13  Other Response Distributions

Agresti (2015), sections 1.1 and 4.1-4.4.

Logistic regression is just one type of generalized linear model (GLM). Again, generalized linear models have two parts: a systematic part, linking the mean of $$Y$$ to some function of $$\beta\T X$$, and a random part describing the variation of $$Y$$ around that mean.

Specifically, the systematic part is of the form $g(\E[Y \mid X = x]) = \beta\T X,$ where $$g$$ is known as the link function. For example, in logistic regression, $$g$$ was the logit function. The random part gives the distribution of $$Y$$ as an exponential family distribution, as we will describe below—for now it suffices to know that we can only use certain distributions, not any arbitrary distribution.

We use a generalized linear model when $$Y$$ takes on limited values or is known to be non-normal. We used logistic regression, for instance, for problems where $$Y \in \{0, 1\}$$; in other settings, when $$Y \in \{0, 1, 2, \dots\}$$, it might be reasonable to model $$Y$$ has being Poisson-distributed, and we can use a GLM for that too.

13.1 Exponential families

Generalized linear models work when the random part, the response distribution, is an exponential family distribution, which we will define momentarily. The exponential family is not magic, and there is nothing stopping you from creating a new kind of model that uses a different distribution; the key is that exponential family distributions all share useful properties that help us fit GLMs and do inference.

Definition 13.1 (Exponential dispersion family) The exponential dispersion family consists of distributions whose probability density or probability mass function can be written in the form $f(x; \theta, \phi) = \exp\left(\frac{x \theta - b(\theta)}{a(\phi)} + c(x, \phi)\right),$ where $$\phi$$ is the dispersion parameter and $$\theta$$ is the natural parameter. Different distributions in the family can be created by choosing different functions for $$a$$, $$b$$, and $$c$$.

There are many ways to rearrange and rewrite this definition that you may encounter in other books and sources. We will follow Agresti (2015), section 4.1, for our formulation.

Many useful distributions are exponential family distributions.

Example 13.1 (Normal distribution) If $$X \sim \normal(\mu, \sigma^2)$$, it has the probability density function \begin{align*} f(x; \mu, \sigma^2) &= \frac{1}{\sqrt{2 \pi} \sigma} \exp\left(- \frac{(x - \mu)^2}{2 \sigma^2} \right)\\ &= \exp\left( \frac{x \mu - \mu^2 / 2}{\sigma^2} - \frac{\log(2 \pi \sigma^2)}{2} - \frac{x^2}{2 \sigma^2}\right). \end{align*} This is an exponential dispersion family distribution with natural parameter $$\theta = \mu$$, dispersion parameter $$\phi = \sigma^2$$, and \begin{align*} a(\phi) &= \sigma^2 \\ b(\theta) &= \frac{\mu^2}{2} \\ c(x, \phi) &= - \frac{\log(2 \pi \sigma^2)}{2} - \frac{x^2}{2 \sigma^2}. \end{align*} Notice this definition treats $$\sigma^2$$ as fixed and known, and only $$\mu$$ is a parameter to be estimated.

Example 13.2 (Poisson distribution) If $$X \sim \poisson(\lambda)$$, it has the probability mass function \begin{align*} f(x; \lambda) &= \frac{e^{-\lambda} \lambda^x}{x!}\\ &= \exp\left(x \log (\lambda) - \lambda - \log(x!)\right). \end{align*} This is an exponential dispersion family distribution with natural parameter $$\theta = \log \lambda$$, dispersion parameter $$\phi = 1$$, and \begin{align*} a(\phi) &= 1 \\ b(\theta) &= \exp(\theta) = \lambda \\ c(x, \phi) &= - \log(x!). \end{align*}

Example 13.3 (Binomial distribution) If $$nX \sim \binomial(n, p)$$, so that $$x$$ is the sample proportion of successes and $$n$$ is fixed, $$X$$ has the probability mass function \begin{align*} f(x; n, p) &= \binom{n}{nx} p^{nx} (1 - p)^{n - nx} \\ &= \exp\left( \frac{x \theta - \log(1 + \exp(\theta))}{1 / n} + \log \binom{n}{nx} \right). \end{align*} This is an exponential dispersion family distribution with natural parameter $$\theta = \logit(p) = \log(p/ (1 - p))$$, dispersion parameter $$\phi = 1/n$$, and \begin{align*} a(\phi) &= 1 / n\\ b(\theta) &= \log(1 + \exp(\theta))\\ c(x, \phi) &= \log \binom{n}{nx}. \end{align*}

The above examples cover three of the most common distributions we will use, but there are many other examples of exponential dispersion family distributions.

