57 minutes
SR2 :: Solutions for Chapters {9,11,12}
Setup details are described here, and the metapost about these solutions is here.
Materials
The summmer course^{1} is based off of the second edition of Statistical Rethinking by Richard McElreath. This submission covers the following exercise questions:
 Chapter 9
 E{3,4,5,6}
 M{1,2,3}
 Chapter 11
 E{1,2,3,4}
 M{2,3,4,5,6,8}
 Chapter 12
 E{4}
 H{1,2}
Packages
A colophon with details is provided at the end, but the following packages and theme parameters are used throughout.
libsUsed<c("tidyverse","tidybayes","orgutils","dagitty",
"rethinking","tidybayes.rethinking",
"ggplot2","kableExtra","dplyr","glue",
"latex2exp","data.table","printr","devtools")
invisible(lapply(libsUsed, library, character.only = TRUE));
theme_set(theme_grey(base_size=24))
set.seed(1995)
Chapter IX: Markov Chain Monte Carlo
Easy Questions (Ch9)
9E3
Which sort of parameters can Hamiltonian Monte Carlo not handle? Can you explain why?
Solution
Hamiltonian Monte Carlo is derived by adding the concept of momentum which requires that the Hessian is nonnegative, which in term requires a continuous smooth function. Thus HMC cannot handle discrete parameters by construction. More formally, the HMC requires a transform from the Ddimensional parameter space to a 2Ddimensional phase space cite:betancourtConceptualIntroductionHamiltonian2018.
9E4
Explain the difference between the effective number of samples, n_eff
as calculated by Stan, and the actual number of samples.
Solution
We will invoke the precise definition of the effective sample size cite:betancourtConceptualIntroductionHamiltonian2018
\[ ESS = \frac{N}{1+2\sum_{l=1}^{\infty}\rho_{l}} \]
Where we note that \(\rho_{l}\) is the lagl autocorrelation of \(f\) over the Markov chain (in time). In essence, this is the number of independent samples which have equivalent information of the posterior. This is relevant, because the samples from a Marko chain are sequentially correlated (autocorrelated).
9E5
Which value should Rhat
approach, when a chain is sampling the posterior distribution correctly?
Solution
The literature cite:gelmanBayesianDataAnalysis2014 often cites a value of \(1.01\) for convergence. However, newer versions of Stan tend are documented to suggest \(1.05\) since they use newer formulations of the Rhat value cite:vehtariRanknormalizationFoldingLocalization2020. It should also be noted that cite:royConvergenceDiagnosticsMarkov2020 the Rhat value does not necessarily indicate convergence, it is not a necessary and sufficient condition, but a heuristic, and should be understood as such.
HOLD 9E6
Sketch a good trace plot for a Markov chain, one that is effectively sampling from the posterior distribution. What is good about its shape? Then sketch a trace plot for a malfunctioning Markov chain. What about its shape indicates malfunction?
Solution
Recall that the “health” of a chain can be determined by the following qualities in the trace plot.
 Stationarity
 This ensures that the chain is sampling the high probability portion of the posterior distribution
 Mixing
 This ensures that the chain explores the full region
 Convergence
 Convergence implies that independent chains agree on the same region of high probability
We will require a sample model to plot.
data(rugged)
rugDat<rugged
rugDat<rugDat %>% dplyr::mutate(logGDP=log(rgdppc_2000)) %>% tidyr::drop_na() %>% dplyr::mutate(logGDP_std=logGDP/mean(logGDP),
rugged_std=rugged/max(rugged),
cid=ifelse(cont_africa==1,1,2))
datList<list(
logGDP_std=rugDat$logGDP_std,
rugged_std=rugDat$rugged_std,
cid=as.integer(rugDat$cid)
)
m91unif<ulam(
alist(
logGDP_std ~ dnorm(mu,sigma),
mu<a[cid] + b[cid]*(rugged_std0.215),
a[cid]~dnorm(1,0.1),
b[cid]~dnorm(0,0.3),
sigma~dunif(0,1)
), data=datList, chains=4, cores=4
)
We would like to check the trace and trace rank plots.
m91unif %>% traceplot
m91unif %>% trankplot
Clearly this is a good model, with well mixed chains, as can be seen in the trank and trace plots.
We will now check the plots for the unhealthy chain described in the chapter.
m9e4un<ulam(
alist(
y ~ dnorm(mu,sigma),
mu<alpha,
alpha ~ dnorm(0,1000),
sigma~dexp(0.0001)
),data=list(y=c(1,1)),chains=4,cores=4
)
SAMPLING FOR MODEL '726d002e27cec1633082261fcfedb813' NOW (CHAIN 1).
Chain 1:
Chain 1: Gradient evaluation took 8.1e05 seconds
Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.81 seconds.
Chain 1: Adjust your expectations accordingly!
Chain 1:
Chain 1:
Chain 1: Iteration: 1 / 1000 [ 0%] (Warmup)
Chain 1: Iteration: 100 / 1000 [ 10%] (Warmup)
SAMPLING FOR MODEL '726d002e27cec1633082261fcfedb813' NOW (CHAIN 2).
Chain 2:
Chain 2: Gradient evaluation took 3.6e05 seconds
Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.36 seconds.
Chain 2: Adjust your expectations accordingly!
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SAMPLING FOR MODEL '726d002e27cec1633082261fcfedb813' NOW (CHAIN 3).
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Chain 3:
Chain 3: Gradient evaluation took 2.5e05 seconds
Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.25 seconds.
Chain 3: Adjust your expectations accordingly!
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SAMPLING FOR MODEL '726d002e27cec1633082261fcfedb813' NOW (CHAIN 4).
Chain 4:
Chain 4: Gradient evaluation took 3.1e05 seconds
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Chain 4: Adjust your expectations accordingly!
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Warning messages:
1: There were 55 divergent transitions after warmup. Increasing adapt_delta above 0.95 may help. See
http://mcstan.org/misc/warnings.html#divergenttransitionsafterwarmup
2: Examine the pairs() plot to diagnose sampling problems
3: The largest Rhat is 1.08, indicating chains have not mixed.
Running the chains for more iterations may help. See
http://mcstan.org/misc/warnings.html#rhat
4: Bulk Effective Samples Size (ESS) is too low, indicating posterior means and medians may be unreliable.
Running the chains for more iterations may help. See
http://mcstan.org/misc/warnings.html#bulkess
5: Tail Effective Samples Size (ESS) is too low, indicating posterior variances and tail quantiles may be unreliable.
Running the chains for more iterations may help. See
http://mcstan.org/misc/warnings.html#tailess
m9e4un %>% traceplot
m9e4un %>% trankplot
Clearly these plots show a model which is unable to converge.
m9e4un %>% precis
mean sd 5.5% 94.5% n_eff Rhat4
alpha 22.01 267.51 414.95 316.61 154 1.02
sigma 420.97 983.52 5.39 1894.87 151 1.04
This has clear repercussions on the actual predictions as well.
Questions of Medium Complexity (Ch9)
HOLD 9M1
Reestimate the terrain ruggedness model from the chapter, but now using a uniform prior for the standard deviation, sigma
. The uniform prior should be dunif(0,1)
. Use ulam
to estimate the posterior. Does the different prior have any detectable influence on the posterior distribution of sigma
? What or why not?
Solution
Instead of using the complete.cases
formulation in the book, we will instead use a more tidyverse
friendly approach.
data(rugged)
rugDat<rugged
rugDat<rugDat %>% dplyr::mutate(logGDP=log(rgdppc_2000)) %>% tidyr::drop_na() %>% dplyr::mutate(logGDP_std=logGDP/mean(logGDP),
rugged_std=rugged/max(rugged),
cid=ifelse(cont_africa==1,1,2))
datList<list(
logGDP_std=rugDat$logGDP_std,
rugged_std=rugDat$rugged_std,
cid=as.integer(rugDat$cid)
)
We can now formulate a model with a uniform prior on sigma.
m91unif<ulam(
alist(
logGDP_std ~ dnorm(mu,sigma),
mu<a[cid] + b[cid]*(rugged_std0.215),
a[cid]~dnorm(1,0.1),
b[cid]~dnorm(0,0.3),
sigma~dunif(0,1)
), data=datList, chains=4, cores=4
)
m91exp<ulam(
alist(
logGDP_std ~ dnorm(mu,sigma),
mu<a[cid] + b[cid]*(rugged_std0.215),
a[cid]~dnorm(1,0.1),
b[cid]~dnorm(0,0.3),
sigma~dexp(1)
), data=datList, chains=4, cores=4
)
SAMPLING FOR MODEL '9b462775c5cc2badb2b667c53f2020c8' NOW (CHAIN 1).
Chain 1:
Chain 1: Gradient evaluation took 2.8e05 seconds
Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.28 seconds.
Chain 1: Adjust your expectations accordingly!
Chain 1:
Chain 1:
Chain 1: Iteration: 1 / 1000 [ 0%] (Warmup)
SAMPLING FOR MODEL '9b462775c5cc2badb2b667c53f2020c8' NOW (CHAIN 2).
Chain 2:
Chain 2: Gradient evaluation took 3.3e05 seconds
Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.33 seconds.
Chain 2: Adjust your expectations accordingly!
