Chapter 4 - Geocentric Models

This chapter introduced the simple linear regression model, a framework for estimating the association between a predictor variable and an outcome variable. The Gaussian distribution comprises the likelihood in such models, because it counts up the relative numbers of ways different combinations of means and standard deviations can produce an observation. To fit these models to data, the chapter introduced quadratic approximation of the posterior distribution and the tool quap. It also introduced new procedures for visualizing prior and posterior distributions.

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. Problems are labeled Easy (E), Medium (M), and Hard(H).

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.

Questions

4E1. In the model definition below, which line is the likelihood? \[\begin{align} \ y_i ∼ Normal(μ, σ) \\ \ μ ∼ Normal(0, 10) \\ \ σ ∼ Exponential(1) \\ \end{align}\]

# y_i ∼ Normal(μ, σ)

4E2. In the model definition just above, how many parameters are in the posterior distribution?

# Two: μ, σ

4E3. Using the model definition above, write down the appropriate form of Bayes’ theorem that includes the proper likelihood and priors.

# Pr(μ,σ|y) =(ΠiNormal(yi|μ, σ)Normal(μ|0,10)Uniform(σ|0,10))/ ∫∫ΠiNormal(hi|μ, σ)Normal(μ|0,10)Uniform(σ|0,10)dμdσ

4E4. In the model definition below, which line is the linear model? \[\begin{align} \ y_i ∼ Normal(μ, σ) \\ \ μ_i = α + βx_i \\ \ α ∼ Normal(0, 10) \\ \ β ∼ Normal(0, 1) \\ \ σ ∼ Exponential(2) \\ \end{align}\]

# The linear model is μi=α+βxi

4E5. In the model definition just above, how many parameters are in the posterior distribution?

# There are three parameters in the posterior distribution: α, β, and σ.

4M1. For the model definition below, simulate observed y values from the prior (not the posterior). \[\begin{align} \ y_i ∼ Normal(μ, σ) \\ \ μ ∼ Normal(0, 10) \\ \ σ ∼ Exponential(1) \\ \end{align}\]

sample_mu <- rnorm(1e4, 0, 10)
sample_sigma <- runif(1e4, 0, 10)
prior_y <- rnorm(1e4, sample_mu, sample_sigma)
dens(prior_y)

4M2. Translate the model just above into a quap formula.

formula <- alist(
  y ~ dnorm(mu, sigma),
  mu ~ dnorm(0, 10),
  sigma ~ dunif(0, 10)
)
4M3. Translate the quap model formula below into a mathematical model definition:

y ~ dnorm( mu , sigma ),
mu <- a + b*x,
a ~ dnorm( 0 , 10 ),
b ~ dunif( 0 , 1 ),
sigma ~ dexp( 1 )

flist <- alist(
  y ~ dnorm(mu, sigma),
  mu <- a + b*x,
  a ~ dnorm(0, 50),
  b ~ dunif(0, 10),
  sigma ~ dunif(0, 50)
)

4M4. A sample of students is measured for height each year for 3 years. After the third year, you want to fit a linear regression predicting height using year as a predictor. Write down the mathematical model definition for this regression, using any variable names and priors you choose. Be prepared to defend your choice of priors.

# hi ∼ Normal(μi,σ)
# μi = α + βti
# α ∼ Normal(120,20)
# β ∼ Normal(0,10)
# σ ∼ Uniform(0,50)

4M5. Now suppose I remind you that every student got taller each year. Does this information lead you to change your choice of priors? How?

# No.

4M6. Now suppose I tell you that the variance among heights for students of the same age is never more than 64cm. How does this lead you to revise your priors?