Exercise 13.1 (Gamma distribution) Show that if $$X$$ is gamma, it can be written in exponential dispersion form. Provide the functions $$a$$, $$b$$, and $$c$$, and identify the natural parameter. TODO more explicit

Exponential family distributions share some useful properties that we will exploit when defining generalized linear models.

Theorem 13.1 (Exponential family properties) If $$X$$ is drawn from an exponential dispersion family distribution, as defined in Definition 13.1, then: \begin{align*} \E[X] &= b'(\theta)\\ \var(X) &= b''(\theta) a(\phi). \end{align*}

Proof. We can write the log-likelihood function $$\ell(\theta)$$ as $\ell(\theta) = \frac{x \theta - b(\theta)}{a(\phi)} + c(x, \phi).$ Consequently, \begin{align*} \ell'(\theta) &= \frac{x - b'(\theta)}{a(\phi)} \\ \ell''(\theta) &= - \frac{b''(\theta)}{a(\phi)}, \end{align*} where the prime represents derivatives with respect to $$\theta$$.

TODO

Now we must connect the exponential family distribution to the covariates to produce a regression model. This is the job of the link function $$g$$. In principle, since we are modeling $$g(\E[Y \mid X = x]) = \beta\T x$$, we can use any link function such that $$g^{-1}(x)$$ is always in the domain of $$Y$$ for the chosen response distribution. But certain link functions will make the mathematics easier.

For example, if $$nX \sim \binomial(n, p)$$ as in Example 13.3, then $$\E[X] = p$$ and $$\theta = \logit(p)$$, and so $$g(p) = \logit(p)$$ is the canonical link function. Table 13.1 gives the canonical link function for the most commonly used exponential family distributions.

Table 13.1: Exponential family distributions, the range of the response variable $$Y$$, the canonical link function, and the mean $$\E[Y \mid X = x]$$ in terms of the linear predictor in a GLM.
$$Y \sim \normal(\mu, \sigma^2)$$ $$(-\infty, \infty)$$ $$g(x) = x$$ $$\beta\T X$$
$$Y \sim \bernoulli(p)$$ $$[0, 1]$$ $$g(x) = \logit(x)$$ $$\ilogit(\beta\T X)$$
$$nY \sim \binomial(n, p)$$ $$[0, 1]$$ $$g(x) = \logit(x)$$ $$\ilogit(\beta\T X)$$
$$Y \sim \poisson(\lambda)$$ $$[0, 1, \dots, \infty)$$ $$g(x) = \log(x)$$ $$\exp(\beta\T X)$$

13.2 Fitting a generalized linear model

Just as with logistic regression, we will fit generalized linear models using maximum likelihood. Each observation $$i$$ may have a different mean, so we will write $$\theta_i$$ for the natural parameter. The log-likelihood function is \begin{align*} \ell(\beta) &= \sum_{i=1}^N \log f(y_i; \theta_i, \phi) \\ &= \frac{y_i \theta_i - b(\theta_i)}{a(\phi)} + c(y_i, \phi). \end{align*} When we use the canonical link function, $$\theta_i = \beta\T x_i$$, so this becomes $\ell(\beta) = \sum_{i=1}^N \frac{y_i \beta\T x_i - b(\beta\T x_i)}{a(\phi)} + c(y_i, \phi).$

To maximize the likelihood, we take its gradient with respect to $$\beta$$: \begin{align*} D\ell(\beta) &= \sum_{i=1}^N \frac{y_i x_i - b'(\beta\T x_i) x_i}{a(\phi)} \\ &= \sum_{i=1}^N \frac{(y_i - b'(\beta\T x_i)) x_i}{a(\phi)} \\ &= \frac{\X\T(\Y - P)}{a(\phi)}, \end{align*} where $$P \in \R^n$$ is a vector with entries $$P_i = b'(\beta\T x_i)$$. If we set this equal to zero to try to find a maximum, we will unfortunately find that there is no analytical solution. We can instead proceed numerically, again using Newton’s method (Definition 12.2). We will need the Hessian of the log-likelihood: $D^2 \ell(\beta) = - \X\T W \X,$ where $$W$$ is a diagonal matrix with entries $W_{ii} = b''(\beta\T x_i).$ This should look familiar: in Section 12.2.1, the Newton–Raphson procedure for fitting logistic regression had very similar gradients and Hessians, and could be done by a sequence of iteratively reweighted least squares problems. The same is true here, by exactly the same procedure. The estimator has the same properties and allows the same inference procedures as described in Section 12.4 and Section 12.5. The gradient and Hessian above were derived for the canonical link function, but similar expressions can be obtained for other link functions with extra bookkeeping to keep track of the link function’s derivative.

