Chapter 3 - Sampling the Imaginary

This chapter introduced the basic procedures for manipulating posterior distributions. Our fundamental tool is samples of parameter values drawn from the posterior distribution. These samples can be used to produce intervals, point estimates, posterior predictive checks, as well as other kinds of simulations. Posterior predictive checks combine uncertainty about parameters, as described by the posterior distribution, with uncertainty about outcomes, as described by the assumed likelihood function. These checks are useful for verifying that your software worked correctly. They are also useful for prospecting for ways in which your models are inadequate.

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

Simulate the posterior distribution

p_grid <- seq( from=0 , to=1 , length.out=1000 )
prior <- rep( 1 , 1000 )
likelihood <- dbinom( 6 , size=9 , prob=p_grid )
posterior <- likelihood * prior
posterior <- posterior / sum(posterior)
set.seed(100)
samples <- sample( p_grid , prob=posterior , size=1e4 , replace=TRUE )

Questions

Use the values in samples to answer the questions that follow.

3E1. How much posterior probability lies below p = 0.2?

mean(samples < 0.2)
## [1] 4e-04
##0.04% of the posterior probability lies below  p = 0.2.

3E2. How much posterior probability lies above p = 0.8?

mean(samples > 0.8)
## [1] 0.1116
##11.16% of the posterior probability lies above p = 0.8.

3E3. How much posterior probability lies between p = 0.2 and p = 0.8?

mean(samples > 0.2 & samples < 0.8)
## [1] 0.888
##88.8% of the posterior probability lies between p = 0.2 and p = 0.8.

3E4. 20% of the posterior probability lies below which value of p?

quantile(samples, 0.20)
##       20% 
## 0.5185185
#20% of the posterior probability lies below 0.52.

3E5. 20% of the posterior probability lies above which value of p?

quantile(samples, 1 - 0.20)
##       80% 
## 0.7557558
#20% of the posterior probability lies above 0.76.

3E6. Which values of p contain the narrowest interval equal to 66% of the posterior probability?

library(rethinking)
## Loading required package: rstan
## Loading required package: StanHeaders
## Loading required package: ggplot2
## rstan (Version 2.21.2, GitRev: 2e1f913d3ca3)
## For execution on a local, multicore CPU with excess RAM we recommend calling
## options(mc.cores = parallel::detectCores()).
## To avoid recompilation of unchanged Stan programs, we recommend calling
## rstan_options(auto_write = TRUE)
## Do not specify '-march=native' in 'LOCAL_CPPFLAGS' or a Makevars file
## Loading required package: parallel
## rethinking (Version 2.12)
## 
## Attaching package: 'rethinking'
## The following object is masked from 'package:stats':
## 
##     rstudent
HPDI(samples, prob = 0.66)
##     |0.66     0.66| 
## 0.5085085 0.7737738
##The narrowest interval equal to 66% of the posterior probability is between 0.51 and 0.77.

3E7. Which values of p contain 66% of the posterior probability, assuming equal posterior probability both below and above the interval?

PI(samples, prob = 0.66)
##       17%       83% 
## 0.5025025 0.7697698
#66% of the posterior probability is between 0.50 and 0.77, assuming equal posterior probability both below and above the interval.

3M1. Suppose the globe tossing data had turned out to be 8 water in 15 tosses. Construct the posterior distribution, using grid approximation. Use the same flat prior as before.

p_grid <- seq(from = 0, to = 1, length.out = 1000)
prior <- rep(1, 1000)
likelihood <- dbinom(8, size = 15, prob = p_grid)
posterior <- likelihood * prior
posterior <- posterior / sum(posterior)
plot(x = p_grid, y = posterior, type = "l")

3M2. Draw 10,000 samples from the grid approximation from above. Then use the samples to calculate the 90% HPDI for p.

samples <- sample(p_grid, prob = posterior, size = 10000, replace = TRUE)
HPDI(samples, prob = .90)
##      |0.9      0.9| 
## 0.3293293 0.7167167
##90% HPDI is bwtween 0.33 and 0.72.

3M3. Construct a posterior predictive check for this model and data. This means simulate the distribution of samples, averaging over the posterior uncertainty in p. What is the probability of observing 8 water in 15 tosses?

water <- rbinom(10000, size = 15, prob = samples)
simplehist(water)

mean(water==8)
## [1] 0.1444
##The probability of observing 8 water in 15 tosses is about 14%.

3M4. Using the posterior distribution constructed from the new (8/15) data, now calculate the probability of observing 6 water in 9 tosses.

water <- rbinom(10000, size = 9, prob = samples)
simplehist(water)

mean(water==6)
## [1] 0.1751
##The probability of observing 8 water in 15 tosses is about 18%.

3M5. Start over at 3M1, but now use a prior that is zero below p = 0.5 and a constant above p = 0.5. This corresponds to prior information that a majority of the Earth’s surface is water. Repeat each problem above and compare the inferences. What difference does the better prior make? If it helps, compare inferences (using both priors) to the true value p = 0.7.

#Repeat 3M1
p_grid <- seq(from = 0, to = 1, length.out = 1000)
prior <- ifelse(p_grid < 0.5, 0, 1)
likelihood <- dbinom(8, size = 15, prob = p_grid)
posterior <- likelihood * prior
posterior <- posterior / sum(posterior)
plot(x = p_grid, y = posterior, type = "l")

#Repeat 3M2
samples <- sample(p_grid, prob = posterior, size = 10000, replace = TRUE)
HPDI(samples, prob = .90)
##      |0.9      0.9| 
## 0.5005005 0.7117117
##90% HPDI is bwtween 0.50 and 0.71. The prior impacts the lower bound of the HPDI, increasing from 0.33 to 0.5, but it does not impact the upper bound a lot. 

#Repeat 3M3
water <- rbinom(10000, size = 15, prob = samples)
simplehist(water)

mean(water==8)
## [1] 0.1589
##The probability of observing 8 water in 15 tosses has increased slightly from 14% to 16%.

#Repeat 3M4
water <- rbinom(10000, size = 9, prob = samples)
simplehist(water)

mean(water==6)
## [1] 0.2415
##The probability of observing 8 water in 15 tosses has increased from 18% to 24%.
###Hence, the better prior can increased the posterior probability of the true value p=0.7.

3M6. Suppose you want to estimate the Earth’s proportion of water very precisely. Specifically, you want the 99% percentile interval of the posterior distribution of p to be only 0.05 wide. This means the distance between the upper and lower bound of the interval should be 0.05. How many times will you have to toss the globe to do this?

compute_pi_width <- function(N, true_p) {
  likelihood <- dbinom(round(N*true_p), size=N, prob=p_grid)
  posterior <- likelihood * prior
  posterior <- posterior / sum(posterior)
  samples <- sample(p_grid, prob=posterior, size=10000, replace=TRUE)
  interval <- PI(samples, prob=0.99)
  names(interval) <- NULL
  diff(interval)
}

true_p <- 0.7
#with 10 observations
N <- 10
compute_pi_width(N, true_p)
## [1] 0.4364364
#with 100 observations
N <- 100
compute_pi_width(N, true_p)
## [1] 0.2322372
#with 1000 observations
N <- 1000
compute_pi_width(N, true_p)
## [1] 0.07507508
#with 2000 observations
N <- 2000
compute_pi_width(N, true_p)
## [1] 0.05305305
#with 2300 observations
N <- 2300
compute_pi_width(N, true_p)
## [1] 0.05005005
##With 2300 observations, we then get the 99% percentile interval to be 0.05 wide.