# Variance term can be adjust by introducing an uniform distribution

4M7. Refit model m4.3 from the chapter, but omit the mean weight xbar this time. Compare the new model’s posterior to that of the original model. In particular, look at the covariance among the parameters. What is different? Then compare the posterior predictions of both models.

data(Howell1)
d <- Howell1
d2 <- d[d$age>=18,]
m4.3 <- quap(
  alist(
    height~dnorm(mu,sigma),
    mu<-a+b*(weight),
    a~dnorm(178,20),
    b~dlnorm(0,1),
    sigma~dunif(0,50)
  ),
  data=d2)
precis(m4.3)
##              mean         sd        5.5%       94.5%
## a     114.5337335 1.89773962 111.5007791 117.5666880
## b       0.8907413 0.04175782   0.8240042   0.9574784
## sigma   5.0727009 0.19124722   4.7670509   5.3783509

4M8. In the chapter, we used 15 knots with the cherry blossom spline. Increase the number of knots and observe what happens to the resulting spline. Then adjust also the width of the prior on the weights—change the standard deviation of the prior and watch what happens. What do you think the combination of knot number and the prior on the weights controls?

data(cherry_blossoms)
df <- cherry_blossoms
precis(df)
##                   mean          sd      5.5%      94.5%       histogram
## year       1408.000000 350.8845964 867.77000 1948.23000   ▇▇▇▇▇▇▇▇▇▇▇▇▁
## doy         104.540508   6.4070362  94.43000  115.00000        ▁▂▅▇▇▃▁▁
## temp          6.141886   0.6636479   5.15000    7.29470        ▁▃▅▇▃▂▁▁
## temp_upper    7.185151   0.9929206   5.89765    8.90235 ▁▂▅▇▇▅▂▂▁▁▁▁▁▁▁
## temp_lower    5.098941   0.8503496   3.78765    6.37000 ▁▁▁▁▁▁▁▃▅▇▃▂▁▁▁
q4m8 <- df[complete.cases(cherry_blossoms$doy),]
numknots15 <- 15
knot15 <- quantile(q4m8$year, probs = seq(0, 1, length.out=numknots15))
knot15
##        0% 7.142857% 14.28571% 21.42857% 28.57143% 35.71429% 42.85714%       50% 
##       812      1036      1174      1269      1377      1454      1518      1583 
## 57.14286% 64.28571% 71.42857% 78.57143% 85.71429% 92.85714%      100% 
##      1650      1714      1774      1833      1893      1956      2015
numknots30 <- 30
knot30 <- quantile(q4m8$year, probs = seq(0, 1, length.out = numknots30))
knot30
##        0% 3.448276% 6.896552% 10.34483%  13.7931% 17.24138% 20.68966% 24.13793% 
##  812.0000  958.4828 1032.9655 1115.3448 1168.8621 1213.4138 1261.6897 1315.3793 
## 27.58621% 31.03448% 34.48276% 37.93103% 41.37931% 44.82759% 48.27586% 51.72414% 
## 1366.8621 1410.6897 1442.6552 1477.3103 1505.7931 1534.2759 1563.7586 1597.2414 
## 55.17241% 58.62069% 62.06897% 65.51724% 68.96552% 72.41379% 75.86207% 79.31034% 
## 1631.1724 1664.2069 1692.6897 1724.1724 1753.6552 1782.1379 1810.6207 1839.1034 
## 82.75862%  86.2069% 89.65517% 93.10345% 96.55172%      100% 
## 1867.5862 1898.0690 1928.5517 1958.0345 1986.5172 2015.0000

4H2. Select out all the rows in the Howell1 data with ages below 18 years of age. If you do it right, you should end up with a new data frame with 192 rows in it.

  1. Fit a linear regression to these data, using quap. Present and interpret the estimates. For every 10 units of increase in weight, how much taller does the model predict a child gets?

  2. Plot the raw data, with height on the vertical axis and weight on the horizontal axis. Superimpose the MAP regression line and 89% interval for the mean. Also superimpose the 89% interval for predicted heights.

  3. What aspects of the model fit concern you? Describe the kinds of assumptions you would change, if any, to improve the model. You don’t have to write any new code. Just explain what the model appears to be doing a bad job of, and what you hypothesize would be a better model.

#(a)
qhdata <- Howell1[Howell1$age<18,]
nrow(qhdata)
## [1] 192
# (b)
plot(height ~ weight, data = qhdata)

# (c)
# The skewness of the data. Might need to transform it such as introducing log or other ways