normdp {Bolstad} | R Documentation |
Evaluates and plots the posterior density for mu, the mean of a normal distribution, with a discrete prior on mu
normdp(x, sigma.x = NULL, mu = NULL, mu.prior = NULL, n.mu = 50, ...)
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 |
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 |
## 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)