma^2\)

• Variance function \(V(\mu) = 1\)

In other words, for the normal distribution, there is no mean-variance relationship. (Hence the normal-distribution-based assumption of constant variance.)

Binomial distribution

For the binomial distribution for proportions \(y\), \(Var[y]=\frac{\mu(1-\mu)}{m}\), where \(\mu\) is the mean parameter (expected proportion) and \(m\) is the number of trials. So

• Dispersion parameter \(\phi = 1/m\)

• Variance function \(V(\mu) = \mu(1-\mu)\)

Effect of number of trials (\(m\)) on variance: As m increases the variance decreases

Effect of expected proportion (\(\mu\)) on variance: Variance is highest at \(\mu = 0.5\) and then decreases as it gets to 0 or 1.

Poisson distribution

For the Poisson distribution for counts \(y\), \(Var[y]=\mu\), where \(\mu\) is the mean parameter (expected number of counts). So

• Dispersion parameter \(\phi = 1\)

• Variance function \(V(\mu) = \mu\)

Gamma distribution

• Dispersion parameter \(\phi\) (let’s just call it \(\phi\) and forget that it came from 1/scale parameter.)

• Variance function \(V(\mu) = \mu^2\)

The fact that the variance is proportional to the mean squared also means that the standard deviation is proportion to the mean.

Summary table

Note that the dispersion parameters

• must be estimated for the Normal and Gamma distributions

• are known for the Poisson and Binomial distributions

Divide the data into groups

Section 5.3.6, Example 5.9

For a more detailed assessment, we can divide the data into groups according to Eucs. The cut function will make groups. You should try to have the groups be roughly the same size.

nminer <- nminer |> 
  mutate(
    euc_gp = cut(Eucs, breaks = c(-Inf, 4, 11, 15, 19, Inf) + .5)
  )
nminer |> 
  count(euc_gp)

##        euc_gp n
## 1  (-Inf,4.5] 9
## 2  (4.5,11.5] 6
## 3 (11.5,15.5] 5
## 4 (15.5,19.5] 6
## 5 (19.5, Inf] 5

This plot shows the results of the grouping:

ggplot(
  data = nminer, 
  mapping = aes(x = Eucs, y = Minerab, color = euc_gp)
) + 
  geom_jitter()

log variance vs. log mean for groups

Recall that

\[Var[y] = \phi V(\mu)\].

For Poisson, we want \(V(\mu) = \mu^b\) for b = 1. Taking the log of both sides gives us

\[\log(Var[y]) = \log(\phi) + b\log(\mu)\]

Thus, \(b\) is the slope in the simple linear relationship between the group means and group variances:

\[log(group variance) = a + b log(group mean)\]

Calculating the log(mean) and log(variance) of the responses for the observations in each group:

grouped <- nminer |> 
  group_by(euc_gp) |> 
  summarize(
    n = n(),
    logmean = mean(Minerab) |> log(), 
    logvar = var(Minerab) |> log()
  )

grouped

## # A tibble: 5 × 4
##   euc_gp          n logmean logvar
##   <fct>       <int>   <dbl>  <dbl>
## 1 (-Inf,4.5]      9  -2.20  -2.20 
## 2 (4.5,11.5]      6  -0.693  0.405
## 3 (11.5,15.5]     5   1.34   2.42 
## 4 (15.5,19.5]     6   1.47   2.06 
## 5 (19.5, Inf]     5   1.95   3.88

Plotting log(variance) vs. log(mean):

ggplot(
  data = grouped, 
  aes(x = logmean, y = logvar)
) + 
  geom_point() + 
  geom_smooth(method = "lm", se = FALSE)

And fitting the linear model with \(n\), number of observations per group, as weights.

fit <- lm(logvar ~ logmean, data = grouped, weights = n)
tidy(fit, conf.int = TRUE)

## # A tibble: 2 × 7
##   term        estimate std.error statistic p.value conf.low conf.high
##   <chr>          <dbl>     <dbl>     <dbl>   <dbl>    <dbl>     <dbl>
## 1 (Intercept)    0.803     0.250      3.21 0.0488   0.00781      1.60
## 2 logmean        1.30      0.149      8.69 0.00320  0.821        1.77

This estimates \(b\approx 1.3\), which supports the \(V(\mu) = \mu^1\) of the Poisson distribution. (There is no evidence against it based on the confidence interval.)