This chapter has been an introduction to the motivation, implementation, and interpretation of basic multilevel models. It focused on varying intercepts, which achieve better estimates of baseline differences among clusters in the data. They achieve better estimates, because they simultaneously model the population of clusters and use inferences about the population to pool information among parameters. From another perspective, varying intercepts are adaptively regularized parameters, relying upon a prior that is itself learned from the data.
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.
13E1. Which of the following priors will produce more shrinkage in the estimates? Show the prior plot. \[\begin{align} \ α_{TANK} ∼ Normal(0, 1) \tag{a} \\ \ α_{TANK} ∼ Normal(0, 2) \tag{b} \\ \end{align}\]
#(a) produces more shrinkage in estimates.13E2. Rewrite the following model as a multilevel model. \[\begin{align} y_i ∼ Binomial(1, p_i) \\ logit(p_i) = α_{group[i]} + βx_i \\ α_{group} ∼ Normal(0, 1.5) \\ β ∼ Normal(0, 0.5) \\ \end{align}\]
# y_i ~ Binomial(1,p_i)
# logit(p_i) = a[i] + beta*x_i
# a[i] ~ Normal(a_l2, sigma_l2)
# beta ~ Normal(0,0.5)
# a_l2 ~ Normal(0, 1.5)
# sigma_l2 ~ Exponential(1)13E3. Rewrite the following model as a multilevel model. \[\begin{align} y_i ∼ Normal(μ_i, σ) \\ μ_i = α_{group[i]} + βx_i \\ α_{group} ∼ Normal(0, 5) \\ β ∼ Normal(0, 1) \\ σ ∼ Exponential(1) \\ \end{align}\]
# y_i ∼ Normal(μ_i, σ)
# σ ∼ Exponential(1)
# μ_i = a[i] + βx_i
# β ∼ Normal(0, 1)
# a[i] ~ Normal(a_l2, sigma_l2)
# a_l2 ~ Normal(0, 5)
# sigma_l2 ~ HalfCauchy(0,1)13E4. Write a mathematical model formula for a Poisson regression with varying intercepts.
# y_i ~ Poisson(lambda_i)
# log(lambda_i) = a[i] + beta*x_i
# a[i] ~ Normal(a_l2, sigma_l2)
# beta ~ Normal(0,1)
# a_l2 ~ Normal(0, 1)
# sigma_l2 ~ HalfCauchy(0,1)13E5. Write a mathematical model formula for a Poisson regression with two different kinds of varying intercepts, a cross-classified model.
# y_i ~ Poisson(lambda_i)
# log(lambda_i) = a_l1 + a1[i] + a2[i] + beta*x_i
# a_l1 ~ Normal(0, 10)
# beta ~ Normal(0,1)
# a1[i] ~ Normal(0, sigma1_l2)
# a2[i] ~ Normal(0, sigma2_l2)
# sigma1_l2 ~ HalfCauchy(0,1)
# sigma2_l2 ~ HalfCauchy(0,1)13M1. Revisit the Reed frog survival data, data(reedfrogs), and add the predation and size treatment variables to the varying intercepts model. Consider models with either main effect alone, both main effects, as well as a model including both and their interaction. Instead of focusing on inferences about these two predictor variables, focus on the inferred variation across tanks. Explain why it changes as it does across models. Plot the sigma estimates.
library(rethinking)
data(reedfrogs)
df <- reedfrogs
#predation
df$pred <- ifelse(df$pred == "yes", 1,0)
#size dummy
df$sizes <- ifelse(df$size == "small", 0, 1)
#intercept
df$tanks <- 1:nrow(df)
#model
m <- map2stan(
alist(
surv ~ dbinom(density, c),
logit(c) <- a_tank[tanks],
a_tank[tanks] ~ dnorm(x, s),
x ~ dnorm(0, 10),
s ~ dexp(1)
),
data = df, chains = 4
)##
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## x 1.382915 0.2655376 0.9639071 1.810357 4923.777 0.9993249
## s 1.624148 0.2140572 1.3091939 1.995797 3014.359 1.0002539m1 <- map2stan(
alist(
surv ~ dbinom(density, c),
logit(c) <- a_tank[tanks] + dp*pred,
dp ~ dnorm(0,1),
a_tank[tanks] ~ dnorm(x, s),
x ~ dnorm(0, 10),
s ~ dexp(1)
),
data = df, chains = 4
)##
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## Chain 4:precis(m1)## mean sd 5.5% 94.5% n_eff Rhat4
## dp 0.00767944 1.0576484 -1.6624606 1.713339 8191.769 0.9991612
## x 1.38864798 0.2595262 0.9820116 1.812969 6037.128 0.9999841
## s 1.62283604 0.2106192 1.3158565 1.983098 3269.056 0.9997748m2 <- map2stan(
alist(
surv ~ dbinom(density, c),
logit(c) <- a_tank[tanks] + ds*sizes,
ds ~ dnorm(0,1),
a_tank[tanks] ~ dnorm(x, s),
x ~ dnorm(0, 10),
s ~ dexp(1)
),
data = df, chains = 4
)##
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## ds -0.3373148 0.4775461 -1.079215 0.443458 472.5423 1.002730
## x 1.5589884 0.3581254 0.991156 2.146302 787.4397 1.000720
## s 1.6246653 0.2223812 1.306638 2.010759 2240.0449 1.001215m3 <- map2stan(
alist(
