laplace method for one parameter posterior

Jim Albert
October 1, 2013

Outline

  • Observe sample from zero-truncated Poisson distribution
  • Assign prior \[ g(\lambda) = 1 / \lambda \]
  • Illustrate the use of the Laplace method to summarize a posterior density

Observe data

y <- c(3, 4, 1, 2, 2, 1, 3, 3, 3, 4, 3, 2, 
       1, 1, 3, 2, 1, 2, 2, 1, 4, 2)

Construct a R function to compute the log posterior of log lambda

zero.poisson <- function(theta, y){
  lambda <- exp(theta)
  n <- length(y)
  sum(y * log(lambda)) - n * lambda - n * log(1 - exp(-lambda))
}

Plot the exact posterior

lam <- seq(0, 1.4, by=0.01)
g.lam <- exp(sapply(lam, zero.poisson, y))
g.lam <- g.lam / sum(g.lam) / .01
plot(lam, g.lam, type="l")

plot of chunk unnamed-chunk-3

Compute the Laplace approximation

library(LearnBayes)
fit <- laplace(zero.poisson, 1, y)
fit

$mode
[1] 0.6674

$var
        [,1]
[1,] 0.02957

$int
[1] -6.974

$converge
[1] TRUE

Normal approximation

paste("log Lambda is distributed approximately normal with mean", 
      round(fit$mode[1], 2), "and standard deviation", 
      round(sqrt(fit$var), 2))

[1] "log Lambda is distributed approximately normal with mean 0.67 and standard deviation 0.17"

Add approximation to graph

plot of chunk unnamed-chunk-6