Chapter 9 - Markov Chain Monte Carlo

This chapter has been an informal introduction to Markov chain Monte Carlo (MCMC) estimation. The goal has been to introduce the purpose and approach MCMC algorithms. The major algorithms introduced were the Metropolis, Gibbs sampling, and Hamiltonian Monte Carlo algorithms. Each has its advantages and disadvantages. The ulam function in the rethinking package was introduced. It uses the Stan (mc-stan.org) Hamiltonian Monte Carlo engine to fit models as they are defined in this book. General advice about diagnosing poor MCMC fits was introduced by the use of a couple of pathological examples.

Place each answer inside the code chunk (grey box). The code chunks should contain a text response or a code that completes/answers the question or activity requested. Make sure to include plots if the question requests them. Problems are labeled Easy (E), Medium (M), and Hard(H).

Finally, upon completion, name your final output .html file as: YourName_ANLY505-Year-Semester.html and publish the assignment to your R Pubs account and submit the link to Canvas. Each question is worth 5 points.

Questions

9E1. Which of the following is a requirement of the simple Metropolis algorithm?

  1. The parameters must be discrete.
  2. The likelihood function must be Gaussian.
  3. The proposal distribution must be symmetric.
# 3 The proposal distribution must be symmetric.

9E2. Gibbs sampling is more efficient than the Metropolis algorithm. How does it achieve this extra efficiency? Are there any limitations to the Gibbs sampling strategy?

# Gibbs sampling is more efficient because distribution of proposed parameter 
# values is adjusted to current parameter values. Gibbs sampler uses pairs of 
# priors and likelihoods that have analytic solutions for the posterior of an 
# individual parameter.

9E3. Which sort of parameters can Hamiltonian Monte Carlo not handle? Can you explain why?

# HMC cannot handle discrete parameters because there is no slope. 
# We can only use continuous parameters

9E4. Explain the difference between the effective number of samples, n_eff as calculated by Stan, and the actual number of samples.

# n_eff is the number of independent samples 
# Samples here means the number of iterations of the Markov chains, not 
# data points. Markov chains are typically auto-correlated, making sequential 
# samples not independent.

9E5. Which value should Rhat approach, when a chain is sampling the posterior distribution correctly?

# Rhat should approach 1 from the top.

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?

library(rethinking)
data(rugged)
d <- rugged

d$log_gdp <- log(d$rgdppc_2000)
dd <- d[ complete.cases(d$rgdppc_2000),]
dd$log_gdp_std <- dd$log_gdp/mean(dd$log_gdp)
dd$rugged_std <- dd$rugged/max(dd$rugged)
dd$cid <- ifelse( dd$cont_africa == 1, 1, 2)

m9e6 <- quap(alist(log_gdp_std ~ dnorm(mu, sigma),
                mu <- a[cid] + b[cid]*(rugged_std - 0.215),
                a[cid] ~ dnorm(1, 0.1),
                b[cid] ~ dnorm(0, 0.3),
                sigma ~ dexp(1)), data = dd)
precis(m9e6, depth = 2)
##             mean          sd        5.5%       94.5%
## a[1]   0.8865702 0.015675558  0.86151762  0.91162275
## a[2]   1.0505675 0.009936512  1.03468708  1.06644802
## b[1]   0.1324700 0.074203795  0.01387797  0.25106196
## b[2]  -0.1425876 0.054748875 -0.23008684 -0.05508829
## sigma  0.1094930 0.005935156  0.10000752  0.11897857
dat_slim <- list(log_gdp_std = dd$log_gdp_std,
                rugged_std = dd$rugged_std,
                cid = as.integer(dd$cid))
str(dat_slim)
## List of 3
##  $ log_gdp_std: num [1:170] 0.88 0.965 1.166 1.104 0.915 ...
##  $ rugged_std : num [1:170] 0.138 0.553 0.124 0.125 0.433 ...
##  $ cid        : int [1:170] 1 2 2 2 2 2 2 2 2 1 ...
m9.1 <- ulam(alist(log_gdp_std ~ dnorm(mu , sigma),
        mu <- a[cid] + b[cid]*(rugged_std - 0.215),
        a[cid] ~ dnorm(1, 0.1),
        b[cid] ~ dnorm(0, 0.3),
        sigma ~ dexp(1)), data=dat_slim, chains=4, cores=4)

show(m9.1)
## Hamiltonian Monte Carlo approximation
## 2000 samples from 4 chains
## 
## Sampling durations (seconds):
##         warmup sample total
## chain:1   0.13    0.1  0.22
## chain:2   0.13    0.1  0.23
## chain:3   0.14    0.1  0.25
## chain:4   0.13    0.1  0.22
## 
## Formula:
## log_gdp_std ~ dnorm(mu, sigma)
## mu <- a[cid] + b[cid] * (rugged_std - 0.215)
## a[cid] ~ dnorm(1, 0.1)
## b[cid] ~ dnorm(0, 0.3)
## sigma ~ dexp(1)
traceplot(m9.1)
## [1] 1000
## [1] 1
## [1] 1000
y <- c(-1,1)
set.seed(11)

