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. Make sure to include plots if the question requests them. 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? \[\begin{align} \ y_i ∼ Normal(μ, σ) \\ \ μ ∼ Normal(0, 10) \\ \ σ ∼ Exponential(1) \\ \end{align}\]

# The first line y_i ∼ Normal(μ, σ) is the liklihood.

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

# There are two parameters to be estimated in this model: μ and 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) = ∏𝑖 Normal(yi|μ,σ)Normal(μ|0,10)Uniform(σ|0,10) / ∫∫ ∏iNormal(hi|μ,σ)Normal(μ|0,10)Uniform(σ|0,10)dμdσ

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

# μ_i = α + βx_i

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). Make sure to plot the simulation. \[\begin{align} \ y_i ∼ Normal(μ, σ) \\ \ μ ∼ Normal(0, 10) \\ \ σ ∼ Exponential(1) \\ \end{align}\]

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. Simulate from the priors that you chose to see what the model expects before it sees the data. Do this by sampling from the priors. Then consider 50 students, each simulated with a different prior draw. For each student simulate 3 years. Plot the 50 linear relationships with height(cm) on the y-axis and year on the x-axis. What can we do to make these priors more likely?

# If the same sample of students are repeatedly sampled each year, then the observations are not independent and we should use a linear mixed model. I will instead use a linear model. The question talks about “students” without specifying age, so I am going to start with a weak prior for the intercept, α, that will capture likely heights for students all the way from school age children to college age young adults (from around 110 cm for a 5 year old female to around 180 cm for a 20 year old male). Similarly, I will use a weak prior for the slope, β, that will capture likely yearly growth rates for this wide age range (from around 7.0 cm/year for a 5 year old to around 0.5 cm/year for a 20 year old). Finally, I will use a uniform prior for the standard deviation of heights which can cover the full range if students from all ages are included.
# hi∼Normal(μ,σ)    
# μi = α+βxi
# α ∼Normal(150,25)
# β∼Normal(4,2)
# σ ∼Uniform(0,50)
# I chose a linear model without any polynomial terms or transformations because I noticed that a later question will ask for log transformation and I want an untransformed point of comparison. For the alpha prior, I chose a normal distribution centered on 150 cm with an SD of 25 cm, 150 cm is in the middle of the expected distribution if both school and college students are included and 25 cm is enough variability that two SDs around the mean (i.e., 100 cm to 200 cm) should include most students at the high and low end of the age distribution. For the beta prior, I chose a normal distribution centered on 4 cm/year with an SD of 2 cm/year, 4 cm/year is in the middle of the expected distribution if both school and college students are included and 2 cm/year is enough variability that two SDs around the mean (i.e., 0 cm/year to 8 cm/year) should include most students at the high and low end of the age distribution. This also captures prior knowledge that students should only very rarely grow less tall over time. Finally, for the sigma prior, I chose a uniform distribution from 0 cm to 50 cm, this range includes both a tight distribution of students around the same age/height and a wide range of students at both school and college ages/heights. 50 cm is a bit high, but I want a conservative prior to begin with.

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? Again, simulate from the priors and plot.

# Knowing that the average height at the first year was 120 cm and that every student got taller each year makes me more confident that we are talking about school age students ( around 7 years old). Then, I can narrow the range of my prior distributions to make heights and growth rates from older ages less plausible. My expectation for 𝜎 is also much lower now too as I no longer expect a balanced mix of young and old students.
#
# hi∼Normal(μ,σ)    
# μi = α+βxi
# α ∼Normal(120,10)
# β∼Normal(7,1)
# σ ∼Uniform(0,20)

# No we already assumed every student got taller.

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?

# The variance is the square of σ, so if variance is never more than 64 cm, then σ is never more than 8 cm. So we can adjust the maximum of the σ prior. This information about σ may also have implications for the 𝛼 prior, but I am not confident enough about this relationship to update that prior.
# hi∼Normal(μ,σ)    
# μi = α+βxi
# α ∼Normal(120,10)
# β∼Normal(7,1)
# σ ∼Uniform(0,8)

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. Show the pairs() plot. What is different? Then compare the posterior predictions of both models.

data(Howell1)
d <- Howell1
d2 <- d[ d$age >= 18, ]
xbar <- mean(d2$weight)
m4.3 <- quap(alist(
      height ~ dnorm( mu , sigma ) , 
      mu <- a + b*( weight - xbar ) , 
      a ~ dnorm( 178 , 20 ) ,
      b ~ dlnorm( 0 , 1 ) ,
      sigma ~ dunif( 0 , 50 )), data=d2 )
precis(m4.3)
##              mean         sd        5.5%       94.5%
## a     154.6013675 0.27030764 154.1693637 155.0333713
## b       0.9032809 0.04192363   0.8362788   0.9702829
## sigma   5.0718805 0.19115474   4.7663783   5.3773827
#Ommitting xbar
m4.3_new <- 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_new)
##              mean         sd        5.5%       94.5%
## a     114.5343214 1.89774734 111.5013546 117.5672882
## b       0.8907299 0.04175799   0.8239926   0.9574673
## sigma   5.0727185 0.19124890   4.7670658   5.3783712

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) 
d <- cherry_blossoms 
d2 <- d[ complete.cases(d$temp) , ]
num_knots <- 30
knot_list <- quantile( d2$year , probs=seq(0,1,length.out=num_knots))
library(splines) 
B <- bs(d2$year, knots=knot_list[-c(1,num_knots)] , degree=3 , intercept=TRUE )
plot( NULL , xlim=range(d2$year) , ylim=c(0,1) , xlab="year" , ylab="value" )
for ( i in 1:ncol(B) ) lines( d2$year , B[,i])

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.

data(Howell1)


d2 <- Howell1[Howell1$age < 18, ]
nrow(d2)
## [1] 192
# Part a

formula <- alist(
  height ~ dnorm(mu, sigma),
  mu <- a + b * weight,
  a ~ dnorm(110, 30),
  b ~ dnorm(0, 10),
  sigma ~ dunif(0, 60)
)
m <- map(formula, data = d2)
precis(m, corr = TRUE)
##            mean         sd      5.5%     94.5%
## a     58.344984 1.39573566 56.114328 60.575639
## b      2.715034 0.06823386  2.605983  2.824084
## sigma  8.437225 0.43057337  7.749085  9.125364
# Part b
plot(height ~ weight, data = d2, col = col.alpha(rangi2, 0.3))

weight.seq <- seq(from = min(d2$weight), to = max(d2$weight), by = 1)
mu <- link(m, data = data.frame(weight = weight.seq))

mu.mean <- apply(mu, 2, mean)
mu.HPDI <- apply(mu, 2, HPDI, prob = 0.89)
lines(weight.seq, mu.mean)
shade(mu.HPDI, weight.seq)

sim.height <- sim(m, data = list(weight = weight.seq))

height.HPDI <- apply(sim.height, 2, HPDI, prob = 0.89)
shade(height.HPDI, weight.seq)

# Part c

#we need to assess the model's fit. The linear model seems to be doing a poor job predicting height at most weights. It overestimates height at both low (<10) and high (>30) weights and underestimates height for most middling (10-30) weights. I might start with a polynomial regression.