The chapter began with the problem of overfitting, a universal phenomenon by which models with more parameters fit a sample better, even when the additional parameters are meaningless. Two common tools were introduced to address overfitting: regularizing priors and estimates of out-of-sample accuracy (WAIC and PSIS). Regularizing priors reduce overfitting during estimation, and WAIC and PSIS help estimate the degree of overfitting. Practical functions compare in the rethinking package were introduced to help analyze collections of models fit to the same data. If you are after causal estimates, then these tools will mislead you. So models must be designed through some other method, not selected on the basis of out-of-sample predictive accuracy. But any causal estimate will still overfit the sample. So you always have to worry about overfitting, measuring it with WAIC/PSIS and reducing it with regularization.
Place each answer inside the code chunk (grey box). The code chunks should contain a text response or a code that completes/answers the question or activity requested. Make sure to include plots if the question requests them.
Finally, upon completion, name your final output .html file as: YourName_ANLY505-Year-Semester.html and publish the assignment to your R Pubs account and submit the link to Canvas. Each question is worth 5 points.
7-1. When comparing models with an information criterion, why must all models be fit to exactly the same observations? What would happen to the information criterion values, if the models were fit to different numbers of observations? Perform some experiments.
# Information criterion is based on deviance which is a sum over the sample. A model with more observations will have a higher deviance and less accuracy in information criteria compared with others with less observations. This is not a robust comparison between models, will lead to unfair comparison between models.7-2. What happens to the effective number of parameters, as measured by PSIS or WAIC, as a prior becomes more concentrated? Why? Perform some experiments.
# Prior and effective number of parameters are indirectly proportional.When prior becomes more concentrated, the effective number of parameters decreases. The reason can be observed in WAIC formula. pWAIC is log-likelihood variance sum. When priors become more concentrated, that likelihood will also become more concentrated and the variance will decrease(pWAIC).7-3. Consider three fictional Polynesian islands. On each there is a Royal Ornithologist charged by the king with surveying the bird population. They have each found the following proportions of 5 important bird species:
| Island | Species A | Species B | Species C | Species D | Species E |
| 1 | 0.2 | 0.2 | 0.2 | 0.2 | 0.2 |
| 2 | 0.8 | 0.1 | 0.05 | 0.025 | 0.025 |
| 3 | 0.05 | 0.15 | 0.7 | 0.05 | 0.05 |
Notice that each row sums to 1, all the birds. This problem has two parts. It is not computationally complicated. But it is conceptually tricky. First, compute the entropy of each island’s bird distribution. Interpret these entropy values. Second, use each island’s bird distribution to predict the other two. This means to compute the KL divergence of each island from the others, treating each island as if itwere a statistical model of the other islands. You should end up with 6 different KL divergence values. Which island predicts the others best? Why?
library(rethinking)
# We can see that the more variance in the species, the higher the entropy. Island 1 has equal amount of each species, so it has the highest entropy, island 2 has the lowest, since most species are species A.
#
# By computing the K-L divergence, we can see Island 1 predicts other Islands' species the best, since it has the lowest entropy, its species distribution is not as extreme as the other two.
getIsland = function(island) {
cur = as.numeric(c(d[island + 1, ]))
cur
}
getIslandEntropy = function(island) {
row = getIsland(island)
-sum(row * log(row))
}
klDivergence = function(islandP, islandQ) {
p = getIsland(islandP)
q = getIsland(islandQ)
sum(p * log(p/q))
}
print("Entropy for Island 1:")## [1] "Entropy for Island 1:"getIslandEntropy(1)## [1] 1.609438print("Entropy for Island 2:")## [1] "Entropy for Island 2:"getIslandEntropy(2)## [1] 0.7430039print("Entropy for Island 3:")## [1] "Entropy for Island 3:"getIslandEntropy(3)## [1] 0.9836003print("K-L divergence between 1 and 2:")## [1] "K-L divergence between 1 and 2:"klDivergence(1, 2)## [1] 0.9704061print("K-L divergence between 2 and 1:")## [1] "K-L divergence between 2 and 1:"klDivergence(2, 1)## [1] 0.866434print("K-L divergence between 2 and 3:")## [1] "K-L divergence between 2 and 3:"klDivergence(2, 3)## [1] 2.010914print("K-L divergence between 3 and 2:")## [1] "K-L divergence between 3 and 2:"klDivergence(3, 2)## [1] 1.838845print("K-L divergence between 1 and 3:")## [1] "K-L divergence between 1 and 3:"klDivergence(1, 3)## [1] 0.6387604print("K-L divergence between 3 and 1:")## [1] "K-L divergence between 3 and 1:"klDivergence(3, 1)## [1] 0.62583767-4. Recall the marriage, age, and happiness collider bias example from Chapter 6. Run models m6.9 and m6.10 again (page 178). Compare these two models using WAIC (or PSIS, they will produce identical results). Which model is expected to make better predictions? Which model provides the correct causal inference about the influence of age on happiness? Can you explain why the answers to these two questions disagree?
