Question

a. Assuming a low-informative prior probability distribution, find the posterior distribution of left-handed students in the population. Summarize your results with an estimation of the mean, median, mode, and a 95% credible interval. Plot your posterior probability distribution.

b. According to the literature, 5 to 20% of people are left-handed. Take this information into account in your prior probability and calculate a new posterior probability distribution. Summarize your results with an estimation of the mean, median, mode, and a 95% credible interval. Plot your posterior probability distribution.

b. According to the literature, 5 to 20% of people are left-handed. Take this information into account in your prior probability and calculate a new posterior probability distribution. Summarize your results with an estimation of the mean, median, mode, and a 95% credible interval. Plot your posterior probability distribution.

Answer 1

PART A

This is the case of Binomial Likelihood and if the prior and posterior are in the same family then they are called conjugate distributions.

Now, given Binomial Prior the conjugate would be beta distribution. The posterior beta distribution is

Beta(y+alpha,N-y+beta)

where y is the no. of left handed students, N is the sample size, alpha & beta are the parameters of Beta distribution.

Here, y = 2 and N = 18 so, Prior is Uniform(0,1)which is Beta(1,1) and hence, Posterior is Beta(y+alpha,N-y+beta) =  Beta(3,17)

Posterior mean = (y+alpha)/(N+alpha+beta) = 0.15

Posterior Median =(y+alpha-1/3)/(N+alpha+beta-2/3) = 0.137931

Posterior Mode = (y+alpha-1)/(N+alpha+beta-2) = 0.11111

R-Script for Part A

> #---Part A
> a<-1
> b<-1
> y<-2
> N<-18
> mean <-(y+a)/(N+a+b)
> mean
[1] 0.15
> median<-(y+a-(1/3))/(N+a+b-(2/3))
> median
[1] 0.137931
> mode<-(y+a-1)/(N+a+b-2)
> mode
[1] 0.1111111
> p = seq(0,1, length=500)
> plot(p, dbeta(p, 3,17), ylab="density", type ="l", col=4,main = "Beta(3,17)")

Part B

Now, let the prior beta distribution has mean equal to 0.125 which is the mean of 0.05 and .20. For choosing new alpha & beta parameters with mean 0.125 let the pseudo sample size be n, so the alpha and beta parameters become n*0.125,n*(1-0.125). Let us suppose n=1000 so alpha and beta are 125 and 875 respectively.

Posterior is Beta(y+alpha,N-y+beta) =  Beta(127,891)

Posterior mean = (y+alpha)/(N+alpha+beta) = 0.1247544

Posterior Median =(y+alpha-1/3)/(N+alpha+beta-2/3) = 0.1245085

Posterior Mode = (y+alpha-1)/(N+alpha+beta-2) = 0.1417323

R-Script for Part B

> #---Part B
> aa<-127
> bb<-891
> amean<-aa/(aa+bb)
> amean
[1] 0.1247544
> amedian<-(aa-(1/3))/(aa+bb-(2/3))
> amedian
[1] 0.1245085
> amode<-(aa-1)/(bb-2)
> amode
[1] 0.1417323
> p = seq(0,1, length=500)
> plot(p, dbeta(p, 127,891), ylab="density", type ="l", col=4,main = "Beta(127,891)")