m9.2 <- ulam(alist(y ~ dnorm(mu, sigma),
    mu <- alpha,
    alpha ~ dnorm(0, 1000),
    sigma ~ dexp(0.0001)), data=list(y=y), chains=3)
## 
## SAMPLING FOR MODEL '726d002e27cec1633082261fcfedb813' NOW (CHAIN 1).
## Chain 1: 
## Chain 1: Gradient evaluation took 1.6e-05 seconds
## Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.16 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)
## Chain 1: Iteration: 200 / 1000 [ 20%]  (Warmup)
## Chain 1: Iteration: 300 / 1000 [ 30%]  (Warmup)
## Chain 1: Iteration: 400 / 1000 [ 40%]  (Warmup)
## Chain 1: Iteration: 500 / 1000 [ 50%]  (Warmup)
## Chain 1: Iteration: 501 / 1000 [ 50%]  (Sampling)
## Chain 1: Iteration: 600 / 1000 [ 60%]  (Sampling)
## Chain 1: Iteration: 700 / 1000 [ 70%]  (Sampling)
## Chain 1: Iteration: 800 / 1000 [ 80%]  (Sampling)
## Chain 1: Iteration: 900 / 1000 [ 90%]  (Sampling)
## Chain 1: Iteration: 1000 / 1000 [100%]  (Sampling)
## Chain 1: 
## Chain 1:  Elapsed Time: 0.094121 seconds (Warm-up)
## Chain 1:                0.017242 seconds (Sampling)
## Chain 1:                0.111363 seconds (Total)
## Chain 1: 
## 
## SAMPLING FOR MODEL '726d002e27cec1633082261fcfedb813' NOW (CHAIN 2).
## Chain 2: 
## Chain 2: Gradient evaluation took 4e-06 seconds
## Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.04 seconds.
## Chain 2: Adjust your expectations accordingly!
## Chain 2: 
## Chain 2: 
## Chain 2: Iteration:   1 / 1000 [  0%]  (Warmup)
## Chain 2: Iteration: 100 / 1000 [ 10%]  (Warmup)
## Chain 2: Iteration: 200 / 1000 [ 20%]  (Warmup)
## Chain 2: Iteration: 300 / 1000 [ 30%]  (Warmup)
## Chain 2: Iteration: 400 / 1000 [ 40%]  (Warmup)
## Chain 2: Iteration: 500 / 1000 [ 50%]  (Warmup)
## Chain 2: Iteration: 501 / 1000 [ 50%]  (Sampling)
## Chain 2: Iteration: 600 / 1000 [ 60%]  (Sampling)
## Chain 2: Iteration: 700 / 1000 [ 70%]  (Sampling)
## Chain 2: Iteration: 800 / 1000 [ 80%]  (Sampling)
## Chain 2: Iteration: 900 / 1000 [ 90%]  (Sampling)
## Chain 2: Iteration: 1000 / 1000 [100%]  (Sampling)
## Chain 2: 
## Chain 2:  Elapsed Time: 0.114454 seconds (Warm-up)
## Chain 2:                0.021433 seconds (Sampling)
## Chain 2:                0.135887 seconds (Total)
## Chain 2: 
## 
## SAMPLING FOR MODEL '726d002e27cec1633082261fcfedb813' NOW (CHAIN 3).
## Chain 3: 
## Chain 3: Gradient evaluation took 8e-06 seconds
## Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.08 seconds.
## Chain 3: Adjust your expectations accordingly!
## Chain 3: 
## Chain 3: 
## Chain 3: Iteration:   1 / 1000 [  0%]  (Warmup)
## Chain 3: Iteration: 100 / 1000 [ 10%]  (Warmup)
## Chain 3: Iteration: 200 / 1000 [ 20%]  (Warmup)
## Chain 3: Iteration: 300 / 1000 [ 30%]  (Warmup)
## Chain 3: Iteration: 400 / 1000 [ 40%]  (Warmup)
## Chain 3: Iteration: 500 / 1000 [ 50%]  (Warmup)
## Chain 3: Iteration: 501 / 1000 [ 50%]  (Sampling)
## Chain 3: Iteration: 600 / 1000 [ 60%]  (Sampling)
## Chain 3: Iteration: 700 / 1000 [ 70%]  (Sampling)
## Chain 3: Iteration: 800 / 1000 [ 80%]  (Sampling)
## Chain 3: Iteration: 900 / 1000 [ 90%]  (Sampling)
## Chain 3: Iteration: 1000 / 1000 [100%]  (Sampling)
## Chain 3: 
## Chain 3:  Elapsed Time: 0.109743 seconds (Warm-up)
## Chain 3:                0.010902 seconds (Sampling)
## Chain 3:                0.120645 seconds (Total)
## Chain 3:
## Warning: There were 82 divergent transitions after warmup. See
## http://mc-stan.org/misc/warnings.html#divergent-transitions-after-warmup
## to find out why this is a problem and how to eliminate them.
## Warning: Examine the pairs() plot to diagnose sampling problems
## Warning: The largest R-hat is 1.07, indicating chains have not mixed.
## Running the chains for more iterations may help. See
## http://mc-stan.org/misc/warnings.html#r-hat
## Warning: 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://mc-stan.org/misc/warnings.html#bulk-ess
## Warning: 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://mc-stan.org/misc/warnings.html#tail-ess
show(m9.2)
## Hamiltonian Monte Carlo approximation
## 1500 samples from 3 chains
## 
## Sampling durations (seconds):
##         warmup sample total
## chain:1   0.09   0.02  0.11
## chain:2   0.11   0.02  0.14
## chain:3   0.11   0.01  0.12
## 
## Formula:
## y ~ dnorm(mu, sigma)
## mu <- alpha
## alpha ~ dnorm(0, 1000)
## sigma ~ dexp(1e-04)
traceplot(m9.2)