This simplicity illustrates why we chose to consider only exponential dispersion family distributions: the same procedure can fit a GLM using any of them, merely by knowing the functions $$a$$ and $$b$$ characterizing the distribution. This will also help us prove properties and describe inference procedures that work on the entire class of models, instead of having to derive new results for each model we use.

Let’s now consider some common kinds of GLMs and how they can be used.

13.3 Binomial outcomes

The binomial distribution is a common response distribution whenever outcomes are binary, or whenever we count a certain binary outcome out of a total number of trials. As one special case, $$\binomial(1, p) = \bernoulli(p)$$, so logistic regression is just a GLM with a binomially distributed response with $$n = 1$$. (This is why we set family = binomial() when calling glm().) Often we refer to the binary outcomes as “successes” and “failures”, so a binomial random variable is the number of successes out of a fixed number of trials.

The binomial distribution is suitable when there is a fixed and known total number of trials. The probability mass function shown in Example 13.3, and hence the likelihood function, depends on the number of trial $$n$$, so it must be known. Each observation hence consists of a number of successes and a total number of trials; the number of trials may differ between observations. With $$n$$ fixed, the model relates the success probability $$p$$ to the predictors, so the interpretation of coefficients is much like the interpretation in logistic regression (Section 12.3).

For a binomial response, the logit is the canonical link function, though sometimes the standard normal CDF is used; this is called the “probit” link. The two link functions are very similar in shape, so they tend to produce very similar fitted models, though the logit link has the advantage of easier interpretation in terms of log-odds, odds, and odds-ratios. With either link function, the model is $n_i Y_i \mid X_i = x_i \sim \binomial\left(n_i, g^{-1}(\beta\T x_i)\right),$ where $$g$$ is the link function. The sample proportion $$Y_i / n_i$$ should hence be proportional to $$g^{-1}(\beta\T x_i)$$.

In R, a binomial response variable can be provided as a binary factor (as we saw in logistic regression) or, for $$n > 1$$, a two-column matrix providing the number of successes and the number of failures.

Example 13.4 (Seed germination) TODO get source

This data comes from a factorial experiment on seeds. There are two varieties of seed, 1 and 2, and two types of extract the seeds can be treated with. The goal of the experiment was to figure out which combination of seed and extract leads to the highest rate of germination of the seeds.

A factorial design means the researchers tested each possible combination in (nearly) equal numbers. Each row of the data set represents one group of seeds grown together, getting the same seeds and extract; we have the total number of seeds in that group and the number that successfully germinated.

library(dplyr)

mutate(seed = factor(seed),
extract = factor(extract))

Figure 13.1 shows the germination rate for each combination of seed and extract. It looks like seed 1, extract 2 has the highest success rate, but we can use a model to explore the differences in more detail.

For each row $$i$$ of data, let $$N_i$$ be the number of seeds tested, $$Y_i$$ the number that germinated, and $$X_i$$ be the vector of covariates (seed and extract type). A binomial model would make sense here, and if we use the default link function, our model would be $Y_i \sim \binomial(N_i, \logit(\beta\T X_i)).$

Let’s explore whether the effect of extract is identical for both seed varieties, or if instead the extract’s effect differs by seed type. To do so, we fit two models, one with an interaction and one without.

seed_fit_1 <- glm(cbind(germinated, total - germinated) ~
extract + seed,
data = seeds, family = binomial())
seed_fit_2 <- glm(cbind(germinated, total - germinated) ~
extract * seed,
data = seeds, family = binomial())

We can again use a deviance test (Theorem 12.3) to determine if the interaction is statistically significant here.

seed_test <- anova(seed_fit_1, seed_fit_2, test = "Chisq")
seed_test
Analysis of Deviance Table

Model 1: cbind(germinated, total - germinated) ~ extract + seed
Model 2: cbind(germinated, total - germinated) ~ extract * seed
Resid. Df Resid. Dev Df Deviance Pr(>Chi)
1        18     39.686
2        17     33.278  1   6.4081  0.01136 *
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

We conclude that the extract effect differs by seed type, $$\chi^2(1) = 6.41$$, $$p = 0.011$$.

