## 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

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)