normmixp {Bolstad}

Bayesian inference on a normal mean with a mixture of normal priors

Description

Evaluates and plots the posterior density for \mu, the mean of a normal distribution, with a mixture of normal priors on \mu

Usage

normmixp(
  x,
  sigma.x,
  prior0,
  prior1,
  p = 0.5,
  mu = NULL,
  n.mu = max(100, length(mu)),
  ...
)

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 observations.

prior0

the vector of length 2 which contains the means and standard deviation of your precise prior.

prior1

the vector of length 2 which contains the means and standard deviation of your vague prior.

p

the mixing proportion for the two component normal priors.

mu

a vector of prior possibilities for the mean. If it is NULL, then a vector centered on the sample mean is created.

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

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

binomixp normdp normgcp

Examples

## 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 N(0, 1) prior on mu - a 50:50 mix of
## two N(0, 1) densities
normmixp(x, 1, c(0, 1), c(0, 1))

## find the posterior density with 50:50 mix of a N(0.5, 3) prior and a
## N(0, 1) prior on mu
normmixp(x, 1, c(0.5, 3), c(0, 1))

## Find the posterior density for mu, given a random sample of 4
## observations from N(mu, 1), y = [2.99, 5.56, 2.83, 3.47],
## and a 80:20 mix of a N(3, 2) prior and a N(0, 100) prior for mu
x = c(2.99, 5.56, 2.83, 3.47)
normmixp(x, 1, c(3, 2), c(0, 100), 0.8)