Assignment #8

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.

# The proposal distribution must be symmetric is a requirement of the simple metropolis alogorithm

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 achieve this extra efficiency from adaptive proposals in which the distribution of proposed parameter values adjusts itself intelligently, depending upon the parameter values at the moment

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

# Hamiltonian Monte Carlo cannot handle discrete parameters as it requires continuous parameters because they cannot glide thorough discrete ones.

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

# Effective number of samples, n_eff, is an estimate of the number of independent draws from the posterior distribution in terms of estimating functions like posterior mean. n_eff is usually smaller than actual sample size

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

# Rhat should approach 1

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?

data(rugged)
d1 <- rugged
d1$log_gdp <- log(d1$rgdppc_2000)
dd <- d1[ complete.cases(d1$rgdppc_2000) , ]
dd$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 )
m8.3 <- 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( m8.3 , depth=2 )

##             mean          sd        5.5%       94.5%
## a[1]   0.8865631 0.015674896  0.86151157  0.91161459
## a[2]   1.0505698 0.009936092  1.03469005  1.06644964
## b[1]   0.1325058 0.074200804  0.01391863  0.25109306
## b[2]  -0.1425762 0.054746637 -0.23007195 -0.05508055
## sigma  0.1094884 0.005934523  0.10000388  0.11897291

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   0.04   0.03  0.06
## chain:2   0.03   0.02  0.06
## chain:3   0.04   0.02  0.06
## chain:4   0.04   0.02  0.06
## 
## 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 )

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 'd26c527083e7eda89b17a8c2eccd6019' 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)
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## 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: 0.056 seconds (Warm-up)
## Chain 1:                0.033 seconds (Sampling)
## Chain 1:                0.089 seconds (Total)
## Chain 1: 
## 
## SAMPLING FOR MODEL 'd26c527083e7eda89b17a8c2eccd6019' NOW (CHAIN 2).
## Chain 2: 
## Chain 2: Gradient evaluation took 0 seconds
## Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0 seconds.
## Chain 2: Adjust your expectations accordingly!
## Chain 2: 
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## Chain 2: Iteration: 100 / 1000 [ 10%]  (Warmup)
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## Chain 2: Iteration: 1000 / 1000 [100%]  (Sampling)
## Chain 2: 
## Chain 2:  Elapsed Time: 0.07 seconds (Warm-up)
## Chain 2:                0.012 seconds (Sampling)
## Chain 2:                0.082 seconds (Total)
## Chain 2: 
## 
## SAMPLING FOR MODEL 'd26c527083e7eda89b17a8c2eccd6019' NOW (CHAIN 3).
## Chain 3: 
## Chain 3: Gradient evaluation took 0 seconds
## Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0 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)
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## Chain 3: Iteration: 1000 / 1000 [100%]  (Sampling)
## Chain 3: 
## Chain 3:  Elapsed Time: 0.078 seconds (Warm-up)
## Chain 3:                0.017 seconds (Sampling)
## Chain 3:                0.095 seconds (Total)
## Chain 3:

## Warning: There were 42 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 1.11, 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   0.06   0.03  0.09
## chain:2   0.07   0.01  0.08
## chain:3   0.08   0.02  0.10
## 
## Formula:
## y ~ dnorm(mu, sigma)
## mu <- alpha
## alpha ~ dnorm(0, 1000)
## sigma ~ dexp(1e-04)

traceplot( m9.2 )

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

trankplot(m9.1)

## Warning in if (class(x) == "numeric") x <- array(x, dim = c(length(x), 1)): the
## condition has length > 1 and only the first element will be used

## Warning in if (class(x) == "numeric") x <- array(x, dim = c(length(x), 1)): the
## condition has length > 1 and only the first element will be used

## Warning in if (class(x) == "numeric") x <- array(x, dim = c(length(x), 1)): the
## condition has length > 1 and only the first element will be used

## Warning in if (class(x) == "numeric") x <- array(x, dim = c(length(x), 1)): the
## condition has length > 1 and only the first element will be used

## Warning in if (class(x) == "numeric") x <- array(x, dim = c(length(x), 1)): the
## condition has length > 1 and only the first element will be used

trankplot(m9.2)

## Warning in if (class(x) == "numeric") x <- array(x, dim = c(length(x), 1)): the
## condition has length > 1 and only the first element will be used

## Warning in if (class(x) == "numeric") x <- array(x, dim = c(length(x), 1)): the
## condition has length > 1 and only the first element will be used

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)
d2 <- rugged
d2$log_gdp <- log(d2$rgdppc_2000)
dd <- d2[ complete.cases(d2$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)

m8.3 <- 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(m8.3 , 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(m8.3)

m8.3_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(m8.3_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(m8.3_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?

m9.1 <- 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] ~ dexp(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   0.14   0.09  0.23
## chain:2   0.24   0.06  0.31
## chain:3   0.23   0.06  0.30
## chain:4   0.23   0.09  0.32
## 
## Formula:
## log_gdp_std ~ dnorm(mu, sigma)
## mu <- a[cid] + b[cid] * (rugged_std - 0.215)
## a[cid] ~ dnorm(1, 0.1)
## b[cid] ~ dexp(0.3)
## sigma ~ dexp(1)

precis( m9.1 , 2 )

##             mean          sd         5.5%      94.5%    n_eff    Rhat4
## a[1]  0.88640736 0.016606866 0.8595054558 0.91284743 1771.968 1.000143
## a[2]  1.04843773 0.010597943 1.0314914783 1.06579238 1682.711 1.000424
## b[1]  0.14665605 0.072736881 0.0317260735 0.26705101 1071.298 1.001974
## b[2]  0.01792566 0.017048437 0.0009877327 0.04963585 1978.213 1.001016
## sigma 0.11375596 0.006208079 0.1042213927 0.12393636 1421.608 1.000756

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 )

precis( m9.3 , 2 )

##             mean          sd         5.5%       94.5%    n_eff     Rhat4
## a[1]   0.8864965 0.016171300  0.861237129  0.91221293 2220.142 0.9983686
## a[2]   1.0503031 0.009663780  1.035163351  1.06553874 2900.406 0.9989061
## b[1]   0.1336247 0.076577887  0.009316302  0.25483619 2137.094 1.0017141
## b[2]  -0.1417819 0.056287013 -0.231071096 -0.05401132 2092.135 0.9992578
## sigma  0.1114578 0.006078674  0.102081667  0.12162563 2606.657 0.9987522

pairs( m9.1 )

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)
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## Chain 1: Iteration: 501 / 1000 [ 50%]  (Sampling)
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## 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: 0.011 seconds (Warm-up)
## Chain 1:                0.011 seconds (Sampling)
## Chain 1:                0.022 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

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?

# Plot a looks like a normal distribution where as plot b has some extreme values so it is a Cauchy distribution.