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? \[\begin{align} \ y_i ∼ Normal(μ, σ) \\ \ μ ∼ Normal(0, 10) \\ \ σ ∼ Exponential(1) \\ \end{align}\]

# \ y_i ∼ Normal(μ, σ) \\ is the likelihood

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

# 2 parameters in the posterior

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? \[\begin{align} \ y_i ∼ Normal(μ, σ) \\ \ μ_i = α + βx_i \\ \ α ∼ Normal(0, 10) \\ \ β ∼ Normal(0, 1) \\ \ σ ∼ Exponential(2) \\ \end{align}\]

# Linear Model is \ μ_i = α + βx_i \\

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

# 3 parameters in the posterior: α, β, σ. And the µ can be estimated by using the equation

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

##Sampling from the priors
mu_prior <- rnorm( 1e4 , 0 , 10 )
sigma_prior <- runif( 1e4 , 0, 10 )

#Simulation and visualizing the results
h_sim <- rnorm( 1e4 , mu_prior , sigma_prior )
dens( h_sim )

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

#quap formula of 4M1 model
quap_formula <- alist(
  y ~ dnorm(mu, sigma),
  mu ~ dnorm(0, 10),
  sigma ~ dexp(1))
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?

#Model and samplings with Y simulation
set.seed(2971)
sample_alpha = rnorm(1e4, 150, 25)
sample_beta = rnorm(1e4, 0, 10)
sample_sigma = runif(1e4, 0, 50)
sample_x = runif(1e4, 0,3)
prior_y <- rnorm(1e4, sample_alpha + sample_beta*sample_x,sample_sigma )
dens(prior_y)

# Prior Predictive Simulations with 50 students
N <- 50
a <- rnorm(N, 165, 15)
b <- rnorm(N, 0, 10)
plot(NULL, xlim=c(0,3), ylim=c(-100,400),
     xlab="Year", ylab="Height")

mtext("b ~ dnorm(0,10)")
for(i in 1:N) curve(a[i]+b[i]*x,
                    from=0, to =3, add=TRUE,
                    col=col.alpha("black",0.2))

##Simulated values of 50 student heights are between 80 and 240 cm approximately. However, the simulation needs to be adjusted so that there is no reduction in student heights which is not possible in real life.

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.

# to fix the simulated negative increment in student heights, updating the prior beta to log normal will modify the simulation of student heights to only improve or stay same instead of having the downside change in 3 years. 

# Prior Predictive Simulations
set.seed(2971)
N <- 50
a <- rnorm(N, 165, 15)
b <- rlnorm(N, 0, 2.5)
plot(NULL, xlim=c(0,3), ylim=c(-100,400),
     xlab="Year", ylab="Height")

mtext("b ~ dlnorm(0,2.5)")
for(i in 1:N) curve(a[i]+b[i]*x,
                    from=0, to =3, add=TRUE,
                    col=col.alpha("black",0.2))

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?

# Updating the uniform function from variance 64 that is standard deviation 8 would be be suitable.

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.

# Refit the model by excluding the mean weight xbar
set.seed(2971)
data(Howell1); d <- Howell1; d2 <- d[ d$age >= 18 , ]
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     114.5343664 1.89775066 111.5013943 117.5673385
## b       0.8907286 0.04175806   0.8239912   0.9574661
## sigma   5.0727268 0.19124969   4.7670729   5.3783807
# Compare new and old model's posterior beta and sigma values are approximately similar but the α in refit ~ N(114.53, 1.9) vs. α in original ~ N(154.60, 0.27) has significant difference 

round(vcov(m4.3_new),3)
##            a      b sigma
## a      3.601 -0.078 0.009
## b     -0.078  0.002 0.000
## sigma  0.009  0.000 0.037
# # the covariance between β and σ stayed same. However, the variance of new model α increased. And the covariance between α and β/σ has changed

pairs(m4.3_new)

#pair wise plots of new model shows the strong correlation between a and b which is almost opposite in the old model.

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="basis value")
for (i in 1:ncol(B)) lines(d2$year, B[,i])

#m4.8 <- quap(alist(
 #                   T ~ dnorm(mu, sigma), mu <- a + B %*% w,
  #                  a ~ dnorm(6, 50),
   #                 w ~ dnorm(0, 1),
    #                sigma ~ dexp(1)),
     #         data=list(T=d2$temp, B=B),
      #        start=list(w=rep(0, ncol(B))))

#post <- extract.samples(m4.8)
#w <- apply(post$w, 2, mean)
#plot(NULL, xlim=range(d2$year), ylim=c(-2, 2), xlab="year", ylab="basis * weight")
#for (i in 1:ncol(B)) lines(d2$year, w[i]*B[,i])


#mu <- link(m4.8)
#mu_PI <- apply(mu, 2, PI, 0.97)
#plot(d2$year, d2$temp, col=col.alpha(rangi2, 0.3), pch=16) 
#shade(mu_PI, d2$year, col=col.alpha("black", 0.5))

##Observing the resulting splines after increasing the number of knots, the chart waves look more compressed than before whereas increasing the width has made the plot look visibly clear.

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)

d <- Howell1
d1 <- d[d$age < 18, ]
nrow(d1)
## [1] 192
xbar <- mean(d1$weight)
xbar
## [1] 18.41419
modeld1 <- quap(
  alist(
    height ~ dnorm(mu, sigma),
    mu <- a + b * (weight - xbar),
    a ~ dnorm(156, 20),
    b ~ dlnorm(0, 1),
    sigma ~ dunif(0, 50)
  ),
  data=d1
)
precis(modeld1)
##             mean         sd       5.5%      94.5%
## a     108.363010 0.60864230 107.390282 109.335738
## b       2.716657 0.06831626   2.607474   2.825839
## sigma   8.437272 0.43058004   7.749122   9.125422
plot(height ~ weight, data = d1, col = col.alpha(rangi2, 0.3))

weight.seq <- seq(from = min(d1$weight), to = max(d1$weight), by = 1)


# (A) From the result, the beta mean for the model is 2.72, which means that for every 10 units, the child is expected to grow 10*2.72 = 27.2cm.

mu <- link(modeld1, 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(modeld1, data = list(weight = weight.seq))

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

#The linear model predicts the height well when weight is between 10 and 30 but not outside these ranges.