本次荷兰代写主要为风险报告相关

Exercise 1 Generalized Linear Models (11 points)
We estimate a Poisson GLM to explain policyholders’ average annual claim frequency from
various rating factors. The car insurance data used to estimate the model have been provided by
the insurance company RIP-OFF BV. We consider the following risk factors:
• Gender [gender]: 1 = female, 2 = male.
• Type of driver [driver]: 1 = normal drivers, 2 = dynamic drivers, 3 = older drivers, 4 =
Sunday drivers.
• Type of car [type]: 1 = small, 2 = big, 3 = middle.
• Number of previous accidents [prev.acc]: 1 = no previous accidents, 2 = one previous
accident, or 3 = more than 2 previous accidents.
Furthermore, the total number of claims filed and the number of policyholders (for each com-
bination of rating factors) are denoted by nCl and Expo, respectively. We get the following R
output:
Call:
glm(formula = nCl ~ as.factor(gender) + as.factor(drivers) +
as.factor(type) + as.factor(prev.acc) + offset(log(Expo)),
family = poisson(link = log))
Deviance Residuals:
Min 1Q Median 3Q Max
-0.8756 -0.1005 -0.0093 0.1146 2.4177
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -1.2071055 0.0266421 -45.308 < 2e-16 ***
as.factor(gender)2 0.0009215 0.0303621 0.030 0.975788
as.factor(driver)2 0.0122349 0.0365654 0.335 0.737924
as.factor(driver)3 -0.1690421 0.0565676 -2.988 0.002805 **
as.factor(driver)4 -0.2180865 0.0629450 -3.465 0.000531 ***
as.factor(type)2 0.0041287 0.0244117 0.169 0.865695
as.factor(type)3 0.0029646 0.0244630 0.121 0.903542
as.factor(prev.acc)2 1.1356274 0.0530632 21.401 < 2e-16 ***

as.factor(prev.acc)3 1.2304196 0.0416330 29.554 < 2e-16 ***

Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1
(Dispersion parameter for poisson family taken to be 1)
Null deviance: 2599.0 on 999 degrees of freedom
Residual deviance: 44.5 on 991 degrees of freedom
AIC: 4173.6
Number of Fisher Scoring iterations: 4
In our R code, we use the as.factor command to make dummy variables from the categorical
risk factors.

(i) [3 points] Interpret the coefficients of as.factor(driver)3 and as.factor(prev.acc)2.
Detail on sign, magnitude and significance.
(ii) [3 points] Consider two policyholders that differ in exactly three ways. The first is male,
has a big car and had one previous accident, while the second is female, has a small
car and had no previous accidents, ceteris paribus. How much higher or lower is the
expected annual claim frequency of the first policyholder relative to the expected annual
claim frequency of the second one?
(iii) [2 points] We also run the following R command:
glm(formula = nCl ~ as.factor(gender) + as.factor(drivers) +
as.factor(type) + as.factor(prev.acc) + offset(log(Expo)),
family = quasipoisson(link = log))
As part of the resulting output, R tells us:
Dispersion parameter for quasipoisson family taken to be 0.8866867.
Is there underdispersion or overdispersion in this model? Explain your answer by refer-
ring to the variance function.
(iv) [3 points] For the quasi-Poisson model, calculate the mean and variance of the average
annual claim frequency of a female policyholder, who is a normal driver in a small car
and with no previous accidents. In your calculations, assume that the estimation output
of the quasi-Poisson model is the same as for the Poisson model as displayed above and
use 0.8866867 as the value of the dispersion parameter.


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