# The first model has lower WAIC score, so it made a better fit to the data, better at causal inference. The second model had higher WAIC score, but it is not overfitting, so it is better at prediction. This is reasonable because the first model directly used a linear relationship between married and A, which isolated one path to happiness, so it is worse at prediction due to overfitting but better at causual inference.
d = sim_happiness(seed=1977, N_years = 1000)
d2 = d[d$age > 17,]
d2$A = (d2$age - 18) / (65 - 18)
d2$mid = d2$married + 1
m6.9 = quap(
alist(
happiness ~ dnorm(mu, sigma),
mu <- a[mid] + bA * A,
a[mid] ~ dnorm(0, 1),
bA ~ dnorm(0, 2),
sigma ~ dexp(1)
), data=d2
)
m6.10 = quap(
alist(
happiness ~ dnorm(mu, sigma),
mu <- a + bA * A,
a ~ dnorm(0, 1),
bA ~ dnorm(0, 2),
sigma ~ dexp(1)
), data=d2
)
WAIC(m6.9)## WAIC lppd penalty std_err
## 1 2713.971 -1353.247 3.738532 37.54465WAIC(m6.10)## WAIC lppd penalty std_err
## 1 3101.906 -1548.612 2.340445 27.743797-5. Revisit the urban fox data, data(foxes), from the previous chapter’s practice problems. Use WAIC or PSIS based model comparison on five different models, each using weight as the outcome, and containing these sets of predictor variables: - avgfood + groupsize + area - avgfood + groupsize - groupsize + area - avgfood - area
Can you explain the relative differences in WAIC scores, using the fox DAG from the previous chapter? Be sure to pay attention to the standard error of the score differences (dSE).
# The first three models provid better prediction, because they all have lower WAIC scores. This makes sense because they have the most information. However, they are overfitting, in reality avgfood is the causal factor of both group size and area, so the last two have the lowest dSE.
data(foxes)
# avgfood + groupsize + area
model_1 = quap(
alist(
weight ~ dnorm(mu, sigma),
mu <- a + bf * avgfood + bg * groupsize + ba * area,
sigma ~ dexp(1),
a ~ dnorm(4, 1),
bf ~ dnorm(2, 1),
ba ~ dnorm(0, 1),
bg ~ dnorm(0, 1)
), data=foxes
)
WAIC(model_1)## WAIC lppd penalty std_err
## 1 361.0342 -176.1679 4.349176 16.63425# avgfood + groupsize
model_2 = quap(
alist(
weight ~ dnorm(mu, sigma),
mu <- a + bf * avgfood + bg * groupsize,
sigma ~ dexp(1),
a ~ dnorm(4, 1),
bf ~ dnorm(2, 1),
bg ~ dnorm(0, 1)
), data=foxes
)
WAIC(model_2)## WAIC lppd penalty std_err
## 1 362.4916 -177.8684 3.377455 15.86947# groupsize + area
model_3 = quap(
alist(
weight ~ dnorm(mu, sigma),
mu <- a + bg * groupsize + ba * area,
sigma ~ dexp(1),
a ~ dnorm(4, 1),
ba ~ dnorm(0, 1),
bg ~ dnorm(0, 1)
), data=foxes
)
WAIC(model_3)## WAIC lppd penalty std_err
## 1 363.1302 -177.7101 3.854931 15.97327# avgfood
model_4 = quap(
alist(
weight ~ dnorm(mu, sigma),
mu <- a + bf * avgfood,
sigma ~ dexp(1),
a ~ dnorm(4, 1),
bf ~ dnorm(2, 1)
), data=foxes
)
WAIC(model_4)## WAIC lppd penalty std_err
## 1 373.3557 -184.2908 2.387063 13.63095# area
model_5 = quap(
alist(
weight ~ dnorm(mu, sigma),
mu <- a + ba * area,
sigma ~ dexp(1),
a ~ dnorm(4, 1),
ba ~ dnorm(0, 1)
), data=foxes
)
WAIC(model_5)## WAIC lppd penalty std_err
## 1 372.9585 -183.8168 2.662476 13.71016