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Chain 2: Iteration: 1 / 1000 [ 0%] (Warmup)
SAMPLING FOR MODEL '9b462775c5cc2badb2b667c53f2020c8' NOW (CHAIN 3).
Chain 3:
Chain 3: Gradient evaluation took 2.6e05 seconds
Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.26 seconds.
Chain 3: Adjust your expectations accordingly!
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SAMPLING FOR MODEL '9b462775c5cc2badb2b667c53f2020c8' NOW (CHAIN 4).
Chain 4:
Chain 4: Gradient evaluation took 2.5e05 seconds
Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.25 seconds.
Chain 4: Adjust your expectations accordingly!
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The posterior distributions are simply:
m91exp %>% extract.samples %>% .$sigma %>% dens(.,xlab="sigma")
m91unif %>% extract.samples %>% .$sigma %>% dens(.,add=TRUE,col="blue")
mtext("posterior")
With the priors being:
m91exp %>% extract.prior %>% .$sigma %>% dens(.,xlab="sigma")
m91unif %>% extract.prior %>% .$sigma %>% dens(.,add=TRUE,col="blue")
mtext("prior")
This makes sense, since we know that uniform prior is essentially a step function between 0 and 1 with a value of 1, while the exponential function decays normally, but should actually be spiked upwards to 1 as well.
HOLD 9M2
Modify the terrain ruggedness model again. This times, change the prior for b[cid]
to dexp(0.3)
. What does this do to the posterior distribution? Can you explain it?
Solution
m92exp<ulam(
alist(
logGDP_std ~ dnorm(mu,sigma),
mu<a[cid] + b[cid]*(rugged_std0.215),
a[cid]~dnorm(1,0.1),
b[cid]~dexp(0.3),
sigma~dexp(1)
),data=datList, chains=4, cores=4
)
Priors:
m92exp %>% extract.prior %>% .$sigma %>% dens(.,xlab="sigma",col="blue")
mtext("prior")
Posterior:
m92exp %>% extract.samples %>% .$sigma %>% dens(.,xlab="sigma",col="blue")
mtext("posterior")
m92exp %>% precis(.,depth = 2)
m91exp %>% precis(.,depth = 2)
m91unif %>% precis(.,depth = 2)
mean sd 5.5% 94.5% n_eff Rhat4
a[1] 1.01 0.02 0.98 1.03 1161 1
b[1] 0.13 0.08 0.02 0.27 821 1
sigma 0.12 0.01 0.10 0.15 1097 1
mean sd 5.5% 94.5% n_eff Rhat4
a[1] 1.01 0.02 0.98 1.04 1609 1
b[1] 0.09 0.10 0.06 0.24 1518 1
sigma 0.12 0.01 0.10 0.15 1493 1
mean sd 5.5% 94.5% n_eff Rhat4
a[1] 1.01 0.02 0.97 1.04 1744 1
b[1] 0.10 0.09 0.05 0.25 1874 1
sigma 0.12 0.01 0.10 0.15 1824 1
We can see that there isn’t much difference, however, the main difference is in the b
parameter, which seems to have fewer samples, and is also no longer takes any negative values.
HOLD 9M3
Reestimate one of the Stan models from the chapter, but at different numbers of warmup
iterations. Be sure to use the same number of sampling iterations in each case. Compare the n_eff
values. How much warmup is enough?
Solution
For brevity, we will reuse the same data and model as used in the previous questions.
warmTrial<seq.int(10,10000,length.out = 10)
nSampleEff<matrix(NA,nrow=length(warmTrial),ncol=3)
nSampleEffExp<matrix(NA,nrow=length(warmTrial),ncol=3)

Uniform Model
for(i in 1:length(warmTrial)){ tmp<ulam(m91unif,chains=4,cores=4,refresh=1,warmup=warmTrial[i],iter=1000+warmTrial[i]) nSampleEff[i,]<precis(tmp,2)$n_eff }
Chain 1: Chain 1: Gradient evaluation took 9.7e05 seconds Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.97 seconds. Chain 1: Adjust your expectations accordingly! Chain 1: Chain 1: Chain 1: WARNING: No variance estimation is Chain 1: performed for num_warmup < 20 Chain 1: Chain 2: Chain 2: Gradient evaluation took 9.6e05 seconds Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.96 seconds. Chain 2: Adjust your expectations accordingly! Chain 2: Chain 2: Chain 2: WARNING: No variance estimation is Chain 2: performed for num_warmup < 20 Chain 2: Chain 3: Chain 3: Gradient evaluation took 0.000107 seconds Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 1.07 seconds. Chain 3: Adjust your expectations accordingly! Chain 3: Chain 3: Chain 3: WARNING: No variance estimation is Chain 3: performed for num_warmup < 20 Chain 3: Chain 1: Chain 1: Elapsed Time: 0.00162 seconds (Warmup) Chain 1: 0.1738 seconds (Sampling) Chain 1: 0.17542 seconds (Total) Chain 1: Chain 3: Chain 3: Elapsed Time: 0.003755 seconds (Warmup) Chain 3: 0.059869 seconds (Sampling) Chain 3: 0.063624 seconds (Total) Chain 3: Chain 4: Chain 4: Gradient evaluation took 9.2e05 seconds Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.92 seconds. Chain 4: Adjust your expectations accordingly! Chain 4: Chain 4: Chain 4: WARNING: No variance estimation is Chain 4: performed for num_warmup < 20 Chain 4: Chain 2: Chain 2: Elapsed Time: 0.003924 seconds (Warmup) Chain 2: 0.257815 seconds (Sampling) Chain 2: 0.261739 seconds (Total) Chain 2: Chain 4: Chain 4: Elapsed Time: 0.003735 seconds (Warmup) Chain 4: 0.127719 seconds (Sampling) Chain 4: 0.131454 seconds (Total) Chain 4: Chain 1: Chain 1: Gradient evaluation took 0.000209 seconds Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 2.09 seconds. Chain 1: Adjust your expectations accordingly! Chain 1: Chain 1: Chain 2: Chain 2: Gradient evaluation took 7.8e05 seconds Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.78 seconds. Chain 2: Adjust your expectations accordingly! Chain 2: Chain 2: Chain 3: Chain 3: Gradient evaluation took 6.2e05 seconds Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.62 seconds. Chain 3: Adjust your expectations accordingly! Chain 3: Chain 3: Chain 4: Chain 4: Gradient evaluation took 8.2e05 seconds Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.82 seconds. Chain 4: Adjust your expectations accordingly! Chain 4: Chain 4: Chain 1: Chain 1: Elapsed Time: 0.118609 seconds (Warmup) Chain 1: 0.075057 seconds (Sampling) Chain 1: 0.193666 seconds (Total) Chain 1: Chain 2: Chain 2: Elapsed Time: 0.122167 seconds (Warmup) Chain 2: 0.05402 seconds (Sampling) Chain 2: 0.176187 seconds (Total) Chain 2: Chain 3: Chain 3: Elapsed Time: 0.118156 seconds (Warmup) Chain 3: 0.035528 seconds (Sampling) Chain 3: 0.153684 seconds (Total) Chain 3: Chain 4: Chain 4: Elapsed Time: 0.073505 seconds (Warmup) Chain 4: 0.040445 seconds (Sampling) Chain 4: 0.11395 seconds (Total) Chain 4: Chain 1: Chain 1: Gradient evaluation took 9.8e05 seconds Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.98 seconds. Chain 1: Adjust your expectations accordingly! Chain 1: Chain 1: Chain 2: Chain 2: Gradient evaluation took 7.3e05 seconds Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.73 seconds. Chain 2: Adjust your expectations accordingly! Chain 2: Chain 2: Chain 3: Chain 3: Gradient evaluation took 0.000107 seconds Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 1.07 seconds. Chain 3: Adjust your expectations accordingly! Chain 3: Chain 3: Chain 4: Chain 4: Gradient evaluation took 0.000109 seconds Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 1.09 seconds. Chain 4: Adjust your expectations accordingly! Chain 4: Chain 4: Chain 1: Chain 1: Elapsed Time: 0.23286 seconds (Warmup) Chain 1: 0.032903 seconds (Sampling) Chain 1: 0.265763 seconds (Total) Chain 1: Chain 2: Chain 2: Elapsed Time: 0.196139 seconds (Warmup) Chain 2: 0.032946 seconds (Sampling) Chain 2: 0.229085 seconds (Total) Chain 2: Chain 3: Chain 3: Elapsed Time: 0.15298 seconds (Warmup) Chain 3: 0.042238 seconds (Sampling) Chain 3: 0.195218 seconds (Total) Chain 3: Chain 4: Chain 4: Elapsed Time: 0.109015 seconds (Warmup) Chain 4: 0.041119 seconds (Sampling) Chain 4: 0.150134 seconds (Total) Chain 4: Chain 1: Chain 1: Gradient evaluation took 7.4e05 seconds Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.74 seconds. Chain 1: Adjust your expectations accordingly! Chain 1: Chain 1: Chain 2: Chain 2: Gradient evaluation took 7.9e05 seconds Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.79 seconds. Chain 2: Adjust your expectations accordingly! Chain 2: Chain 2: Chain 3: Chain 3: Gradient evaluation took 0.000109 seconds Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 1.09 seconds. Chain 3: Adjust your expectations accordingly! Chain 3: Chain 3: Chain 4: Chain 4: Gradient evaluation took 0.000101 seconds Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 1.01 seconds. Chain 4: Adjust your expectations accordingly! Chain 4: Chain 4: Chain 1: Chain 1: Elapsed Time: 0.304209 seconds (Warmup) Chain 1: 0.038004 seconds (Sampling) Chain 1: 0.342213 seconds (Total) Chain 1: Chain 2: Chain 2: Elapsed Time: 0.274087 seconds (Warmup) Chain 2: 0.034831 seconds (Sampling) Chain 2: 0.308918 seconds (Total) Chain 2: Chain 3: Chain 3: Elapsed Time: 0.231719 seconds (Warmup) Chain 3: 0.038131 seconds (Sampling) Chain 3: 0.26985 seconds (Total) Chain 3: Chain 4: Chain 4: Elapsed Time: 0.177718 seconds (Warmup) Chain 4: 0.038546 seconds (Sampling) Chain 4: 0.216264 seconds (Total) Chain 4: Chain 1: Chain 1: Gradient evaluation took 0.000117 seconds Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 1.17 seconds. Chain 1: Adjust your expectations accordingly! Chain 1: Chain 1: Chain 2: Chain 2: Gradient evaluation took 7.3e05 seconds Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.73 seconds. Chain 2: Adjust your expectations accordingly! Chain 2: Chain 2: Chain 3: Chain 3: Gradient evaluation took 7.2e05 seconds Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.72 seconds. Chain 3: Adjust your expectations accordingly! Chain 3: Chain 3: Chain 4: Chain 4: Gradient evaluation took 7.3e05 seconds Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.73 seconds. Chain 4: Adjust your expectations accordingly! Chain 4: Chain 4: Chain 1: Chain 1: Elapsed Time: 0.322041 seconds (Warmup) Chain 1: 0.048995 seconds (Sampling) Chain 1: 0.371036 seconds (Total) Chain 1: Chain 2: Chain 2: Elapsed Time: 0.293325 seconds (Warmup) Chain 2: 0.032541 seconds (Sampling) Chain 2: 0.325866 seconds (Total) Chain 2: Chain 3: Chain 3: Elapsed Time: 0.264383 seconds (Warmup) Chain 3: 0.04051 seconds (Sampling) Chain 3: 0.304893 seconds (Total) Chain 3: Chain 4: Chain 4: Elapsed Time: 0.220301 seconds (Warmup) Chain 4: 0.040218 seconds (Sampling) Chain 4: 0.260519 seconds (Total) Chain 4: Chain 1: Chain 1: Gradient evaluation took 4.9e05 seconds Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.49 seconds. Chain 1: Adjust your expectations accordingly! Chain 1: Chain 1: Chain 2: Chain 2: Gradient evaluation took 5.1e05 seconds Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.51 seconds. Chain 2: Adjust your expectations accordingly! Chain 2: Chain 2: Chain 3: Chain 3: Gradient evaluation took 3.9e05 seconds Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.39 seconds. Chain 3: Adjust your expectations accordingly! Chain 3: Chain 3: Chain 4: Chain 4: Gradient evaluation took 3.9e05 seconds Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.39 seconds. Chain 4: Adjust your expectations accordingly! Chain 4: Chain 4: Chain 1: Chain 1: Elapsed Time: 0.306918 seconds (Warmup) Chain 1: 0.048194 seconds (Sampling) Chain 1: 0.355112 seconds (Total) Chain 1: Chain 2: Chain 2: Elapsed Time: 0.299256 seconds (Warmup) Chain 2: 0.052776 seconds (Sampling) Chain 2: 0.352032 seconds (Total) Chain 2: Chain 3: Chain 3: Elapsed Time: 0.300093 seconds (Warmup) Chain 3: 0.052132 seconds (Sampling) Chain 3: 0.352225 seconds (Total) Chain 3: Chain 4: Chain 4: Elapsed Time: 0.278897 seconds (Warmup) Chain 4: 0.053532 seconds (Sampling) Chain 4: 0.332429 seconds (Total) Chain 4: Chain 1: Chain 1: Gradient evaluation took 9.4e05 seconds Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.94 seconds. Chain 1: Adjust your expectations accordingly! Chain 1: Chain 1: Chain 2: Chain 2: Gradient evaluation took 8.4e05 seconds Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.84 seconds. Chain 2: Adjust your expectations accordingly! Chain 2: Chain 2: Chain 3: Chain 3: Gradient evaluation took 7.1e05 seconds Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.71 seconds. Chain 3: Adjust your expectations accordingly! Chain 3: Chain 3: Chain 4: Chain 4: Gradient evaluation took 6.2e05 seconds Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.62 seconds. Chain 4: Adjust your expectations accordingly! Chain 4: Chain 4: Chain 1: Chain 1: Elapsed Time: 0.437519 seconds (Warmup) Chain 1: 0.04052 seconds (Sampling) Chain 1: 0.478039 seconds (Total) Chain 1: Chain 3: Chain 3: Elapsed Time: 0.3479 seconds (Warmup) Chain 3: 0.033007 seconds (Sampling) Chain 3: 0.380907 seconds (Total) Chain 3: Chain 2: Chain 2: Elapsed Time: 0.412403 seconds (Warmup) Chain 2: 0.052948 seconds (Sampling) Chain 2: 0.465351 seconds (Total) Chain 2: Chain 4: Chain 4: Elapsed Time: 0.315601 seconds (Warmup) Chain 4: 0.038006 seconds (Sampling) Chain 4: 0.353607 seconds (Total) Chain 4: Chain 1: Chain 1: Gradient evaluation took 8.8e05 seconds Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.88 seconds. Chain 1: Adjust your expectations accordingly! Chain 1: Chain 1: Chain 2: Chain 2: Gradient evaluation took 5.8e05 seconds Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.58 seconds. Chain 2: Adjust your expectations accordingly! Chain 2: Chain 2: Chain 3: Chain 3: Gradient evaluation took 5.9e05 seconds Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.59 seconds. Chain 3: Adjust your expectations accordingly! Chain 3: Chain 3: Chain 4: Chain 4: Gradient evaluation took 4.4e05 seconds Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.44 seconds. Chain 4: Adjust your expectations accordingly! Chain 4: Chain 4: Chain 1: Chain 1: Elapsed Time: 0.387624 seconds (Warmup) Chain 1: 0.033915 seconds (Sampling) Chain 1: 0.421539 seconds (Total) Chain 1: Chain 2: Chain 2: Elapsed Time: 0.385989 seconds (Warmup) Chain 2: 0.044089 seconds (Sampling) Chain 2: 0.430078 seconds (Total) Chain 2: Chain 3: Chain 3: Elapsed Time: 0.409957 seconds (Warmup) Chain 3: 0.039642 seconds (Sampling) Chain 3: 0.449599 seconds (Total) Chain 3: Chain 4: Chain 4: Elapsed Time: 0.363549 seconds (Warmup) Chain 4: 0.042624 seconds (Sampling) Chain 4: 0.406173 seconds (Total) Chain 4: Chain 1: Chain 1: Gradient evaluation took 7.8e05 seconds Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.78 seconds. Chain 1: Adjust your expectations accordingly! Chain 1: Chain 1: Chain 2: Chain 2: Gradient evaluation took 5.1e05 seconds Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.51 seconds. Chain 2: Adjust your expectations accordingly! Chain 2: Chain 2: Chain 3: Chain 3: Gradient evaluation took 4.7e05 seconds Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.47 seconds. Chain 3: Adjust your expectations accordingly! Chain 3: Chain 3: Chain 4: Chain 4: Gradient evaluation took 5.1e05 seconds Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.51 seconds. Chain 4: Adjust your expectations accordingly! Chain 4: Chain 4: Chain 2: Chain 2: Elapsed Time: 0.355205 seconds (Warmup) Chain 2: 0.051537 seconds (Sampling) Chain 2: 0.406742 seconds (Total) Chain 2: Chain 1: Chain 1: Elapsed Time: 0.394264 seconds (Warmup) Chain 1: 0.037432 seconds (Sampling) Chain 1: 0.431696 seconds (Total) Chain 1: Chain 3: Chain 3: Elapsed Time: 0.383287 seconds (Warmup) Chain 3: 0.036322 seconds (Sampling) Chain 3: 0.419609 seconds (Total) Chain 3: Chain 4: Chain 4: Elapsed Time: 0.335487 seconds (Warmup) Chain 4: 0.048271 seconds (Sampling) Chain 4: 0.383758 seconds (Total) Chain 4: Chain 1: Chain 1: Gradient evaluation took 5e05 seconds Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.5 seconds. Chain 1: Adjust your expectations accordingly! Chain 1: Chain 1: Chain 2: Chain 2: Gradient evaluation took 5.5e05 seconds Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.55 seconds. Chain 2: Adjust your expectations accordingly! Chain 2: Chain 2: Chain 3: Chain 3: Gradient evaluation took 6.3e05 seconds Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.63 seconds. Chain 3: Adjust your expectations accordingly! Chain 3: Chain 3: Chain 4: Chain 4: Gradient evaluation took 5.1e05 seconds Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.51 seconds. Chain 4: Adjust your expectations accordingly! Chain 4: Chain 4: Chain 1: Chain 1: Elapsed Time: 0.454863 seconds (Warmup) Chain 1: 0.037816 seconds (Sampling) Chain 1: 0.492679 seconds (Total) Chain 1: Chain 2: Chain 2: Elapsed Time: 0.415499 seconds (Warmup) Chain 2: 0.037093 seconds (Sampling) Chain 2: 0.452592 seconds (Total) Chain 2: Chain 3: Chain 3: Elapsed Time: 0.363295 seconds (Warmup) Chain 3: 0.061002 seconds (Sampling) Chain 3: 0.424297 seconds (Total) Chain 3: Chain 4: Chain 4: Elapsed Time: 0.462541 seconds (Warmup) Chain 4: 0.043532 seconds (Sampling) Chain 4: 0.506073 seconds (Total) Chain 4: Warning messages: 1: There were 470 divergent transitions after warmup. Increasing adapt_delta above 0.95 may help. See http://mcstan.org/misc/warnings.html#divergenttransitionsafterwarmup 2: There were 1 chains where the estimated Bayesian Fraction of Missing Information was low. See http://mcstan.org/misc/warnings.html#bfmilow 3: Examine the pairs() plot to diagnose sampling problems 4: The largest Rhat is 1.05, indicating chains have not mixed. Running the chains for more iterations may help. See http://mcstan.org/misc/warnings.html#rhat 5: Bulk Effective Samples Size (ESS) is too low, indicating posterior means and medians may be unreliable. Running the chains for more iterations may help. See http://mcstan.org/misc/warnings.html#bulkess 6: Tail Effective Samples Size (ESS) is too low, indicating posterior variances and tail quantiles may be unreliable. Running the chains for more iterations may help. See http://mcstan.org/misc/warnings.html#tailess