## [1] 1000
## [1] 1
## [1] 1000

9E7. Repeat the problem above, but now for a trace rank plot.

trankplot(m9.1)

trankplot(m9.2)

9M1. Re-estimate 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). Visualize the priors. Use ulam to estimate the posterior. Visualize the posteriors for both models. Does the different prior have any detectible influence on the posterior distribution of sigma? Why or why not?

data(rugged)
d <- rugged

d$log_gdp <- log(d$rgdppc_2000)
dd <- d[ complete.cases(d$rgdppc_2000),]

dd$log_gdp_std <- dd$log_gdp/mean(dd$log_gdp)
dd$rugged_std<- dd$rugged/max(dd$rugged)

dd$cid<-ifelse(dd$cont_africa==1,1,2)

m9.3 <- quap(alist(
    log_gdp_std ~ dnorm(mu, sigma ),
    mu <- a[cid] + b[cid]* (rugged_std-0.215),
    a[cid] ~ dnorm(1,0.1),
    b[cid] ~ dnorm(0,0.3),
    sigma ~ dexp(1)), data=dd)

precis(m9.3, depth=2)
##             mean          sd        5.5%       94.5%
## a[1]   0.8865660 0.015675078  0.86151419  0.91161779
## a[2]   1.0505679 0.009936208  1.03468791  1.06644787
## b[1]   0.1325350 0.074201585  0.01394649  0.25112342
## b[2]  -0.1425568 0.054747270 -0.23005354 -0.05506012
## sigma  0.1094897 0.005934696  0.10000487  0.11897445
pairs(m9.3)

m9.3_unif <- quap(alist(
    log_gdp_std ~ dnorm(mu, sigma),
    mu <- a[cid] + b[cid]*(rugged_std-0.215),
    a[cid] ~ dnorm(1,0.1),
    b[cid] ~ dnorm(0,0.3),
    sigma ~ dunif(0,1)), data=dd)

precis(m9.3_unif, depth=2)
##             mean          sd        5.5%       94.5%
## a[1]   0.8865646 0.015680645  0.86150390  0.91162530
## a[2]   1.0505685 0.009939796  1.03468276  1.06645419
## b[1]   0.1325028 0.074227013  0.01387368  0.25113189
## b[2]  -0.1425733 0.054766564 -0.23010089 -0.05504579
## sigma  0.1095296 0.005940112  0.10003617  0.11902306
pairs(m9.3_unif)

# Does not have detectable influence on the posterior distribution of sigma

9M2. Modify the terrain ruggedness model again. This time, change the prior for b[cid] to dexp(0.3). What does this do to the posterior distribution? Can you explain it?

m9.3_exp <- quap(alist(
    log_gdp_std ~ dnorm(mu, sigma),
    mu <- a[cid] + b[cid]*(rugged_std-0.215),
    a[cid] ~ dnorm(1,0.1),
    b[cid] ~ dnorm(0,0.3),
    sigma ~ dexp(0.3)), data=dd)

precis(m9.3_exp, depth=2)
##             mean          sd        5.5%       94.5%
## a[1]   0.8865649 0.015679232  0.86150647  0.91162335
## a[2]   1.0505696 0.009938884  1.03468537  1.06645388
## b[1]   0.1325026 0.074220566  0.01388385  0.25112145
## b[2]  -0.1425740 0.054761661 -0.23009371 -0.05505429
## sigma  0.1095195 0.005938737  0.10002822  0.11901072
pairs(m9.3_exp)

