Q55. Gamma GLM for reaction times

Topic: Underdispersion, positive responses, and quasi-likelihood · more questions on this topic
‹ Q47 C23 ›

[6 marks]

A researcher measures the time Y_i>0 taken for a chemical reaction to complete at temperature x_i. The reaction times are positively skewed, and their variability appears to increase as the mean reaction time increases. They fit

glm(time ~ temperature, family = Gamma(link = "log"))

(a) Write down explicitly the statistical model corresponding to this R code. Include:

(b) the conditional distribution of Y_i;

(c) the mean \mu_i;

(d) the variance of Y_i;

(e) the link between \mu_i and the linear predictor.

[2 marks]

(f) Suppose the fitted coefficient of temperature is \hat\beta_1=-0.08. Interpret this coefficient on the mean response scale. [2 marks]

(g) Why might a Gamma response distribution be more sensible here than a normal response distribution? [1 mark]

(h) Under this model, how does \operatorname{Var}(Y_i\mid x_i) change as the conditional mean \mu_i increases? [1 mark]

Worked answer (click to show)

Definitions. A \operatorname{Gamma}(a,b) variable with shape a and rate b has mean a/b and variance a/b^2. Here \mu_i=\mathbb{E}(Y_i\mid x_i) and \phi>0 is the common dispersion. The log link models the mean through \eta_i=\log\mu_i=\beta_0+\beta_1x_i; it does not assert a distribution for \log Y_i. Temperatures are fixed and the responses conditionally independent.

(a) Explicit model statement. Treat the temperatures as fixed. For i=1,\ldots,n, assume conditionally independent responses with \boxed{ Y_i\mid x_i\overset{\mathrm{ind}}{\sim} \operatorname{Gamma}\!\left( \text{shape}=\frac1\phi,\quad \text{rate}=\frac1{\phi\mu_i}\right),\qquad Y_i>0, } where \phi>0 is a common dispersion parameter. The log link gives \boxed{\log\mu_i=\eta_i=\beta_0+\beta_1x_i, \qquad\mu_i=\exp(\beta_0+\beta_1x_i).} Here \eta_i is the linear predictor and \beta_0,\beta_1\in\mathbb R. A Gamma distribution with shape a and rate b has mean a/b and variance a/b^2. Substituting the parameters above gives \boxed{\mathbb{E}[Y_i\mid x_i]=\mu_i,\qquad \operatorname{Var}(Y_i\mid x_i)=\phi\mu_i^2.} Equivalently, the Gamma scale is \phi\mu_i. State the parameterisation: the mean \mu_i is not the Gamma shape parameter.

(b) Temperature effect on the mean response scale. For a one-unit increase in the recorded temperature, \frac{\hat\mu(x+1)}{\hat\mu(x)} =\frac{\exp\{\hat\beta_0+\hat\beta_1(x+1)\}} {\exp(\hat\beta_0+\hat\beta_1x)} =e^{-0.08}\approx0.9231. Thus each additional temperature unit is associated with a fitted mean reaction time that is multiplied by 0.9231, or is approximately \boxed{100(1-e^{-0.08})\%\approx7.69\%\text{ lower}.} This is a proportional change in the expected time, not a decrease of 0.08 time units. An 8% decrease is only a first-order approximation.

(c) Why a Gamma response? Reaction time is positive and continuous. A Gamma response has this support, permits right skewness, and accommodates larger variance at larger means. An ordinary constant-variance normal model is symmetric, allows negative times, and does not capture the stated mean–variance pattern. These features motivate Gamma regression; they do not by themselves prove that it fits adequately.

(d) How the variance changes. For fixed \phi, variance increases quadratically with the mean: \operatorname{Var}(Y_i\mid x_i)=\phi\mu_i^2. Multiplying the mean by k>0 multiplies the variance by k^2; in particular, doubling the mean quadruples the variance. The conditional standard deviation is \sqrt\phi\,\mu_i, so the coefficient of variation is constant at \sqrt\phi.