nSampleEff %>% tibble(nWarmup=warmTrial)
[90m# A tibble: 10 x 2[39m .[,"a[1]"] [,"b[1]"] [,"sigma"] nWarmup [3m[90m<dbl>[39m[23m [3m[90m<dbl>[39m[23m [3m[90m<dbl>[39m[23m [3m[90m<int>[39m[23m [90m 1[39m [4m1[24m051. 399. 45.8 10 [90m 2[39m [4m3[24m101. [4m3[24m085. [4m3[24m135. [4m1[24m120 [90m 3[39m [4m3[24m515. [4m3[24m529. [4m3[24m214. [4m2[24m230 [90m 4[39m [4m3[24m122. [4m3[24m277. [4m3[24m522. [4m3[24m340 [90m 5[39m [4m3[24m145. [4m3[24m382. [4m3[24m322. [4m4[24m450 [90m 6[39m [4m3[24m378. [4m3[24m193. [4m3[24m701. [4m5[24m560 [90m 7[39m [4m3[24m299. [4m3[24m539. [4m3[24m149. [4m6[24m670 [90m 8[39m [4m3[24m570. [4m3[24m050. [4m3[24m079. [4m7[24m780 [90m 9[39m [4m3[24m247. [4m3[24m148. [4m3[24m340. [4m8[24m890 [90m10[39m [4m3[24m159. [4m2[24m929. [4m2[24m960. [4m1[24m[4m0[24m000

Exponential Model
for(i in 1:length(warmTrial)){ tmp<ulam(m91exp,chains=4,cores=4,refresh=1,warmup=warmTrial[i],iter=1000+warmTrial[i]) nSampleEffExp[i,]<precis(tmp,2)$n_eff }
Chain 1: Chain 1: Gradient evaluation took 3e05 seconds Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.3 seconds. Chain 1: Adjust your expectations accordingly! Chain 1: Chain 1: Chain 1: WARNING: No variance estimation is Chain 1: performed for num_warmup < 20 Chain 1: Chain 2: Chain 2: Gradient evaluation took 2.8e05 seconds Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.28 seconds. Chain 2: Adjust your expectations accordingly! Chain 2: Chain 2: Chain 2: WARNING: No variance estimation is Chain 2: performed for num_warmup < 20 Chain 2: Chain 3: Chain 3: Gradient evaluation took 2.7e05 seconds Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.27 seconds. Chain 3: Adjust your expectations accordingly! Chain 3: Chain 3: Chain 3: WARNING: No variance estimation is Chain 3: performed for num_warmup < 20 Chain 3: Chain 4: Chain 4: Gradient evaluation took 2.5e05 seconds Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.25 seconds. Chain 4: Adjust your expectations accordingly! Chain 4: Chain 4: Chain 4: WARNING: No variance estimation is Chain 4: performed for num_warmup < 20 Chain 4: Chain 1: Chain 1: Elapsed Time: 0.000825 seconds (Warmup) Chain 1: 0.074412 seconds (Sampling) Chain 1: 0.075237 seconds (Total) Chain 1: Chain 3: Chain 3: Elapsed Time: 0.000433 seconds (Warmup) Chain 3: 0.05124 seconds (Sampling) Chain 3: 0.051673 seconds (Total) Chain 3: Chain 4: Chain 4: Elapsed Time: 0.001827 seconds (Warmup) Chain 4: 0.059028 seconds (Sampling) Chain 4: 0.060855 seconds (Total) Chain 4: Chain 2: Chain 2: Elapsed Time: 0.005092 seconds (Warmup) Chain 2: 0.111894 seconds (Sampling) Chain 2: 0.116986 seconds (Total) Chain 2: Chain 1: Chain 1: Gradient evaluation took 2.7e05 seconds Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.27 seconds. Chain 1: Adjust your expectations accordingly! Chain 1: Chain 1: Chain 2: Chain 2: Gradient evaluation took 2.7e05 seconds Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.27 seconds. Chain 2: Adjust your expectations accordingly! Chain 2: Chain 2: Chain 3: Chain 3: Gradient evaluation took 2.7e05 seconds Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.27 seconds. Chain 3: Adjust your expectations accordingly! Chain 3: Chain 3: Chain 4: Chain 4: Gradient evaluation took 3.3e05 seconds Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.33 seconds. Chain 4: Adjust your expectations accordingly! Chain 4: Chain 4: Chain 1: Chain 1: Elapsed Time: 0.048625 seconds (Warmup) Chain 1: 0.03552 seconds (Sampling) Chain 1: 0.084145 seconds (Total) Chain 1: Chain 2: Chain 2: Elapsed Time: 0.045476 seconds (Warmup) Chain 2: 0.03068 seconds (Sampling) Chain 2: 0.076156 seconds (Total) Chain 2: Chain 3: Chain 3: Elapsed Time: 0.059723 seconds (Warmup) Chain 3: 0.034697 seconds (Sampling) Chain 3: 0.09442 seconds (Total) Chain 3: Chain 4: Chain 4: Elapsed Time: 0.055658 seconds (Warmup) Chain 4: 0.030508 seconds (Sampling) Chain 4: 0.086166 seconds (Total) Chain 4: Chain 1: Chain 1: Gradient evaluation took 3.7e05 seconds Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.37 seconds. Chain 1: Adjust your expectations accordingly! Chain 1: Chain 1: Chain 2: Chain 2: Gradient evaluation took 2.8e05 seconds Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.28 seconds. Chain 2: Adjust your expectations accordingly! Chain 2: Chain 2: Chain 3: Chain 3: Gradient evaluation took 2.6e05 seconds Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.26 seconds. Chain 3: Adjust your expectations accordingly! Chain 3: Chain 3: Chain 4: Chain 4: Gradient evaluation took 2.6e05 seconds Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.26 seconds. Chain 4: Adjust your expectations accordingly! Chain 4: Chain 4: Chain 1: Chain 1: Elapsed Time: 0.10462 seconds (Warmup) Chain 1: 0.035886 seconds (Sampling) Chain 1: 0.140506 seconds (Total) Chain 1: Chain 3: Chain 3: Elapsed Time: 0.08377 seconds (Warmup) Chain 3: 0.036938 seconds (Sampling) Chain 3: 0.120708 seconds (Total) Chain 3: Chain 2: Chain 2: Elapsed Time: 0.096442 seconds (Warmup) Chain 2: 0.0419 seconds (Sampling) Chain 2: 0.138342 seconds (Total) Chain 2: Chain 4: Chain 4: Elapsed Time: 0.091513 seconds (Warmup) Chain 4: 0.062068 seconds (Sampling) Chain 4: 0.153581 seconds (Total) Chain 4: Chain 1: Chain 1: Gradient evaluation took 3.1e05 seconds Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.31 seconds. Chain 1: Adjust your expectations accordingly! Chain 1: Chain 1: Chain 2: Chain 2: Gradient evaluation took 3.1e05 seconds Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.31 seconds. Chain 2: Adjust your expectations accordingly! Chain 2: Chain 2: Chain 3: Chain 3: Gradient evaluation took 2.6e05 seconds Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.26 seconds. Chain 3: Adjust your expectations accordingly! Chain 3: Chain 3: Chain 4: Chain 4: Gradient evaluation took 2.4e05 seconds Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.24 seconds. Chain 4: Adjust your expectations accordingly! Chain 4: Chain 4: Chain 1: Chain 1: Elapsed Time: 0.12662 seconds (Warmup) Chain 1: 0.033816 seconds (Sampling) Chain 1: 0.160436 seconds (Total) Chain 1: Chain 2: Chain 2: Elapsed Time: 0.119929 seconds (Warmup) Chain 2: 0.048545 seconds (Sampling) Chain 2: 0.168474 seconds (Total) Chain 2: Chain 4: Chain 4: Elapsed Time: 0.118087 seconds (Warmup) Chain 4: 0.040127 seconds (Sampling) Chain 4: 0.158214 seconds (Total) Chain 4: Chain 3: Chain 3: Elapsed Time: 0.131859 seconds (Warmup) Chain 3: 0.053882 seconds (Sampling) Chain 3: 0.185741 seconds (Total) Chain 3: Chain 1: Chain 1: Gradient evaluation took 2.8e05 seconds Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.28 seconds. Chain 1: Adjust your expectations accordingly! Chain 1: Chain 1: Chain 2: Chain 2: Gradient evaluation took 3.1e05 seconds Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.31 seconds. Chain 2: Adjust your expectations accordingly! Chain 2: Chain 2: Chain 3: Chain 3: Gradient evaluation took 3.8e05 seconds Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.38 seconds. Chain 3: Adjust your expectations accordingly! Chain 3: Chain 3: Chain 4: Chain 4: Gradient evaluation took 2.8e05 seconds Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.28 seconds. Chain 4: Adjust your expectations