Assignment #11
Jie Xiong
2020-12-09
Chapter 13 - Models With Memory
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. 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
13E1. Which of the following priors will produce more shrinkage in the estimates? \[\begin{align} \ α_{TANK} ∼ Normal(0, 1) \tag{a} \\ \ α_{TANK} ∼ Normal(0, 2) \tag{b} \\ \end{align}\]
curve( dnorm(x,0,1) , from=-5 , to=5 , ylab="Density" )
curve( dnorm(x,0,2) , add=TRUE , lty=2 ) 
#The first one with solid line is more concentrated around 0, and will produce more shrinkage.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) = α_group[i] + βx_i
#α_group ∼ Normal(α_bar, σ_α)
#σ_α ~ Exponential(1)
#α_bar ~ Normal(0, 1.5)
#β ∼ Normal(0, 0.5)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)
#μ_i = α_group[i] + βx_i
#α_group[i] ∼ Normal(α_bar, σ_α)
#σ ∼ Exponential(1)
#β ∼ Normal(0, 1)
#α_bar ~ Normal(0, 1)
#σ_α ~ Exponential(1)13E4. Write a mathematical model formula for a Poisson regression with varying intercepts.
#y_i ∼ poisson(p_i)
#log(p_i) = α_{group[i]} + βx_i
#α_{group} ∼ Normal(α, α_α)
#α = Normal(0,1)
#α_α = Exponential(1)
#β ∼ Normal(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(p_i)
#log(p_i) = α_{group[i]} α1_{group[i]}+ βx_i
#α_{group} ∼ Normal(α, α_α)
#α1_{group} ∼ Normal(α1, α1_α1)
#α = Normal(0,1)
#α1 = Normal(0,1)
#α_α = Exponential(1)
#α1_α1 = Exponential(1)
#β ∼ Normal(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.
data(reedfrogs)
data <- reedfrogs
data$pred <- ifelse(data$pred == "yes", 1,0)
data$sizes <- ifelse(data$size == "big", 1, 0)
data$tanks <- 1:nrow(data)
#model 1
m1 <- 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 = data, chains = 4 )##
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## Chain 4:## Computing WAIC#model 2
m2 <- 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 = data, chains = 4 )##
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## Chain 4:## Computing WAIC#model 3
m3 <- 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 = data, chains = 4 )##
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## Chain 4:## Computing WAIC #model 4
m4 <- 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 = data, chains = 4 )##
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## Chain 4:## Computing WAIC#Model 5
m5 <- 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 = data, chains = 4 )##
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## http://mc-stan.org/misc/warnings.html#bulk-ess## Computing WAICprecis (m1)## 48 vector or matrix parameters hidden. Use depth=2 to show them.## mean sd 5.5% 94.5% n_eff Rhat4
## x 1.384711 0.2636367 0.9642894 1.804768 4870.421 0.9993133
## s 1.624439 0.2140445 1.3162096 2.000676 3236.246 0.9997997precis (m2)## 48 vector or matrix parameters hidden. Use depth=2 to show them.## mean sd 5.5% 94.5% n_eff Rhat4
## dp 0.00529477 1.0063472 -1.5993011 1.596061 8904.963 0.9996261
## x 1.39082444 0.2679910 0.9764649 1.828283 4553.208 1.0000300
## s 1.63297661 0.2210655 1.3147239 2.015525 2748.418 1.0014545precis (m3)## 48 vector or matrix parameters hidden. Use depth=2 to show them.## mean sd 5.5% 94.5% n_eff Rhat4
## ds -0.373145 0.4519600 -1.089378 0.3639623 469.6751 1.005763
## x 1.575980 0.3464575 1.028199 2.1317007 745.7184 1.002583
## s 1.626714 0.2167655 1.320934 1.9953314 2280.2087 1.000501precis (m4)## 48 vector or matrix parameters hidden. Use depth=2 to show them.## mean sd 5.5% 94.5% n_eff Rhat4
## dp 0.008733628 1.0483908 -1.638514 1.6336066 4668.4149 1.000221
## ds -0.340618895 0.4661732 -1.066902 0.3896198 507.5182 1.019433
## x 1.554517504 0.3460232 1.006086 2.1019577 892.9973 1.012718
## s 1.622587032 0.2129676 1.315102 1.9932205 2281.2081 1.000486precis (m5)## 48 vector or matrix parameters hidden. Use depth=2 to show them.## mean sd 5.5% 94.5% n_eff Rhat4
## dp 0.00994832 1.0073716 -1.595104 1.5923096 5048.5505 0.9993252
## ds -0.33680101 0.4625324 -1.043770 0.4158547 347.4370 1.0185126
## dps 0.02729951 0.9926937 -1.563747 1.6486121 4996.3892 0.9994832
## x 1.55512957 0.3465119 1.011993 2.1089216 772.4697 1.0096269
## s 1.62057535 0.2205728 1.300528 1.9950903 2079.4645 1.002173713M2. Compare the models you fit just above, using WAIC. Can you reconcile the differences in WAIC with the posterior distributions of the models?