surv ~ dbinom(density, c),
logit(c) <- a_tank[tanks] + dp*pred + ds*sizes,
dp ~ dnorm(0,1),
ds ~ dnorm(0,1),
a_tank[tanks] ~ dnorm(x, s),
x ~ dnorm(0, 10),
s ~ dexp(1)
),
data = df, chains = 4
)##
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## Chain 4:precis(m3)## mean sd 5.5% 94.5% n_eff Rhat4
## dp -0.005760928 1.0193523 -1.590821 1.6594953 5437.6488 0.9993186
## ds -0.341330163 0.4566063 -1.071238 0.3927167 588.6118 1.0092082
## x 1.556482086 0.3488849 1.014587 2.1281196 944.8047 1.0029362
## s 1.623521676 0.2214245 1.295695 2.0049497 2250.5088 1.0016171m4 <- map2stan(
alist(
surv ~ dbinom(density, c),
logit(c) <- a_tank[tanks] + dp*pred + ds*sizes + dps*pred*sizes,
dp ~ dnorm(0,1),
ds ~ dnorm(0,1),
dps ~ dnorm(0,1),
a_tank[tanks] ~ dnorm(x, s),
x ~ dnorm(0, 10),
s ~ dexp(1)
),
data = df, chains = 4
)##
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## Chain 4:precis(m4)## mean sd 5.5% 94.5% n_eff Rhat4
## dp 0.01315640 0.9927611 -1.546438 1.6062257 5113.4537 0.9999877
## ds -0.37034545 0.4478582 -1.079959 0.3360704 539.2856 1.0047322
## dps -0.01418517 0.9791160 -1.588295 1.5624070 5517.9367 0.9994906
## x 1.57817564 0.3438729 1.047919 2.1462626 787.7565 1.0017678
## s 1.61619232 0.2151222 1.304628 1.9852461 2596.1509 1.0008358#Considering means of these different models, adding a predictor variable is resulting in variation so tanks is observed after checking models with predictor variables and without predictor variables predation and size.13M2. Compare the models you fit just above, using WAIC. Can you reconcile the differences in WAIC with the posterior distributions of the models? Show the WAIC table.
compare(m, m1, m2, m3, m4)## WAIC SE dWAIC dSE pWAIC weight
## m2 1009.496 38.22505 0.0000000 NA 37.82898 0.2664112
## m1 1009.905 38.12389 0.4090435 0.7106330 38.01980 0.2171350
## m3 1010.121 38.21926 0.6252062 0.3564416 38.14664 0.1948905
## m 1010.209 38.17864 0.7131030 0.6888048 38.21442 0.1865109
## m4 1010.854 38.29700 1.3587551 0.3698422 38.46692 0.1350525#Model with size seems to have very small variation compared to other models.13M3. Re-estimate the basic Reed frog varying intercept model, but now using a Cauchy distribution in place of the Gaussian distribution for the varying intercepts. That is, fit this model: \[\begin{align} s_i ∼ Binomial(n_i, p_i) \\ logit(p_i) = α_{tank[i]} \\ α_{tank} ∼ Cauchy(α, σ) \\ α ∼ Normal(0, 1) \\ σ ∼ Exponential(1) \\ \end{align}\]
(You are likely to see many divergent transitions for this model. Can you figure out why? Can you fix them?) Plot and compare the posterior means of the intercepts, αtank, to the posterior means produced in the chapter, using the customary Gaussian prior. Can you explain the pattern of differences? Take note of any change in the mean α as well.
mc <- map2stan(
alist(
surv ~ dbinom(density, c),
logit(c) <- a_tank[tanks],
a_tank[tanks] ~ dnorm(x, s),
x ~ dnorm(0, 10),
s ~ dexp(1)
),
data = df, chains = 4
)##
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## Chain 4:precis(mc)## mean sd 5.5% 94.5% n_eff Rhat4
## x 1.381434 0.2523615 0.9778204 1.784543 5826.883 0.9995237
## s 1.622163 0.2159911 1.3095614 1.990619 3258.362 1.0001163compare(m, m1, m2, m3, m4, mc)## WAIC SE dWAIC dSE pWAIC weight
## m2 1009.496 38.22505 0.0000000 NA 37.82898 0.2324367
## m1 1009.905 38.12389 0.4090435 0.7106330 38.01980 0.1894445
## m3 1010.121 38.21926 0.6252062 0.3564416 38.14664 0.1700368
## m 1010.209 38.17864 0.7131030 0.6888048 38.21442 0.1627258
## mc 1010.696 38.13228 1.2005865 0.7159788 38.41067 0.1275266
## m4 1010.854 38.29700 1.3587551 0.3698422 38.46692 0.1178297#Not much difference between the Gaussian priors and the Cauchy priors. Cauchy model seems to have higher WAIC score than the ones with Gaussian priors.13M4. Now use a Student-t distribution with ν = 2 for the intercepts: \[\begin{align} α_{tank} ∼ Student(2, α, σ) \end{align}\]
Refer back to the Student-t example in Chapter 7 (page 234), if necessary. Plot and compare the resulting posterior to both the original model and the Cauchy model in 13M3. Can you explain the differences and similarities in shrinkage in terms of the properties of these distributions?