accordingly! Chain 4: Chain 4: Chain 1: Chain 1: Elapsed Time: 0.178377 seconds (Warmup) Chain 1: 0.051505 seconds (Sampling) Chain 1: 0.229882 seconds (Total) Chain 1: Chain 2: Chain 2: Elapsed Time: 0.19769 seconds (Warmup) Chain 2: 0.028853 seconds (Sampling) Chain 2: 0.226543 seconds (Total) Chain 2: Chain 3: Chain 3: Elapsed Time: 0.209976 seconds (Warmup) Chain 3: 0.049889 seconds (Sampling) Chain 3: 0.259865 seconds (Total) Chain 3: Chain 4: Chain 4: Elapsed Time: 0.233406 seconds (Warmup) Chain 4: 0.03065 seconds (Sampling) Chain 4: 0.264056 seconds (Total) Chain 4: Chain 1: Chain 1: Gradient evaluation took 3.1e05 seconds Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.31 seconds. Chain 1: Adjust your expectations accordingly! Chain 1: Chain 1: Chain 2: Chain 2: Gradient evaluation took 3e05 seconds Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.3 seconds. Chain 2: Adjust your expectations accordingly! Chain 2: Chain 2: Chain 3: Chain 3: Gradient evaluation took 2.9e05 seconds Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.29 seconds. Chain 3: Adjust your expectations accordingly! Chain 3: Chain 3: Chain 4: Chain 4: Gradient evaluation took 3.3e05 seconds Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.33 seconds. Chain 4: Adjust your expectations accordingly! Chain 4: Chain 4: Chain 1: Chain 1: Elapsed Time: 0.226122 seconds (Warmup) Chain 1: 0.031202 seconds (Sampling) Chain 1: 0.257324 seconds (Total) Chain 1: Chain 2: Chain 2: Elapsed Time: 0.229103 seconds (Warmup) Chain 2: 0.028798 seconds (Sampling) Chain 2: 0.257901 seconds (Total) Chain 2: Chain 3: Chain 3: Elapsed Time: 0.238441 seconds (Warmup) Chain 3: 0.03459 seconds (Sampling) Chain 3: 0.273031 seconds (Total) Chain 3: Chain 4: Chain 4: Elapsed Time: 0.228212 seconds (Warmup) Chain 4: 0.036574 seconds (Sampling) Chain 4: 0.264786 seconds (Total) Chain 4: Chain 1: Chain 1: Gradient evaluation took 3e05 seconds Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.3 seconds. Chain 1: Adjust your expectations accordingly! Chain 1: Chain 1: Chain 2: Chain 2: Gradient evaluation took 2.7e05 seconds Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.27 seconds. Chain 2: Adjust your expectations accordingly! Chain 2: Chain 2: Chain 3: Chain 3: Gradient evaluation took 2.8e05 seconds Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.28 seconds. Chain 3: Adjust your expectations accordingly! Chain 3: Chain 3: Chain 4: Chain 4: Gradient evaluation took 2.4e05 seconds Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.24 seconds. Chain 4: Adjust your expectations accordingly! Chain 4: Chain 4: Chain 1: Chain 1: Elapsed Time: 0.285094 seconds (Warmup) Chain 1: 0.048246 seconds (Sampling) Chain 1: 0.33334 seconds (Total) Chain 1: Chain 4: Chain 4: Elapsed Time: 0.263814 seconds (Warmup) Chain 4: 0.034411 seconds (Sampling) Chain 4: 0.298225 seconds (Total) Chain 4: Chain 2: Chain 2: Elapsed Time: 0.305613 seconds (Warmup) Chain 2: 0.039627 seconds (Sampling) Chain 2: 0.34524 seconds (Total) Chain 2: Chain 3: Chain 3: Elapsed Time: 0.314974 seconds (Warmup) Chain 3: 0.040995 seconds (Sampling) Chain 3: 0.355969 seconds (Total) Chain 3: Chain 1: Chain 1: Gradient evaluation took 3e05 seconds Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.3 seconds. Chain 1: Adjust your expectations accordingly! Chain 1: Chain 1: Chain 2: Chain 2: Gradient evaluation took 2.8e05 seconds Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.28 seconds. Chain 2: Adjust your expectations accordingly! Chain 2: Chain 2: Chain 3: Chain 3: Gradient evaluation took 3e05 seconds Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.3 seconds. Chain 3: Adjust your expectations accordingly! Chain 3: Chain 3: Chain 4: Chain 4: Gradient evaluation took 2.7e05 seconds Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.27 seconds. Chain 4: Adjust your expectations accordingly! Chain 4: Chain 4: Chain 1: Chain 1: Elapsed Time: 0.293873 seconds (Warmup) Chain 1: 0.028447 seconds (Sampling) Chain 1: 0.32232 seconds (Total) Chain 1: Chain 3: Chain 3: Elapsed Time: 0.288431 seconds (Warmup) Chain 3: 0.04518 seconds (Sampling) Chain 3: 0.333611 seconds (Total) Chain 3: Chain 2: Chain 2: Elapsed Time: 0.36752 seconds (Warmup) Chain 2: 0.027453 seconds (Sampling) Chain 2: 0.394973 seconds (Total) Chain 2: Chain 4: Chain 4: Elapsed Time: 0.383786 seconds (Warmup) Chain 4: 0.032154 seconds (Sampling) Chain 4: 0.41594 seconds (Total) Chain 4: Chain 1: Chain 1: Gradient evaluation took 3.1e05 seconds Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.31 seconds. Chain 1: Adjust your expectations accordingly! Chain 1: Chain 1: Chain 2: Chain 2: Gradient evaluation took 2.7e05 seconds Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.27 seconds. Chain 2: Adjust your expectations accordingly! Chain 2: Chain 2: Chain 3: Chain 3: Gradient evaluation took 2.7e05 seconds Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.27 seconds. Chain 3: Adjust your expectations accordingly! Chain 3: Chain 3: Chain 4: Chain 4: Gradient evaluation took 3.1e05 seconds Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.31 seconds. Chain 4: Adjust your expectations accordingly! Chain 4: Chain 4: Chain 1: Chain 1: Elapsed Time: 0.354637 seconds (Warmup) Chain 1: 0.027996 seconds (Sampling) Chain 1: 0.382633 seconds (Total) Chain 1: Chain 2: Chain 2: Elapsed Time: 0.339298 seconds (Warmup) Chain 2: 0.030313 seconds (Sampling) Chain 2: 0.369611 seconds (Total) Chain 2: Chain 4: Chain 4: Elapsed Time: 0.319312 seconds (Warmup) Chain 4: 0.027904 seconds (Sampling) Chain 4: 0.347216 seconds (Total) Chain 4: Chain 3: Chain 3: Elapsed Time: 0.333814 seconds (Warmup) Chain 3: 0.032747 seconds (Sampling) Chain 3: 0.366561 seconds (Total) Chain 3: Chain 1: Chain 1: Gradient evaluation took 3.8e05 seconds Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.38 seconds. Chain 1: Adjust your expectations accordingly! Chain 1: Chain 1: Chain 2: Chain 2: Gradient evaluation took 3.7e05 seconds Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.37 seconds. Chain 2: Adjust your expectations accordingly! Chain 2: Chain 2: Chain 3: Chain 3: Gradient evaluation took 2.6e05 seconds Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.26 seconds. Chain 3: Adjust your expectations accordingly! Chain 3: Chain 3: Chain 4: Chain 4: Gradient evaluation took 3.1e05 seconds Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.31 seconds. Chain 4: Adjust your expectations accordingly! Chain 4: Chain 4: Chain 1: Chain 1: Elapsed Time: 0.343002 seconds (Warmup) Chain 1: 0.028223 seconds (Sampling) Chain 1: 0.371225 seconds (Total) Chain 1: Chain 3: Chain 3: Elapsed Time: 0.305501 seconds (Warmup) Chain 3: 0.027824 seconds (Sampling) Chain 3: 0.333325 seconds (Total) Chain 3: Chain 2: Chain 2: Elapsed Time: 0.358116 seconds (Warmup) Chain 2: 0.02975 seconds (Sampling) Chain 2: 0.387866 seconds (Total) Chain 2: Chain 4: Chain 4: Elapsed Time: 0.330673 seconds (Warmup) Chain 4: 0.030407 seconds (Sampling) Chain 4: 0.36108 seconds (Total) Chain 4: Warning messages: 1: There were 14 divergent transitions after warmup. Increasing adapt_delta above 0.95 may help. See http://mcstan.org/misc/warnings.html#divergenttransitionsafterwarmup 2: Examine the pairs() plot to diagnose sampling problems
nSampleEffExp %>% tibble(nWarmup=warmTrial)
[90m# A tibble: 10 x 2[39m .[,1] [,2] [,3] nWarmup [3m[90m<dbl>[39m[23m [3m[90m<dbl>[39m[23m [3m[90m<dbl>[39m[23m [3m[90m<int>[39m[23m [90m 1[39m [4m3[24m063. [4m1[24m265. [4m1[24m045. 10 [90m 2[39m [4m3[24m238. [4m2[24m909. [4m3[24m425. [4m1[24m120 [90m 3[39m [4m3[24m268. [4m2[24m949. [4m2[24m966. [4m2[24m230 [90m 4[39m [4m3[24m123. [4m3[24m257. [4m3[24m218. [4m3[24m340 [90m 5[39m [4m3[24m227. [4m3[24m449. [4m3[24m554. [4m4[24m450 [90m 6[39m [4m3[24m494. [4m3[24m743. [4m3[24m343. [4m5[24m560 [90m 7[39m [4m3[24m200. [4m3[24m012. [4m2[24m879. [4m6[24m670 [90m 8[39m [4m3[24m210. [4m3[24m207. [4m3[24m125. [4m7[24m780 [90m 9[39m [4m2[24m919. [4m3[24m329. [4m3[24m060. [4m8[24m890 [90m10[39m [4m2[24m951. [4m3[24m461. [4m3[24m078. [4m1[24m[4m0[24m000
It is important to note that the divergent transitions are probably why the number of effective samples decrease in the last two rows.