compare(m1, m2, m3, m4, m5)## WAIC SE dWAIC dSE pWAIC weight
## m1 1009.594 38.17937 0.0000000 NA 37.97083 0.2667617
## m4 1009.904 38.21583 0.3102183 0.7012334 37.98567 0.2284339
## m5 1010.464 38.23223 0.8696486 0.5730181 38.26083 0.1726958
## m3 1010.513 38.36772 0.9189838 0.6881940 38.37615 0.1684879
## m2 1010.571 38.25517 0.9776104 0.3026617 38.44717 0.163620713M3. 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?) 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.
m_C <- 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 = data, chains = 4
)##
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## Chain 4:## Computing WAICprecis(m_C)## 48 vector or matrix parameters hidden. Use depth=2 to show them.## mean sd 5.5% 94.5% n_eff Rhat4
## x 1.387624 0.2563312 0.9888819 1.795822 4702.097 0.9998669
## s 1.623885 0.2133609 1.3035881 1.986853 3071.704 1.0007370compare(m1,m2,m3,m4,m5,m_C)## WAIC SE dWAIC dSE pWAIC weight
## m1 1009.594 38.17937 0.00000000 NA 37.97083 0.2109276
## m_C 1009.609 38.15278 0.01546044 0.2963405 37.89462 0.2093034
## m4 1009.904 38.21583 0.31021827 0.7012334 37.98567 0.1806219
## m5 1010.464 38.23223 0.86964860 0.5730181 38.26083 0.1365500
## m3 1010.513 38.36772 0.91898376 0.6881940 38.37615 0.1332228
## m2 1010.571 38.25517 0.97761039 0.3026617 38.44717 0.129374313M4. 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. 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?
m6<- map2stan(
alist(
surv ~ dbinom(density , c),
logit(c) <- a_tank[tanks] ,
a_tank[tanks] ~ dstudent(2, a , sigma ),
a ~ dnorm(0, 1) ,
sigma ~ dcauchy(0,1)
),
data=data , chains = 4
)##
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## http://mc-stan.org/misc/warnings.html#tail-ess## Computing WAICprecis (m6)## 48 vector or matrix parameters hidden. Use depth=2 to show them.## mean sd 5.5% 94.5% n_eff Rhat4
## a 1.344656 0.2720123 0.9007529 1.771900 4046.691 1.000998
## sigma 1.271677 0.2260687 0.9478507 1.653098 2871.608 1.000255compare(m_C,m6)## WAIC SE dWAIC dSE pWAIC weight
## m_C 1009.609 38.15278 0.00000 NA 37.89462 0.6793998
## m6 1011.111 37.89449 1.50203 1.375498 38.74027 0.320600213M5. 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 this model to m13.4. What has including \(\bar{γ}\) done?
data(chimpanzees)
data=chimpanzees
m7 <- map2stan(
alist(
pulled_left ~ dnorm(mu , 1),
mu~dnorm(10,1)
),
data=data , chains = 4
)##
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## Chain 4:## Computing WAIC## Error in .local(fit, data, n, ...) :
## There appear to be no linear models hereprecis(m7)## mean sd 5.5% 94.5% n_eff Rhat4
## mu 0.598741 0.04566935 0.5260584 0.6739045 1479.831 1.000213m8 <- map2stan(
alist(
pulled_left ~ dstudent(2, mu , 1),
mu~dnorm(10,1)
),
data=data , chains = 4
)##
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## Chain 4:## Computing WAIC## Error in .local(fit, data, n, ...) :
## There appear to be no linear models here13M6. 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.202 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 the posterior distributions for these models and compare them. Can you explain the results, using the properties of the distributions?
m8 <- map2stan(
alist(
pulled_left ~ dstudent(2, mu , 1),
mu~dnorm(10,1)
),
data=data , chains = 4
)##
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## There appear to be no linear models hereprecis (m8)## mean sd 5.5% 94.5% n_eff Rhat4
## mu 0.6183638 0.04285637 0.5497513 0.6865231 1495.614 1.00416m9 <- map2stan(
alist(
pulled_left ~ dnorm(mu , 1),
mu~dstudent(2,10,1)
),
data=data , chains = 4
)##
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## There appear to be no linear models hereprecis (m9)## mean sd 5.5% 94.5% n_eff Rhat4
## mu 0.5795196 0.04320526 0.5117219 0.6487313 1403.021 1.001711m10 <- map2stan(
alist(
pulled_left ~ dstudent(2,mu , 1),
mu~dstudent(2,10,1)
),
data=data , chains = 4
)##
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## Chain 4:## Computing WAIC## Error in .local(fit, data, n, ...) :
## There appear to be no linear models hereprecis (m10)## mean sd 5.5% 94.5% n_eff Rhat4
## mu 0.6023734 0.04503476 0.5301776 0.6759715 1321.308 1.00072EXTRA 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?