The coefficients of the interaction model are shown in Table 13.2. If we were interested obtaining the probability of germination for each seed/extract combination, and hence find the best-performing combination, these coefficients are inconvenient: we’d have to manually calculate the predicted probabilities and their standard errors. Instead we can make the predictions and let R do the work:

Table 13.2: Interaction model for the seed data.
Term Estimate SE z p
(Intercept) -0.558 0.126 -4.429 0.000
extract2 1.318 0.177 7.428 0.000
seed2 0.146 0.223 0.654 0.513
extract2:seed2 -0.778 0.306 -2.539 0.011
xs <- expand.grid(seed = factor(1:2), extract = factor(1:2))
predictions <- predict(seed_fit_2, newdata = xs,
type = "response", se.fit = TRUE)

The predictions are shown in Table 13.3. We see that seed 1, extract 2 has the highest predicted probability. These standard errors are calculated by calculating the standard error of $$\hat \beta\T x$$, then using the delta method to transform this to a standard error for the probability; to obtain confidence intervals, we could calculate confidence intervals on the link scale and then transform them to the probability scale.

Table 13.3: Predicted germination probabilities and their standard errors.
Seed Extract Germination rate SE
1 1 0.364 0.029
2 1 0.398 0.044
1 2 0.681 0.027
2 2 0.532 0.042

Exercise 13.2 (Seed germination diagnostics) Figure 13.2 shows the residuals from the interaction model fit in Example 13.4.

1. Why are the residuals grouped in four vertical lines?
2. Normally you might use this plot to determine if the modeled mean function is adequate, or if the model needs to include additional terms or flexibility. Explain why that is not necessary or possible here.

13.4 Poisson regression

If a certain event occurs with a fixed rate, and the events are independent (so that the occurrence of one event does not make another more or less likely), then the count of events over a fixed period of time will be Poisson-distributed. This makes Poisson GLMs well-suited for response variables that are counts.

As shown in Table 13.1, the canonical link for a Poisson GLM is the log link. That is, a canonical Poisson GLM assumes the relationship $\log(\E[Y \mid X = x]) = \beta\T x,$ or equivalently, $\E[Y \mid X = x] = \exp(\beta\T x).$ And because the variance of a Poisson-distributed random variable increases with its mean, the variance of $$Y$$ depends on $$X$$. Figure 13.3 illustrates the relationship for a Poisson GLM where $$\log(\E[Y \mid X = x]) = x$$. Notice that $$Y$$ only takes on integer values and has a nonlinear relationship with $$X$$, but when we group $$Y$$ into bins and take the average in each bin, the log of that average is linearly related to $$X$$.

TODO Get a starting example that doesn’t require offsets, so there’s less going on at once

Example 13.5 (Smoking and heart disease) This data comes from a famous study on the health risks of smoking . This was a prospective cohort study, so it followed a number of smokers and non-smokers (who were all male British doctors recruited into the study) for many years and observed their death rates and causes of death. Early results from the study provided some of the earliest evidence that smoking is associated with lung cancer. This data is from later in the study. Each row of the data represents people in a ten-year age range, and records the number of people in that age range who died from coronary heart disease. The full dataset is shown in Table 13.4. (TODO which follow-up period is this data from?)

Table 13.4: Smoking data, for smokers and non-spokers in four age buckets. Note that each age bucket represents a ten-year range, e.g. 40 really represents those age 35–44.
Age Deaths Person-years Smoke?
40 32 52407 yes
40 2 18790 no
50 104 43248 yes
50 12 10673 no
60 206 28612 yes
60 28 5710 no
70 186 12663 yes
70 28 2585 no
80 102 5317 yes
80 31 1462 no

Crucially, note the person-years column. One person-year represents one person who is observed in the study in one year; ten person-years could be ten people observed for one year or one person observed for ten years. Each age bucket has a different number of person-years because different numbers of people in the study were in each age range and were observed for different lengths of time.

Let $$Y$$ be the number of deaths, $$P$$ be the number of person-years of people observed in that bucket, and $$X$$ be a vector of the covariates (age and smoking status). We are interested in the death rate, so we would like a model of $$Y / P$$ as some function of $$X$$. Figure 13.4 shows this death rate plotted against age and smoking status, illustrating the relationship.

We could choose to model the number of deaths as approximately Poisson, given the death rate: $Y \sim \poisson(P f(\beta\T X)),$ so that $$f(\beta\T X)$$ represents the rate. If we choose a Poisson GLM with the log link, we could write this as $Y \sim \poisson(\exp(\beta\T X + \log P)).$ Here $$P$$ is an offset: a term in our model that is fixed to have coefficient 1, rather than having a slope that is estimated. The choice of a Poisson distribution is an approximation here because there is an upper bound to the number of deaths in any age bucket based on the number of people in the study in that age bucket—but as long as the death rate is low, the Poisson distribution will assign very little probability to impossible death numbers.