Results
We can see that the number of effective samples increases almost constantly. This is probably due to correlations in the chain, which are removed during the warmup period.
Chapter XI: God Spiked The Integers
Easy Questions (Ch11)
11E1
If an event has probability \(0.35\), what are the logodds of this event?
Solution
log(0.35/(10.35))
[1] 0.6190392
11E2
If an event has logodds \(3.2\), what is the probability of this event?
Solution
logistic(3.2)
[1] 0.9608343
11E3
Suppose that a coefficient in a logistic regression has value \(1.7\). What does this imply about the proportional change in odds of the outcome?
Solution
exp(1.7)
[1] 5.473947
Note that this is not really the change in the variable, but the proportional odds.
HOLD 11E4
Why do Poisson regressions sometimes require the use of an offset? Provide an example.
Solution
The Poisson distribution is often understood as a limiting distribution of the Binomial where \(λ=np\) as \(n→∞\) and \(p→0\). The single parameter thus expresses the expected value, but is often used to encode different timesteps as well. Essentially, the distribution assumes a constant rate in time or space, and thus the change in exposure, is expressed by the offset, which is the logarithm of the exposure.
For any case where samples are drawn from populations which have different aggregation time periods but are still within the purview of a Poisson distribution, the offset is a natural way of expressing these.
To leverage the example of the book, when constructing a model to account for the fact that one Monastery calculates their averages on a weekly basis, while the other averages by day, this constraint should be modeled by having differing offsets.
Questions of Medium Complexity (Ch11)
HOLD 11M2
If a coefficient in a Poisson regression has values \(1.7\), what does this imply about the change in the outcome?
Solution
exp(1.7)
[1] 5.473947
The coefficient in a Poisson regression implies that the proportional change in the count will be ~5.474 when the predictor variable increase by one unit.
HOLD 11M3
Explain why the logit link is appropriate for a binomial generalized linear model.
Solution
The logit link essentially connects a parameter constrained between zero and one and the real space. The logit function is defined as:
\[\mathrm{logit}(pᵢ)=\log{\frac{pᵢ}{1pᵢ}}\]
Where \(pᵢ\) is a probability mass. This link makes sense for a GLM since the predicted value is a probability distribution parameter, and we would like to obtain this from a linear model which spans the entire set of real numbers.
curve(logit,from=0.5,to=1.5)
The link function maps a parameter onto a linear model.
HOLD 11M4
Explain why the log is appropriate for a Poisson generalized linear model.
Solution
The log function ensures that the parameter cannot take values which are less than zero. This is a natural consequence of the function definition.
curve(log,from=0.5,to=100000)
The log link assumes that the parameter value is the exponentiation of the linear model.
This makes sense for a Poisson GLM as the Poisson distribution does not accept negative values.
HOLD 11M5
What would it imply to use a logit link for the mean of a Poisson generalized linear model? Can you think of a real research problem for which this would make sense?
Solution
We should write this out more explicitly.
This implies that the mean μ lies between zero and one. Since the Poisson distribution is defined by a single parameter, this does limit the model outputs. The premise of a Poisson regression problem is that the GLM models a count with an unknown maximum, so it does seem to be a very severe restriction.
To my mind this is feasible for constrained problems, where the Poisson distribution is to be followed but only within a particular range for some reason, and when the Binomial (of which the Poisson is a special case), decreases too slowly.
It was mentioned on the class forums, that the COVID19 problem was modeled with a two parameter generalized link function, i.e. \(\log{\frac{p}{Sp}}\) which essentially constrains the model to have Poisson dynamics but with an output mean between 0 and S.
HOLD 11M6
State the constraints for which the binomial and Poisson distributions have maximum entropy. Are the constraints different at all for binomial and Poisson? Why or why not?
Solution
The Binomial distribution is defined to be the maximum entropy distributions are:
 Discrete binary outcomes
 Constant probability (or expectation)
This is defined by the number of outcomes (n) as well as the probability (p). The experiment is essentially reduced to a series of independent and identical Bernoulli trials with only two outcomes. The Poisson distribution is derived as a limiting form of the Binomial, where \(n→∞\) and \(p→0\). Since this does not change the underlying constraints, this is still a maximum entropy distribution.
HOLD 11M8
Revisit the data(Kline)
islands example. This time drop Hawaii from the sample and refit the models. What changes do you observe?
Solution
data(Kline)
kDat<Kline
kDat<kDat %>% dplyr::mutate(cid=ifelse(contact=="high",2,1),
stdPop=standardize(log(population))) %>% filter(culture!="Hawaii")
datList<list(
totTools=kDat$total_tools,
stdPop=kDat$stdPop,
cid=as.integer(kDat$cid)
)
We can now fit this.
m11m10res<ulam(
alist(
totTools ~ dpois(lambda),
log(lambda)<a[cid]+b[cid]*stdPop,
a[cid] ~ dnorm(3,0.5),
b[cid] ~ dnorm(0,0.2)
),data=datList, chains=4, cores=4
)
m11m10res %>% precis(2)
mean sd 5.5% 94.5% n_eff Rhat4
a[1] 3.18 0.12 2.99 3.37 1621 1
a[2] 3.61 0.08 3.48 3.73 1962 1
b[1] 0.19 0.13 0.01 0.39 1639 1
b[2] 0.19 0.16 0.06 0.44 1830 1
We see that the slopes are now the same, which makes sense since in this dataset Hawaii was the only outlier.
Chapter XII: Monsters and Mixtures
Easy Questions (Ch12)
HOLD 12E4
Overdispersion is common in count data. Give an example of a natural process that might produce overdispersed counts. Can you also give an example of a process that might produce /under/dispersed counts?
Solution

Overdispersion
Over dispersion is essentially the occurrence of greater variability than accounted for based on the statistical model. The presence of overdispersion is typically due to heterogeneity in populations. This heterogeneity may arise from simple aggregation issues like in the case considered in the text, of Monasteries which accumulate data weekly or daily, inspite of following the same Poisson model.

Underdispersion
Under dispersion is essentially the occurrence of less variability than accounted for based on the statistical model. The clearest example of underdispersion is from the draws of an MCMC sampler. The number of effective samples is typically lower than the number of samples, as the data is highly correlated (autocorrelated) as the sampler draws sequential samples. For a count model, if a hidden rate limiting variable exists and has not been accounted for, then the variation in counts is lowered, and will show up as underdispersion.
Hard Questions (Ch12)
12H1
In 2014, a paper was published that was entitle “Female hurricanes are deadlier than male hurricanes.” As the title suggests, the paper claimed that hurricanes with female names have caused greater loss of life, and the explanation given is that people unconsciously rate female hurricanes as less dangerous and so are less likely to evacuate. Statisticians severely criticized the paper after publication. Here, you’ll explore the complete data used in the paper and consider the hypothesis that hurricanes with female names are deadlier. Load the data with:
library(rethinking)
data(Hurricanes)
Acquaint yourself with the columns by inspecting the help ?Hurricanes
. In this problem, you’ll focus on predicting deaths
using feminity
as a predictor. You can use quap
or ulam
. Compare the model to an interceptonly Poisson model of deaths
. How strong is the association between feminity of name and deaths? Which storms does the model fit (retrodict) well? Which storms does it fit poorly?
