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 )

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

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

mean(samples>0.8)
## [1] 0.1116

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

mean(0.2<samples & samples<0.8)
## [1] 0.888

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

quantile(samples,probs=0.2)
##       20% 
## 0.5185185

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

quantile(samples, probs=0.8)
##       80% 
## 0.7557558

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

HPDI(samples, prob=0.66)
##     |0.66     0.66| 
## 0.5085085 0.7737738

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

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.

likelihood<-dbinom(8,size=15,prob=p_grid)
posterior<-likelihood*prior
posterior<-posterior/sum(posterior)

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=0.9)
##      |0.9      0.9| 
## 0.3293293 0.7167167

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?

predictive<-rbinom(10000,size=15,prob=samples)
mean(predictive==8)
## [1] 0.1444
simplehist(predictive)

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

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

simplehist(predictive)

mean(predictive==6)
## [1] 0.1751

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.

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")

samples<-sample(p_grid,prob=posterior,size=10000,replace=TRUE)
HPDI(samples,prob=0.9)
##      |0.9      0.9| 
## 0.5005005 0.7117117
predictive<-rbinom(10000,size=15,prob=samples)
mean(predictive==8)
## [1] 0.1589
simplehist(predictive)

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

mean(predictive==6)
## [1] 0.2415

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?

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)
}
PI_width(1500,0.7)
## [1] 0.06106106
PI_width(1900,0.7)
## [1] 0.05505506
PI_width(2200,0.7)
## [1] 0.05005005
# toss 2200 times