If this model were true, we’d expect a linear relationship between $$\log(Y/P)$$ and $$X$$, but the right panel of Figure 13.4 shows this is not linear. We can instead consider a polynomial model. Because the gap between smokers and non-smokers even seems to reverse for the oldest group, we’ll also include an interaction between smoking and age. We use the offset argument to tell R to include the term with coefficient 1.

smoke_fit <- glm(deaths ~ (age + I(age^2)) * smoke,
offset = log(py),
data = smokers, family = poisson(link = "log"))
Data points may overlap. Use the jitter argument to add some amount of
random variation to the location of data points and avoid overplotting.

Figure 13.5 gives the effect plot. (Note that ggeffect() sets the offset to its mean by default, meaning it used the average of $$\log(P)$$; in this case that corresponds to 10,657 person-years.) The fit seems good, though with essentially only ten observations, detailed goodness-of-fit checks may not be useful.

Now suppose we want to compare smokers and non-smokers at age 40, calculating a risk ratio: the ratio of risk of death for smokers versus non-smokers. A simple way would be to make a prediction and take the ratio. To predict the rate of death, we can predict the mean number of deaths for one person-year:

new_smokers <- data.frame(smoke = c("no", "yes"),
age = c(40, 40),
py = c(1, 1))
preds <- predict(smoke_fit, newdata = new_smokers, type = "response")
preds[2] / preds[1]
       2
3.927567 

But it’s not easy to make a confidence interval for this. The standard errors are not independent. Let’s more closely examine the regression function—specifically, the ratio between smokers and non-smokers when we set age = 40. The model terms are, in order,

names(coef(smoke_fit))
[1] "(Intercept)"       "age"               "I(age^2)"
[4] "smokeyes"          "age:smokeyes"      "I(age^2):smokeyes"

We can hence write the ratio of predictions as: \begin{align*} \frac{\mathbb{E}[Y \mid \text{age} = 40, \text{smoker}]}{\mathbb{E}[Y \mid \text{age} = 40, \text{non-smoker}]} &= \frac{\exp(\hat \beta_0 + 40 \hat \beta_1 + 1600 \hat \beta_2 + \hat \beta_3 + 40 \hat \beta_4 + 1600 \hat \beta_5)} {\exp(\hat \beta_0 + 40 \hat \beta_1 + 1600 \hat \beta_2)} \\ &= \exp(\hat \beta_3 + 40 \hat \beta_4 + 1600 \hat \beta_5) \end{align*}

Hence the risk ratio can also be calculated as:

coef_vec = c(0, 0, 0, 1, 40, 1600)

exp(sum(coef_vec * coef(smoke_fit)))
[1] 3.927567

We can further create a Wald confidence interval:

se <- sqrt(t(coef_vec) %*% vcov(smoke_fit) %*% coef_vec)[1,1]

bounds <- sum(coef_vec * coef(smoke_fit)) + c(-2, 2) * se
exp(bounds)
[1]  1.49177 10.34059

This is quite a large confidence interval, demonstrating the difficulty in estimating risk ratios when the underlying rate of events is so low.

13.5 Interpretation and inference

Because logistic regression is just a kind of generalized linear model, all the inference techniques we discussed in Section 12.5 apply here. We can conduct Wald tests for coefficients (Theorem 12.2), form confidence intervals for coefficients using the profile likelihood (as in Definition 12.9), and conduct deviance tests to compare nested models (Theorem 12.3).

The key is to keep track of the link function and scales. Let’s consider some common inference tasks.

• Interpreting parameters. In a GLM, the coefficients $$\beta$$ relate $$X$$ to $$g(\E[Y \mid X = x])$$, so interpretation of $$\beta$$ depends on the link function $$g$$. In logistic regression and binomial GLMs, that is the logit link and we must convert between log-odds and odds ratios (Section 12.3). In a Poisson GLM with the log link, $$\beta$$ gives the relationship between $$X$$ and the log of the rate, so we must exponentiate it, as we saw in Example 13.5.
• Confidence intervals for parameters. Generally the profile likelihood confidence intervals for $$\beta$$ are best; these are produced automatically by confint(). Again, you can then transform the confidence interval endpoints using the inverse link function.
• Making predictions. We can use the predict() function, but note that the predict.glm method defaults to predicting on the link scale, i.e. predicting $$\hat \beta\T X$$. Set type = "response" to have R automatically use the inverse link function to get predictions on the response scale.
• Confidence intervals for the mean. If we want a confidence interval for $$\E[Y \mid X = x_0]$$, for a particular $$x_0$$ of interest, we can get a confidence interval for $$\beta\T x_0$$ first. Then we can use the inverse link to transform back. See Example 12.9.