Solution
Since I have no understanding of hurricanes except that it is unlikely to have too much of an effect. I will run through some sample priors. Presumably, most hurricanes do not kill over a thousand people. Furthermore, apriori, I would not like to assume that femininity is positive or negative, so I will instead encode a belief that it shouldn’t matter much either way, ergo a Gaussian.
N<100
a<rnorm(N,1,0.5)
bF<rnorm(N,0.5,2)
seqF<seq(from=2,to=2,length.out=100)
plot(NULL,xlim=c(2,2),ylim=c(0,1000),xlab="Femininity",ylab="deaths")
for(i in 1:N) lines(seqF,exp(a[i]+bF[i]*seqF),col=grau())
This seems to be reasonable to me. It does have a bit of an unreasonable focus on 0, but it does also seem to mostly hug the xaxis in a way indicating my prior belief that it should not matter all that much. There is enough diversity in the priors to allow for stronger trends, but they are by and large unlikely.
Now we can actually use these in a model.
data(Hurricanes)
hurDat<Hurricanes %>% as.data.frame
hurDat<hurDat %>% dplyr::mutate(femStd=standardize(femininity))
datListH<list(deaths=hurDat$deaths,femStd=hurDat$femStd)
m12h1norm<ulam(alist(deaths ~ dpois(lambda),
log(lambda) < a+bF*femStd,
a ~ dnorm(1,0.5),
bF ~ dnorm(0.5,2)
),data=datListH, chains=4, cores=4,log_lik = TRUE)
SAMPLING FOR MODEL 'bd16fb771b491de48a3f8ce09fc68301' NOW (CHAIN 1).
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Chain Chain 12: : 1000 transitions using 10 leapfrog steps per transition would take 0.51 seconds.1000 transitions using 10 leapfrog steps per transition would take 0.39 seconds.
Chain Chain 12: : Adjust your expectations accordingly!Adjust your expectations accordingly!
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m12h1norm %>% precis
mean sd 5.5% 94.5% n_eff Rhat4
a 3.00 0.02 2.96 3.04 1334 1
bF 0.24 0.02 0.20 0.28 1361 1
There are several points to be noted:
 The number of effective samples is far lower than the number simulated (4000)
 The bounds are quite tight for the values
All told, we can see that the model is quite certain of the values, but given the low number of effective samples it would make sense to run this model longer.
Furthermore, the model infers that from our data, there is a positive correlation (quite a high one) between femininity and deaths.
This is best seen by actually visualizing the results.
m12h1norm %>% pairs
We should check the posterior as well.
k<PSIS(m12h1norm,pointwise=TRUE)$k
plot(hurDat$femStd,hurDat$deaths,xlab="Standardized Femininity",ylab="Deaths",col=rangi2,pch=hurDat$female, lwd=2,cex=1+normalize(k))
## Axis for predictions
ns<500
femininity<seq(from=min(hurDat$femStd), to=max(hurDat$femStd),length.out = ns)
## Female
lambda<link(m12h1norm,data=data.frame(femStd=femininity))
lmu<apply(lambda,2,mean)
lci<apply(lambda,2,PI)
lines(femininity,lmu,lty=2,lwd=1.5)
shade(lci,femininity,xpd=TRUE)
We can see that the model prediction does not actually handle the data very well, in that it is evident the model simply cannot account for the high death rate values. We have plotted the 89% interval as well (the default for PI).
We can also inspect the expect PSISk values.
m12h1norm %>% PSISk %>% summary
Some Pareto k values are very high (>1). Set pointwise=TRUE to inspect individual points.
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.1800 0.0500 0.0600 0.1521 0.1500 2.5800
Clearly, there are some points with high leverage.
Finally we can plot the posterior distribution.
posterior<m12h1norm %>% extract.samples
fem<sample(hurDat$femStd,3)
for(i in 1:10){
curve(dgamma(x,exp(posterior$a[i]+posterior$bF[i]*(fem[1]))), from=0,to=100, col="red",ylab="Density",xlab="Average deaths",add=ifelse(i==1,FALSE,TRUE))
curve(dgamma(x,exp(posterior$a[i]+posterior$bF[i]*(fem[2]))), from=0,to=100, col="blue",add=TRUE)
curve(dgamma(x,exp(posterior$a[i]+posterior$bF[i]*(fem[3]))), from=0,to=100, col="black",add=TRUE)}
legend("topright",
legend=c(sprintf("Femininity=%.3f",fem[1]), sprintf("Femininity=%.3f",fem[2]), sprintf("Femininity=%.3f",fem[3])),
col=c("red","blue","black"),
pch=19
)
As is to be expected, for each value of femininity, we have one family of gamma distributions.
HOLD 12H2
Counts are nearly always overdispersed relative to Poisson. So fit a gammaPoisson (aka negativebinomial) model to predict deaths
using feminity
. Show that the overdispersed model no longer shows as precise a positive association between feminity and deaths, with an $89$% interval that overlaps zero. Can you explain why the association diminished in strength?
Solution
Recall that the gammaPoisson has two parameters, one for the rate, and the other for the dispersion of rates. Larger values of the dispersion imply that the distribution is more similar to a pure Poisson process. For ensuring meaningful comparisons, we will keep the same priors as before. We will need a scale parameter, but we will postulate a simple exponential prior for that.
data(Hurricanes)
hurDat<Hurricanes %>% as.data.frame
hurDat<hurDat %>% dplyr::mutate(femStd=standardize(femininity))
datListH<list(deaths=hurDat$deaths,femStd=hurDat$femStd)
m12h2norm<ulam(alist(deaths ~ dgampois(lambda,scale),
log(lambda) < a+bF*femStd,
a ~ dnorm(1,0.5),
bF ~ dnorm(0.5,2),
scale ~ dexp(1)
),data=datListH, chains=4, cores=4,log_lik = TRUE)
SAMPLING FOR MODEL '5dc94bd836781d34a208695cf643c56c' NOW (CHAIN 1).
Chain 1:
Chain 1: Gradient evaluation took 0.00012 seconds
Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 1.2 seconds.
Chain 1: Adjust your expectations accordingly!
Chain 1:
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SAMPLING FOR MODEL '5dc94bd836781d34a208695cf643c56c' NOW (CHAIN 2).
Chain 2:
Chain 2: Gradient evaluation took 9.6e05 seconds
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Chain 2: Adjust your expectations accordingly!
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m12h2norm %>% precis
m12h1norm %>% precis
mean sd 5.5% 94.5% n_eff Rhat4
a 2.86 0.14 2.64 3.08 1705 1
bF 0.21 0.14 0.01 0.44 1946 1
scale 0.45 0.06 0.35 0.55 1957 1
mean sd 5.5% 94.5% n_eff Rhat4
a 3.00 0.02 2.96 3.04 1334 1
bF 0.24 0.02 0.20 0.28 1361 1
We note that the effective number of samples in the second model are greater, which implies that this model is less prone to correlations. We can quantify this with the WAIC as well.
WAIC(m12h2norm) %>% rbind(WAIC(m12h1norm)) %>% tibble(model=c("GammaPoisson","Poisson")) %>% toOrg
 WAIC  lppd  penalty  std_err  model 
++++
 710.78471929582  351.457619486557  3.93474016135277  34.6128979470592  GammaPoisson 
 4427.5667952452  2080.92835360918  132.855044013418  1009.13483879188  Poisson 
The WAIC values show that the gammaPoisson model is less likely to overfit.
We would like to see the models together.
m12h1norm %>% precis(pars=c("a","bF")) %>% plot(col="blue")
m12h2norm %>% precis(pars=c("a","bF")) %>% plot(add=TRUE,col="red")
We can see that there is little to no difference in the means, though the intervals seem wider than before. This is more clear in the coeftab
plot.
plot(coeftab(m12h1norm,m12h2norm))
An important consequence of this is that the model is no longer completely sure that there is any effect of femininity on the death count, as can be seen from the wider uncertainty interval, which includes 0.
We can also visualize the model with pairs
.
pairs(m12h2norm)
k<PSIS(m12h2norm,pointwise=TRUE)$k
plot(hurDat$femStd,hurDat$deaths,xlab="Standardized Femininity",ylab="Deaths",col=rangi2,pch=hurDat$female, lwd=2,cex=1+normalize(k))
## Axis for predictions
ns<500
femininity<seq(from=min(hurDat$femStd),to=max(hurDat$femStd),length.out = ns)
## Gamma Poisson
lambda<link(m12h2norm,data=data.frame(femStd=femininity))
lmu<apply(lambda,2,mean)
lci<apply(lambda,2,PI)
shade(lci,femininity,xpd=TRUE,col="red")
lines(femininity,lmu,lty=2,lwd=1.5,col="white")
## Poisson
lambda<link(m12h1norm,data=data.frame(femStd=femininity))
lmu<apply(lambda,2,mean)
lci<apply(lambda,2,PI)
shade(lci,femininity,xpd=TRUE,col="blue")
lines(femininity,lmu,lty=2,lwd=1.5,col="white")
Clearly, the uncertainty of the newer model is much greater, even though the predictions do not differ much. Unfortunately, both models fail to account for storms with high death counts.
