normmixp: Bayesian inference on a normal mean with a mixture of normal...
In Bolstad: Functions for Elementary Bayesian Inference

Description Usage Arguments Value See Also Examples

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

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

`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`

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

binomixp normdp normgcp

## 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)