The key, as always, is in interpretation. To interpret a coefficient or a hypothesis test, we need to understand the response distribution and the link function. We must always know if we’re working on the link scale or the response scale, and we must keep track of the appropriate link and inverse link functions for converting back and forth as needed.

TODO mention deviance tests and the meaning of the saturated model, giving an example like Example 12.6

13.6 Overdispersion

It is nice the generalized linear models allow us to model responses as coming from any exponential dispersion family. This allows the variance of $$Y$$ to depend on $$X$$: in a Poisson GLM, for instance, the variance is proportional to the mean. This can account for the variance patterns in real data.

But sometimes there is overdispersion: there is more variance in $$Y$$ than the response distribution would predict. This can happen for several reasons:

• Our $$X$$ is not sufficient. The fundamental assumption of a GLM is that each observation $$Y$$ is independent and identically distributed given $$X$$; but maybe there are other factors associated with $$\E[Y]$$ that we did not observe. For a fixed value of $$X$$, these other factors may still vary and cause additional variation in the observed values of $$Y$$.
• There may be correlation we did not account for. For instance, a $$\binomial(n, p)$$ distribution assumes there are $$n$$ independent trials—but what if the trials are dependent, and success in one is correlated with increased success rates in the others?

It would be useful if we could fit a more flexible model that allows more variance. We can, by taking advantage of a useful result.

Theorem 13.2 (Mean/variance form of the likelihood equation) For a generalized linear model with canonical link function $$g$$, the gradient of the log-likelihood can be written as $D\ell(\beta) = \sum_{i=1}^N \frac{(y_i - \E[Y_i \mid X_i = x_i]) x_{ij}}{\var(Y_i \mid X_i = x_i)} Dg^{-1}(\beta\T x_i).$ The maximum likelihood estimate is found by setting this equal to zero, so the MLE depends only on the particular exponential dispersion distribution through its mean and variance.

Proof. In Section 13.2, we showed that $D\ell(\beta) = \sum_{i=1}^N \frac{(y_i - b'(\beta\T x_i)) x_i}{a(\phi)},$ while from Theorem 13.1 we know that if $$Y$$ has an exponential dispersion family distribution given $$X$$, then \begin{align*} \E[Y \mid X = x] &= b'(\theta_i)\\ \var(Y \mid X= x) &= b''(\theta_i) a(\phi), \end{align*} so $$a(\phi) = \var(Y \mid X = x) / b''(\theta_i)$$. If $$g$$ is the canonical link (Definition 13.2), then $$\beta\T x_i = \theta_i$$. Consequently, we can write the gradient as \begin{align*} D\ell(\beta) &= \sum_{i=1}^N \frac{(y_i - \E[Y_i \mid X_i = x_i]) x_{ij}}{a(\phi)}\\ &= \sum_{i=1}^N \frac{(y_i - \E[Y_i \mid X_i = x_i]) x_{ij}}{\var(Y_i \mid X_i = x_i)} b''(\beta\T x_i). \end{align*} Finally, note that since for the canonical link $b'(\beta\T x_i) = \E[Y_i \mid X_i = x_i] = g^{-1}(\beta\T x_i),$ we have $b''(\beta\T x_i) = D g^{-1}(\beta\T x_i).$ This completes the proof.

This result suggests a trick. For any particular GLM with a choice of exponential dispersion distribution, we can write the expectation and variance of $$Y$$ in terms of $$X$$. Suppose that $\var(Y_i \mid X_i = x_i) = v^*(x_i)$ for some function $$v^*$$ for the chosen distribution. But suppose that our data is overdispersed. We might instead assume that $\var(Y_i \mid X_i = x_i) = \phi v^*(x_i),$ for some constant $$\phi$$. If $$\phi > 1$$, the data is overdispersed.

This is no longer a model based on a specific distribution function; we are now fudging it and using only the mean and variance of the distribution, and allowing the variance to be different from what an exponential dispersion family distribution would have. Estimation using this approach is hence known as quasi-likelihood estimation, because it only sort of is using a proper likelihood function.

When we substitute $$\phi v^*(x_i)$$ into the likelihood equation given by Theorem 13.2, then set the gradient to zero to find the MLE, $$\phi$$ drops out: it is a constant in every term, so we can multiply through by $$\phi$$ and remove it. We hence obtain the same MLE $$\hat \beta$$.

If we further obtain the Hessian of the log-likelihood, as in Section 13.2, we will see that the matrix $$W$$ does depend on $$\phi$$. From Theorem 12.1, this means $$\var(\hat \beta)$$ depends on $$\phi$$. Specifically, $$\var(\hat \beta)$$ is $$\phi$$ times larger than the standard GLM.