We would also like to plot the predicted distributions.
posterior<m12h2norm %>% extract.samples
fem<sample(hurDat$femStd,2)
for(i in 1:100){
curve(dgamma2(x,exp(posterior$a[i]+posterior$bF[i]*(fem[1])),posterior$scale[i]), from=0,to=100,col="red",ylab="Density",xlab="Average deaths",add=ifelse(i==1,FALSE,TRUE))
curve(dgamma2(x,exp(posterior$a[i]+posterior$bF[i]*(fem[2])),posterior$scale[i]), from=0,to=100,col="blue",add=TRUE)
}
legend("topright",
legend=c(sprintf("Femininity=%.3f",fem[1]), sprintf("Femininity=%.3f",fem[2])),
col=c("red","blue"),
pch=19
)
This clearly has more spread than the previous predictions. By definition, the dispersion term tends to spread the distribution out, with higher values of the dispersion corresponding to a “true” Poisson distribution.
A: Colophon
To ensure that this document is fully reproducible at a later date, we will record the session info.
devtools::session_info()
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setting value
version R version 4.0.0 (20200424)
os Arch Linux
system x86_64, linuxgnu
ui X11
language (EN)
collate en_US.UTF8
ctype en_US.UTF8
tz Iceland
date 20200621
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package * version date lib source
arrayhelpers 1.10 20200204 [167] CRAN (R 4.0.0)
assertthat 0.2.1 20190321 [34] CRAN (R 4.0.0)
backports 1.1.6 20200405 [68] CRAN (R 4.0.0)
boot 1.324 20191220 [5] CRAN (R 4.0.0)
broom 0.5.6 20200420 [67] CRAN (R 4.0.0)
callr 3.4.3 20200328 [87] CRAN (R 4.0.0)
cellranger 1.1.0 20160727 [55] CRAN (R 4.0.0)
cli 2.0.2 20200228 [33] CRAN (R 4.0.0)
coda 0.193 20190705 [169] CRAN (R 4.0.0)
colorspace 1.41 20190318 [97] CRAN (R 4.0.0)
crayon 1.3.4 20170916 [35] CRAN (R 4.0.0)
curl 4.3 20191202 [26] CRAN (R 4.0.0)
dagitty * 0.22 20160826 [244] CRAN (R 4.0.0)
data.table * 1.12.8 20191209 [27] CRAN (R 4.0.0)
DBI 1.1.0 20191215 [77] CRAN (R 4.0.0)
dbplyr 1.4.3 20200419 [76] CRAN (R 4.0.0)
desc 1.2.0 20180501 [84] CRAN (R 4.0.0)
devtools * 2.3.0 20200410 [219] CRAN (R 4.0.0)
digest 0.6.25 20200223 [42] CRAN (R 4.0.0)
dplyr * 0.8.5 20200307 [69] CRAN (R 4.0.0)
ellipsis 0.3.0 20190920 [30] CRAN (R 4.0.0)
evaluate 0.14 20190528 [82] CRAN (R 4.0.0)
fansi 0.4.1 20200108 [36] CRAN (R 4.0.0)
forcats * 0.5.0 20200301 [29] CRAN (R 4.0.0)
fs 1.4.1 20200404 [109] CRAN (R 4.0.0)
generics 0.0.2 20181129 [71] CRAN (R 4.0.0)
ggplot2 * 3.3.0 20200305 [78] CRAN (R 4.0.0)
glue * 1.4.0 20200403 [37] CRAN (R 4.0.0)
gridExtra 2.3 20170909 [123] CRAN (R 4.0.0)
gtable 0.3.0 20190325 [79] CRAN (R 4.0.0)
haven 2.2.0 20191108 [28] CRAN (R 4.0.0)
hms 0.5.3 20200108 [44] CRAN (R 4.0.0)
htmltools 0.4.0 20191004 [112] CRAN (R 4.0.0)
httr 1.4.1 20190805 [100] CRAN (R 4.0.0)
inline 0.3.15 20180518 [162] CRAN (R 4.0.0)
jsonlite 1.6.1 20200202 [101] CRAN (R 4.0.0)
kableExtra * 1.1.0 20190316 [212] CRAN (R 4.0.0)
knitr 1.28 20200206 [113] CRAN (R 4.0.0)
latex2exp * 0.4.0 20151130 [211] CRAN (R 4.0.0)
lattice 0.2041 20200402 [6] CRAN (R 4.0.0)
lifecycle 0.2.0 20200306 [38] CRAN (R 4.0.0)
loo 2.2.0 20191219 [163] CRAN (R 4.0.0)
lubridate 1.7.8 20200406 [106] CRAN (R 4.0.0)
magrittr 1.5 20141122 [21] CRAN (R 4.0.0)
MASS 7.351.5 20191220 [7] CRAN (R 4.0.0)
matrixStats 0.56.0 20200313 [164] CRAN (R 4.0.0)
memoise 1.1.0 20170421 [229] CRAN (R 4.0.0)
modelr 0.1.6 20200222 [107] CRAN (R 4.0.0)
munsell 0.5.0 20180612 [96] CRAN (R 4.0.0)
mvtnorm 1.10 20200224 [243] CRAN (R 4.0.0)
nlme 3.1147 20200413 [11] CRAN (R 4.0.0)
orgutils * 0.41 20170321 [209] CRAN (R 4.0.0)
pillar 1.4.3 20191220 [39] CRAN (R 4.0.0)
pkgbuild 1.0.6 20191009 [86] CRAN (R 4.0.0)
pkgconfig 2.0.3 20190922 [43] CRAN (R 4.0.0)
pkgload 1.0.2 20181029 [83] CRAN (R 4.0.0)
plyr 1.8.6 20200303 [73] CRAN (R 4.0.0)
prettyunits 1.1.1 20200124 [58] CRAN (R 4.0.0)
printr * 0.1 20170519 [214] CRAN (R 4.0.0)
processx 3.4.2 20200209 [88] CRAN (R 4.0.0)
ps 1.3.2 20200213 [89] CRAN (R 4.0.0)
purrr * 0.3.4 20200417 [50] CRAN (R 4.0.0)
R6 2.4.1 20191112 [48] CRAN (R 4.0.0)
Rcpp 1.0.4.6 20200409 [10] CRAN (R 4.0.0)
readr * 1.3.1 20181221 [45] CRAN (R 4.0.0)
readxl 1.3.1 20190313 [54] CRAN (R 4.0.0)
remotes 2.1.1 20200215 [233] CRAN (R 4.0.0)
reprex 0.3.0 20190516 [108] CRAN (R 4.0.0)
rethinking * 2.01 20200606 [242] local
rlang 0.4.5 20200301 [31] CRAN (R 4.0.0)
rmarkdown 2.1 20200120 [110] CRAN (R 4.0.0)
rprojroot 1.32 20180103 [85] CRAN (R 4.0.0)
rstan * 2.19.3 20200211 [161] CRAN (R 4.0.0)
rstudioapi 0.11 20200207 [91] CRAN (R 4.0.0)
rvest 0.3.5 20191108 [120] CRAN (R 4.0.0)
scales 1.1.0 20191118 [93] CRAN (R 4.0.0)
sessioninfo 1.1.1 20181105 [231] CRAN (R 4.0.0)
shape 1.4.4 20180207 [193] CRAN (R 4.0.0)
StanHeaders * 2.19.2 20200211 [165] CRAN (R 4.0.0)
stringi 1.4.6 20200217 [52] CRAN (R 4.0.0)
stringr * 1.4.0 20190210 [74] CRAN (R 4.0.0)
svUnit 1.0.3 20200420 [168] CRAN (R 4.0.0)
testthat 2.3.2 20200302 [81] CRAN (R 4.0.0)
textutils 0.20 20200107 [210] CRAN (R 4.0.0)
tibble * 3.0.1 20200420 [32] CRAN (R 4.0.0)
tidybayes * 2.0.3 20200404 [166] CRAN (R 4.0.0)
tidybayes.rethinking * 2.0.3.9000 20200607 [246] local
tidyr * 1.0.2 20200124 [75] CRAN (R 4.0.0)
tidyselect 1.0.0 20200127 [49] CRAN (R 4.0.0)
tidyverse * 1.3.0 20191121 [66] CRAN (R 4.0.0)
usethis * 1.6.0 20200409 [238] CRAN (R 4.0.0)
V8 3.0.2 20200314 [245] CRAN (R 4.0.0)
vctrs 0.2.4 20200310 [41] CRAN (R 4.0.0)
viridisLite 0.3.0 20180201 [99] CRAN (R 4.0.0)
webshot 0.5.2 20191122 [213] CRAN (R 4.0.0)
withr 2.2.0 20200420 [90] CRAN (R 4.0.0)
xfun 0.13 20200413 [116] CRAN (R 4.0.0)
xml2 1.3.2 20200423 [122] CRAN (R 4.0.0)
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Summer of 2020 ↩︎