13M5. Modify the cross-classified chimpanzees model m13.4 so that the adaptive prior for blocks contains a parameter \(\bar{γ}\) for its mean: \[\begin{align} γ_j ∼ Normal(\bar{γ}, σ_γ) \\ \bar{γ} ∼ Normal(0, 1.5) \\ \end{align}\]
Compare the precis output of this model to m13.4. What has including \(\bar{γ}\) done?
13M6. Sometimes the prior and the data (through the likelihood) are in conflict, because they concentrate around different regions of parameter space. What happens in these cases depends a lot upon the shape of the tails of the distributions. Likewise, the tails of distributions strongly influence can outliers are shrunk or not towards the mean. I want you to consider four different models to fit to one observation at y = 0. The models differ only in the distributions assigned to the likelihood and prior. Here are the four models:
\[\begin{align} Model \;NN: y &∼ Normal(μ, 1) & Model \;TN: y &∼ Student(2, μ, 1) \\ μ &∼ Normal(10, 1) & μ &∼ Normal(10, 1) \\ Model \;NT: y &∼ Normal(μ, 1) & Model \;TT: y &∼ Student(2, μ, 1) \\ μ &∼ Student(2, 10, 1) & μ &∼ Student(2, 10, 1) \\ \end{align}\]
Estimate and plot the posterior distributions against the likelihoods for these models and compare them. Can you explain the results, using the properties of the distributions?
EXTRA CREDIT (10 POINTS)
13H1. In 1980, a typical Bengali woman could have 5 or more children in her lifetime. By the year 2000, a typical Bengali woman had only 2 or 3. You’re going to look at a historical set of data, when contraception was widely available but many families chose not to use it. These data reside in data(bangladesh) and come from the 1988 Bangladesh Fertility Survey. Each row is one of 1934 women. There are six variables, but you can focus on two of them for this practice problem: (1) district: ID number of administrative district each woman resided in (2) use.contraception: An indicator (0/1) of whether the woman was using contraception The first thing to do is ensure that the cluster variable, district, is a contiguous set of integers. Recall that these values will be index values inside the model. If there are gaps, you’ll have parameters for which there is no data to inform them. Worse, the model probably won’t run. Look at the unique values of the district variable:
data(bangladesh)
d <- bangladesh
sort(unique(d$district)) #District 54 is absent. So district isn’t yet a good index variable, because it’s not contiguous. This is easy to fix. Just make a new variable that is contiguous. This is enough to do it:## [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
## [26] 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
## [51] 51 52 53 55 56 57 58 59 60 61d$district_id <- as.integer(as.factor(d$district))
sort(unique(d$district_id)) # Now there are 60 values, contiguous integers 1 to 60. Now, focus on predicting use.contraception, clustered by district_id.## [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
## [26] 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
## [51] 51 52 53 54 55 56 57 58 59 60Fit both (1) a traditional fixed-effects model that uses an index variable for district and (2) a multilevel model with varying intercepts for district. Plot the predicted proportions of women in each district using contraception, for both the fixed-effects model and the varying-effects model. That is, make a plot in which district ID is on the horizontal axis and expected proportion using contraception is on the vertical. Make one plot for each model, or layer them on the same plot, as you prefer. How do the models disagree? Can you explain the pattern of disagreement? In particular, can you explain the most extreme cases of disagreement, both why they happen where they do and why the models reach different inferences?
mfix <- map2stan( alist( uc ~ dbinom(1, dst), logit(dst) <- a_district[dis], a_district[dis] ~ dnorm(0, 10) ), data = dfl )
mvar <- map2stan( alist( uc ~ dbinom(1, dst), logit(dst) <- x + a_district[dis], x ~ dnorm(0, 10), a_district[dis] ~ dnorm(0, 10) ), data = dfl )
compare(mfix, mvar)
fix2 <- link(mfix, data = dfl)
var2 <- link(mvar, data = dfl)
#mean probabilites for fix2 pfix2 <- apply(fix2, 2, mean) #mean probabilites for var2 pvar2 <- apply(var2, 2, mean)
#fixed effects plot plot(pfix2)
#varying intercepts plot plot(pvar2)
#Both the models are nearly the same except for minor variations of fixed estimates are more closely packed and much closer to 0.4 whereas varying estimates are little more spread out. Both have outlier for district 1 or so. Due to shrinkage, they appear to be pretty close to 0.4. Standard error is lower for fixed effects when compared to varying intercepts model