Generally we do not know $$\phi$$ in advance: we suspect our data is overdispersed, but nature was not kind enough to specify the exact value of $$\phi$$ corresponding to the true distribution. We can use a useful property to estimate it.

Theorem 13.3 (Generalized Pearson statistic) Consider data drawn from an exponential dispersion GLM. Then $X^2 = \sum_{i=1}^N \frac{(y_i - \E[Y_i \mid X_i = x_i])^2}{\var(Y_i \mid X_i = x_i)} \convd \chi^2_{N - p}.$

Proof. A heuristic proof: the $$y$$s are conditionally independent, so this is a sum of squared random variables, each centered at 0 and divided by its variance. If each were standard normal, this would be exactly $$\chi^2$$; because they are not necessarily normal, the sum is approximately $$\chi^2$$.

When using an overdispersed GLM where $$\var(Y_i \mid X_i = x_i) = \phi v^*(x_i)$$, we can apply the method of moments to estimate $$\phi$$. The expectation of $$X^2$$ is $$N - p$$, so we set $$X^2 = N - p$$ and solve for $$\phi$$. We obtain $\hat \phi = \frac{1}{N - p} \sum_{i=1}^N \frac{(y_i - \E[Y_i \mid X_i = x_i])^2}{v^*(x_i)}.$

We can finally summarize the quasi-likelihood approach. Obtain the MLE $$\hat \beta$$ using standard iteratively reweighted least squares, or any other approach; then estimate $$\hat \phi$$ as above, and use it to scale the estimated $$\var(\hat \beta)$$.

Quasi-likelihood methods are commonly used for Poisson and binomial GLMs when the data is ill-behaved. R supports quasi-likelihood estimation for both by using the quasibinomial() and quasipoisson() families in glm().

Example 13.6 (Overdispersed seeds) Let’s revisit the seed data from Example 13.4 to see how to fit and interpret a quasi-binomial model. We can use quasibinomial() to fit the model:

seed_quasi <- glm(cbind(germinated, total - germinated) ~
extract * seed,
data = seeds, family = quasibinomial())

A comparison of this model’s estimates against seed_fit_2 is shown in Table 13.5. Notice that the estimates are identical; only the standard errors differ, by a constant factor related to $$\phi$$. Confidence intervals and tests would be similarly affected.

Table 13.5:

Estimates from a quasibinomial fit to the seeds data.

Term MLE Quasibinomial
Estimate SE Estimate SE
Intercept −0.558 0.126 −0.558 0.172
extract2 1.318 0.177 1.318 0.242
seed2 0.146 0.223 0.146 0.305
extract2:seed2 −0.778 0.306 −0.778 0.418

13.7 Exercises

Exercise 13.3 (Island size and extinction rates) The Sleuth3 package contains the data frame case2101, summarizing data originally collected by Väisänen and Järvinen (1977). The data is based on two censuses of nesting birds taken on the Krunnit Islands archipelago in 1949 and 1959. Each row is one island in the archipelago, and we have the number of bird species present in 1949 (AtRisk) and the number of those that were extinct from the island by 1959 (Extinct). As a covariate, we have the area of the island (in hectares1); many of the islands are quite small. Theories of biodiversity would predict that that extinction rates (Extinct / AtRisk) are higher in small, isolated communities, and lower in larger communities.

1. Conduct an exploratory analysis of the data. Plot extinction rate against area. Also plot the log-odds of extinction against area and the log of area. Comment on the trend in the data.

2. Describe which GLM you will choose to model extinction rate versus island area (including response distribution and the typical link function). Fill in the blanks: your chosen GLM assumes a linear relationship between the [blank] of the extinction rate and [blank]. Based on that and the plots above, what model and transformation would you use?

3. Fit your model. In a sentence (in APA format), describe the observed relationship between extinction rate and area. Your sentence should give the odds ratio for extinction associated with an increase in area, and its confidence interval. If you used area on a log scale, describe your odds ratio in terms of area, not log area.

Hint: A one-unit increase in $$\log_{10}(\text{area})$$ corresponds to what change in area?

4. Plot the deviance residuals against your predictor, including a smoothed trend line. Comment on the shape of the residuals. Do you observe any signs of model misspecification?

Exercise 13.4 (Howard the Duck) Howard the Duck is conducting a study to determine whether eating bread is bad for ducks. (The city council wants to know if they should buy “Please Do Not Feed the Ducks” signs.)

Howard enrolls $$N$$ ducks in his study and observes them. The outcome variable is how many times each duck was hospitalized during the study period. Covariates include:

• The duck’s age and weight at the start of the study
• How much exercise the duck gets, in miles flown per week
• The fraction of time the duck spends at the public park, where people often throw bread to the ducks.

