Bolstad (version 0.2-41)

normdp: Bayesian inference on a normal mean with a discrete prior

Description

Evaluates and plots the posterior density for \(\mu\), the mean of a normal distribution, with a discrete prior on \(\mu\)

Usage

normdp(x, sigma.x = NULL, mu = NULL, mu.prior = NULL, n.mu = 50, ...)

Arguments

x

a vector of observations from a normal distribution with unknown mean and known std. deviation.

sigma.x

the population std. deviation of the normal distribution

mu

a vector of possibilities for the probability of success in a single trial. If mu is NULL then a uniform prior is used.

mu.prior

the associated prior probability mass.

n.mu

the number of possible \(\mu\) values in the prior

…

additional arguments that are passed to Bolstad.control

Value

A list will be returned with the following components:

mu

the vector of possible \(\mu\) values used in the prior

mu.prior

the associated probability mass for the values in \(\mu\)

likelihood

the scaled likelihood function for \(\mu\) given \(x\) and \(\sigma_x\)

posterior

the posterior probability of \(\mu\) given \(x\) and \(\sigma_x\)

See Also

normnp normgcp

Examples

# NOT RUN {
## generate a sample of 20 observations from a N(-0.5,1) population
x = rnorm(20,-0.5,1)

## find the posterior density with a uniform prior on mu
normdp(x,1)

## find the posterior density with a non-uniform prior on mu
mu = seq(-3,3,by=0.1)
mu.prior = runif(length(mu))
mu.prior = sort(mu.prior/sum(mu.prior))
normdp(x,1,mu,mu.prior)

## Let mu have the discrete distribution with 5 possible
## values, 2, 2.5, 3, 3.5 and 4, and associated prior probability of
## 0.1, 0.2, 0.4, 0.2, 0.1 respectively. Find the posterior
## distribution after a drawing random sample of n = 5 observations
## from a N(mu,1) distribution y = [1.52, 0.02, 3.35, 3.49, 1.82]
mu = seq(2,4,by=0.5)
mu.prior = c(0.1,0.2,0.4,0.2,0.1)
y = c(1.52,0.02,3.35,3.49,1.82)
normdp(y,1,mu,mu.prior)

# }