normgcp {Bolstad} | R Documentation |
Bayesian inference on a normal mean with a general continuous prior
Description
Evaluates and plots the posterior density for \mu
, the mean of a
normal distribution, with a general continuous prior on \mu
Usage
normgcp(
x,
sigma.x = NULL,
density = c("uniform", "normal", "flat", "user"),
params = NULL,
n.mu = 50,
mu = NULL,
mu.prior = NULL,
...
)
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 |
density |
distributional form of the prior density can be one of: "normal", "unform", or "user". |
params |
if density = "normal" then params must contain at least a mean
and possible a std. deviation. If a std. deviation is not specified then
sigma.x will be used as the std. deviation of the prior. If density =
"uniform" then params must contain a minimum and a maximum value for the
uniform prior. If a maximum and minimum are not specified then a
|
n.mu |
the number of possible |
mu |
a vector of possibilities for the probability of success in a single trial. Must be set if density="user" |
mu.prior |
the associated prior density. Must be set if density="user" |
... |
additional arguments that are passed to |
Value
A list will be returned with the following components:
likelihood |
the scaled likelihood function for |
posterior |
the posterior
probability of |
mu |
the vector of possible |
mu.prior |
the associated probability mass for the values in
|
See Also
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 uniform U[-3,3] prior on mu
normgcp(x, 1, params = c(-3, 3))
## find the posterior density with a non-uniform prior on mu
mu = seq(-3, 3, by = 0.1)
mu.prior = rep(0, length(mu))
mu.prior[mu <= 0] = 1 / 3 + mu[mu <= 0] /9
mu.prior[mu > 0] = 1 / 3 - mu[mu > 0] / 9
normgcp(x, 1, density = "user", mu = mu, mu.prior = mu.prior)
## find the CDF for the previous example and plot it
## Note the syntax for sintegral has changed
results = normgcp(x,1,density="user",mu=mu,mu.prior=mu.prior)
cdf = sintegral(mu,results$posterior,n.pts=length(mu))$cdf
plot(cdf,type="l",xlab=expression(mu[0])
,ylab=expression(Pr(mu<=mu[0])))
## use the CDF for the previous example to find a 95%
## credible interval for mu. Thanks to John Wilkinson for this simplified code
lcb = cdf$x[with(cdf,which.max(x[y<=0.025]))]
ucb = cdf$x[with(cdf,which.max(x[y<=0.975]))]
cat(paste("Approximate 95% credible interval : ["
,round(lcb,4)," ",round(ucb,4),"]\n",sep=""))
## use the CDF from the previous example to find the posterior mean
## and std. deviation
dens = mu*results$posterior
post.mean = sintegral(mu,dens)$value
dens = (mu-post.mean)^2*results$posterior
post.var = sintegral(mu,dens)$value
post.sd = sqrt(post.var)
## use the mean and std. deviation from the previous example to find
## an approximate 95% credible interval
lb = post.mean-qnorm(0.975)*post.sd
ub = post.mean+qnorm(0.975)*post.sd
cat(paste("Approximate 95% credible interval : ["
,round(lb,4)," ",round(ub,4),"]\n",sep=""))
## repeat the last example but use the new summary functions for the posterior
results = normgcp(x, 1, density = "user", mu = mu, mu.prior = mu.prior)
## use the cdf function to get the cdf and plot it
postCDF = cdf(results) ## note this is a function
plot(results$mu, postCDF(results$mu), type="l", xlab = expression(mu[0]),
ylab = expression(Pr(mu <= mu[0])))
## use the quantile function to get a 95% credible interval
ci = quantile(results, c(0.025, 0.975))
ci
## use the mean and sd functions to get the posterior mean and standard deviation
postMean = mean(results)
postSD = sd(results)
postMean
postSD
## use the mean and std. deviation from the previous example to find
## an approximate 95% credible interval
ciApprox = postMean + c(-1,1) * qnorm(0.975) * postSD
ciApprox