Assignment #3
Zhe Yu
2020-12-02
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 parameters: μ, σ4E3. Using the model definition above, write down the appropriate form of Bayes’ theorem that includes the proper likelihood and priors.
\[{\rm Pr}(\mu, \sigma|y)=\frac{{\rm Normal}(y|\mu, \sigma)\times{\rm Normal}(\mu|0, 10)\times{\rm Uniform}(\sigma|0, 10)}{\iint{\rm Normal}(y|\mu, \sigma)\times{\rm Normal}(\mu|0, 10)\times{\rm Uniform}(\sigma|0, 10){\rm d}\mu{\rm d}\sigma}\]
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_i4E5. In the model definition just above, how many parameters are in the posterior distribution?
# Three parameters: α, β, σ4M1. 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))y ~ dnorm( mu , sigma ),
mu <- a + b*x,
a ~ dnorm( 0 , 10 ),
b ~ dunif( 0 , 1 ),
sigma ~ dexp( 1 )
# y_i ~ Normal(μ, σ)
# μ_i = α + βx_i
# α ~ Normal(0, 10)
# β ~ Uniform(0, 1)
# σ ~ Exponential(1)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.
# y_i ~ Normal(μ, σ)
# μ_i = α + βx_i
# α ~ Normal(150, 20)
# β ~ Normal(4, 2)
# σ ~ Uniform(0, 50)\(\alpha\) prior: Assume the average height of student is 150cm and the standard deviation is 20cm. \(\beta\) prior: Assume the yearly growth is 4cm and the standard deviation is 2cm. \(\sigma\) prior: Uniform distribution 0 to 50cm.
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?
# Yes. I will change β to Normal(6, 2) to make sure β is larger than 0.4M6. 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?
# I will change σ to Uniform(0, 8), because the standard deviation is no more than sqrt(64) = 8cm.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$age >= 18, ]
xbar <- mean(d2$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 154.6013676 0.27030762 154.1693638 155.0333714
## b 0.9032809 0.04192362 0.8362788 0.9702829
## sigma 5.0718801 0.19115471 4.7663780 5.3773823m4.3_u <- 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_u)## mean sd 5.5% 94.5%
## a 114.5359004 1.89779427 111.5028586 117.5689422
## b 0.8906934 0.04175903 0.8239544 0.9574324
## sigma 5.0728365 0.19126011 4.7671659 5.3785071The estimated mean and standard deviation of \(\alpha\) are different.
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$doy), ]
library(splines)
num_knots <- 30
knot_list <- quantile(d2$year, probs = seq(0, 1, length.out = num_knots))
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))
for(i in 1:ncol(B)) lines(d2$year, B[, i])
num_knots2 <- 15
knot_list2 <- quantile(d2$year, probs = seq(0, 1, length.out = num_knots2))
B2 <- bs(d2$year, knots = knot_list2[-c(1, num_knots2)], degree = 3, intercept = TRUE)
plot(NULL, xlim = range(d2$year), ylim = c(0,1))
for(i in 1:ncol(B2)) lines(d2$year, B2[, i])
When the number of knots gets increased, the curve is less smooth and more squeezed. More knots means there are more basis functions and so the curve can fit smaller local details.
4H2. Select out all the rows in the Howell1 data with ages below 18 years of age. If you do it right, you should end up with a new data frame with 192 rows in it.
Fit a linear regression to these data, using quap. Present and interpret the estimates. For every 10 units of increase in weight, how much taller does the model predict a child gets?
Plot the raw data, with height on the vertical axis and weight on the horizontal axis. Superimpose the MAP regression line and 89% interval for the mean. Also superimpose the 89% interval for predicted heights.
What aspects of the model fit concern you? Describe the kinds of assumptions you would change, if any, to improve the model. You don’t have to write any new code. Just explain what the model appears to be doing a bad job of, and what you hypothesize would be a better model.
data(Howell1)
H2 <- Howell1[Howell1$age < 18, ]
nrow(H2)## [1] 192# Part a
fit <- alist(height ~ dnorm(mu, sigma), mu <- a + b*weight, a ~ dnorm(110, 30), b ~ dnorm(0, 10), sigma ~ dunif(0, 60))
fit2 <- quap(fit, data = H2)
precis(fit2)## mean sd 5.5% 94.5%
## a 58.344983 1.39573978 56.114321 60.575645
## b 2.715033 0.06823406 2.605982 2.824085
## sigma 8.437250 0.43057656 7.749105 9.125394\(\beta\) = 2.71, so for 10 units of increase in weight, the predicted height will be 27.1cm taller.
# Part b
plot(height ~ weight, data = H2, col = col.alpha(rangi2, 0.3))
weight.sq <- seq(from = min(H2$weight), to = max(H2$weight), by = 1)
mu <- link(fit2, data = data.frame(weight = weight.sq))
mu.mean = apply(mu, 2, mean)
mu.HPDI = apply(mu, 2, HPDI, prob = 0.89)
lines(weight.sq, mu.mean)
shade(mu.HPDI, weight.sq)
sim.height <- sim(fit2, data = list(weight = weight.sq))
height.HPDI <- apply(sim.height, 2, HPDI, prob = 0.89)
shade(height.HPDI, weight.sq)
# Part cThe linear regression model is not a good fit for the data. For values far from average (for example, values < 10 and values > 30), the model prediction is poor. I would try to introduce a quadratic term to improve the model.