# There is no apparent difference in the posterior distribution

9M3. Re-estimate 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?

m9.3 <- ulam(alist(
        log_gdp_std ~ dnorm(mu, sigma),
        mu <- a[cid] + b[cid]*(rugged_std - 0.215),
        a[cid] ~ dnorm(1, 0.1),
        b[cid] ~ dnorm(0, 0.3),
        sigma ~ dexp(1)), data=dat_slim, chains=4, cores=4, warmup=5, iter=1000)
## Warning: There were 2453 divergent transitions after warmup. See
## http://mc-stan.org/misc/warnings.html#divergent-transitions-after-warmup
## to find out why this is a problem and how to eliminate them.
## Warning: There were 3 chains where the estimated Bayesian Fraction of Missing Information was low. See
## http://mc-stan.org/misc/warnings.html#bfmi-low
## Warning: Examine the pairs() plot to diagnose sampling problems
## Warning: The largest R-hat is 1.93, indicating chains have not mixed.
## Running the chains for more iterations may help. See
## http://mc-stan.org/misc/warnings.html#r-hat
## Warning: 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://mc-stan.org/misc/warnings.html#bulk-ess
## Warning: 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://mc-stan.org/misc/warnings.html#tail-ess
precis(m9.3, depth=2)
##              mean         sd        5.5%     94.5%     n_eff    Rhat4
## a[1]   0.86698875 0.04201230  0.80129303 0.9139275  2.403373 2.670109
## a[2]   1.05345781 0.01077815  1.03671127 1.0653752  4.865112 1.342103
## b[1]   0.16569448 0.26854668 -0.01984105 0.8207163  6.309010 1.635139
## b[2]  -0.01248741 0.26997667 -0.23563664 0.3949069  2.568338 2.323551
## sigma  0.15760468 0.15932286  0.10433841 0.2458100 17.284808 1.068088
pairs(m9.3)
## Warning in KernSmooth::bkde2D(x, bandwidth = bandwidth, gridsize = nbin, :
## Binning grid too coarse for current (small) bandwidth: consider increasing
## 'gridsize'

traceplot(m9.3)
## [1] 1000
## [1] 1
## [1] 1000

9H1. Run the model below and then inspect the posterior distribution and explain what it is accomplishing.

Compare the samples for the parameters a and b. Can you explain the different trace plots? If you are unfamiliar with the Cauchy distribution, you should look it up. The key feature to attend to is that it has no expected value. Can you connect this fact to the trace plot?

mp <- ulam(
 alist(
   a ~ dnorm(0,1),
   b ~ dcauchy(0,1)), data=list(y=1) , chains=1)
## 
## SAMPLING FOR MODEL '3bd3f4d287e9cccab124308e5415245c' NOW (CHAIN 1).
## Chain 1: 
## Chain 1: Gradient evaluation took 8e-06 seconds
## Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.08 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)
## Chain 1: Iteration: 200 / 1000 [ 20%]  (Warmup)
## Chain 1: Iteration: 300 / 1000 [ 30%]  (Warmup)
## Chain 1: Iteration: 400 / 1000 [ 40%]  (Warmup)
## Chain 1: Iteration: 500 / 1000 [ 50%]  (Warmup)
## Chain 1: Iteration: 501 / 1000 [ 50%]  (Sampling)
## Chain 1: Iteration: 600 / 1000 [ 60%]  (Sampling)
## Chain 1: Iteration: 700 / 1000 [ 70%]  (Sampling)
## Chain 1: Iteration: 800 / 1000 [ 80%]  (Sampling)
## Chain 1: Iteration: 900 / 1000 [ 90%]  (Sampling)
## Chain 1: Iteration: 1000 / 1000 [100%]  (Sampling)
## Chain 1: 
## Chain 1:  Elapsed Time: 0.015504 seconds (Warm-up)
## Chain 1:                0.012342 seconds (Sampling)
## Chain 1:                0.027846 seconds (Total)
## Chain 1:
## Warning: 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://mc-stan.org/misc/warnings.html#bulk-ess
## Warning: 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://mc-stan.org/misc/warnings.html#tail-ess
precis(mp, 2)
##          mean        sd      5.5%    94.5%    n_eff    Rhat4
## a -0.01488472 0.9893028 -1.753949 1.604799 158.0203 1.009934
## b -0.37941904 9.4491903 -9.045970 7.714517 263.7751 1.018285
pairs(mp)

traceplot(mp)
## [1] 1000
## [1] 1
## [1] 1000