Howard would like to understand the relationship between being at the public park and the number of hospitalizations for any cause. Howard doesn’t care when the hospitalizations happen, just how many there are.

1. What type of generalized linear model should be used here? Write out its model for $$Y$$ explicitly (e.g. “$$Y$$ is Normal with mean given by $$\X\beta$$” in an ordinary linear model).

2. Suppose that not all ducks are in the study for the same length of time. Some ducks were observed for only a few months, but others were observed for an entire year. Those ducks could obviously be hospitalized more times. How should this be accounted for in Howard’s model? Write out explicitly what you would change in the model.

3. Howard fit the model from part (a) and obtained these results:

Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -0.0820563  0.2145272  -0.382    0.702
fraction     0.8714525  0.1645624   5.296 1.19e-07 ***
exercise    -0.0004261  0.0017543  -0.243    0.808
age          0.0865562  0.0167153   5.178 2.24e-07 ***
weight      -0.0491136  0.0474717  -1.035    0.301
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter taken to be 1)

Null deviance: 147.936  on 99  degrees of freedom
Residual deviance:  81.468  on 95  degrees of freedom
AIC: 402.62

Number of Fisher Scoring iterations: 4

In a sentence or two, give your conclusions to the city council: do ducks that spend a larger fraction of their time at the public park have worse health? Cite appropriate test statistics and $$p$$ values to justify your conclusion in APA format (Chapter 25). Also, state how the rate of hospitalizations changes as the fraction increases by 0.1.

4. Of course, this was an observational study. Sketch a causal diagram of the variables shown, and suggest a variable not included in the study that could have confounded the results.

Exercise 13.5 (Insect attraction to light) Insects can be attracted to bright artificial lights at night. As cities switched from old-fashioned sodium-vapor street lights (with their characteristic orange glow) to more energy-efficient LED lights, Longcore et al. (2015) wanted to know whether different light sources attract more or fewer insects of different species. They conducted a field experiment using light traps: each trap contained a light bulb and captured insects that were attracted to it.

The experiment used six light traps. Five had different types of light bulb, while one used no light bulb, as a control. The traps were set up each evening at sunset and checked in the morning after sunrise, and the insects in each trap were identified and counted. This process was run 32 times: 16 at an urban site (the UCLA Botanical Garden), 8 times at a rural field station (La Kretz), and 8 times at the rural Stunt Ranch Reserve. The repetitions were done at varying moon phases, since the presence of a bright moon in the sky might affect the results.

The full dataset is available at https://cmustatistics.github.io/data-repository/biology/bug-attraction.html.

1. We are interested in the total number of insects collected in each trap (the Total variable) and how it relates to the light bulb type (Light Type). We also know that the location (Location), moon phase (% Moon Visible), mean night temperature (Mean Temp), and mean humidity (Mean Humidity) might affect insect activity. Conduct an EDA of these variables to determine what relationships the predictors may have with the response.

2. Consider the study design. Because six traps were placed out each night, each with a different bulb (or no bulb), the traps were exposed to the same location, moon phase, temperature, and humidity. No bulb type has data from moon phases that are systematically different from the other bulbs, or was used on nights that were warmer than those for the other bulbs, or anything like that. This is known as a balanced design.

Comment on what this design choice means. Can location, moon phase, mean temperature, or mean humidity be confounding variables that would bias our estimates of the effect of bulb type? Do we need to incorporate them into our models to obtain an unbiased estimate of the causal effect of bulb type?

3. Obtain the mean number of insects trapped by each bulb. Display the means in a table. Which bulb attracts the fewest insects?

Note there is one missing count (recorded as NA) that you may need to remove.

4. Select a GLM for the total number of insects collected as a function of bulb type, location, moon phase, temperature, and humidity. Use your EDA to guide your model choice, and produce any necessary diagnostic plots to validate your choice.

Produce a table of coefficients estimated from your final model, including standard errors. Interpret, in words, the meaning of the coefficient for temperature, giving a 95% confidence interval for the size of the effect.

5. Consider a night where the moon is 60% visible, the humidity is 60%, and the mean temperature is 15 °C. (These are roughly the average values in the data.) Produce a table of the predicted number of insects trapped by each bulb type at the La Kretz station on a night like this, giving 95% confidence intervals for each prediction. Does your result match the table from part 3?

1. Ramsey and Schafer (2013), section 21.1.1, claims the units are square kilometers, but the original source makes clear the area is in hectares. There are 100 hectares in a square kilometer, so this is a large difference.↩︎