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. 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.
# 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 uses conjugate priors which allows it to make smarter proposals and is thus more efficient. 
# It explores posterior distribution more efficiently, so there are fewer rejections compared to base Metropolis algorithm. 
# The downside to this, is that it uses conjugate priors which might not be a good or valid prior from a scientific perspective. Also, it becomes inefficient with complex models of hundreds or more parameter.

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

# Hamiltonian Monte Carlo method cannot handle discrete parameters. Method is based on the physical metaphor of the moving particle. This particle moves in the space of parameters with speed proportional to the gradient of likelihood, so the method should be able to differentiate the space. This is not possible with discrete parameters.

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

The effective number of samples gives an estimate of the number of samples that are independent.Ideal samples are entirely uncorrelated. Since Markov chains are autocorrelated, sequential samples are not independent of each other.

# The effective number of samples gives an estimate of the number of samples that are independent.Ideal samples are entirely uncorrelated. 

# Since Markov chains are autocorrelated, sequential samples are not independent of each other.

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

# If a chain is sampling correctly, the `Rhat` value should approach 1. 
# It reflects the fact that inner variance and outer variance between chains is roughly the same, so we could expect that inference is not broken (does't depend on the chain).
# If the values are above 1.00, that may indicate a problem.

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?

# sample model
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)
)

# healthy chain:
m9e6a<-ulam(
  alist(
    logGDP_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=datList, chains=4, cores=4
)

# trace plot
traceplot(m9e6a)

# malfunctioning chain
m9e6b<-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
)


# trace plot
traceplot(m9e6b)

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

# trace rank plot
trankplot(m9e6a)

# trace rank plot
trankplot(m9e6b)

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). Use ulam to estimate the posterior. 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]   0.8865568 0.015675623  0.86150408  0.91160942
## a[2]   1.0505682 0.009936569  1.03468760  1.06644871
## b[1]   0.1324878 0.074204188  0.01389514  0.25108039
## b[2]  -0.1427546 0.054749017 -0.23025407 -0.05525506
## sigma  0.1094937 0.005935237  0.10000804  0.11897935
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 , depth=2)
##             mean          sd        5.5%       94.5%
## a[1]   0.8865568 0.015675623  0.86150408  0.91160942
## a[2]   1.0505682 0.009936569  1.03468760  1.06644871
## b[1]   0.1324878 0.074204188  0.01389514  0.25108039
## b[2]  -0.1427546 0.054749017 -0.23025407 -0.05525506
## sigma  0.1094937 0.005935237  0.10000804  0.11897935
pairs(m9.3_unif)

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]   0.8865660 0.015678309  0.86150901  0.91162294
## a[2]   1.0505690 0.009938289  1.03468568  1.06645229
## b[1]   0.1325141 0.074216337  0.01390205  0.25112614
## b[2]  -0.1425646 0.054758464 -0.23007916 -0.05504996
## sigma  0.1095128 0.005937837  0.10002303  0.11900265
pairs(m9.3_exp)

# there isn't much difference.

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?

dat_slim <- list(
                log_gdp_std = dd$log_gdp_std,
                rugged_std = dd$rugged_std,
                cid = as.integer( dd$cid ))
str(dat_slim)


m9.4 <- 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)

precis(m9.4, depth=2)

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

mp <- ulam(
 alist(
   a ~ dnorm(0,1),
   b ~ dcauchy(0,1)
 ), data=list(y=1) , chains=1 )
## 
## SAMPLING FOR MODEL 'bcf56ee89f6cf2a4224a4139ff01c7d4' NOW (CHAIN 1).
## Chain 1: 
## Chain 1: Gradient evaluation took 0 seconds
## Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0 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: 0.025 seconds (Warm-up)
## Chain 1:                0.026 seconds (Sampling)
## Chain 1:                0.051 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
traceplot(mp)
## [1] 1000
## [1] 1
## [1] 1000

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?