How do I best simulate an arbitrary univariate random variate using its probability function?

Question

In R, what\'s the best way to simulate an arbitrary univariate random variate if only its probability density function is available?

Answer 1

To clarify the "use Metropolis-Hastings" answer above:

suppose ddist() is your probability density function

something like:

n <- 10000
cand.sd <- 0.1
init <- 0
vals <- numeric(n)
vals[1] <- init 
oldprob <- 0
for (i in 2:n) {
    newval <- rnorm(1,mean=vals[i-1],sd=cand.sd)
    newprob <- ddist(newval)
    if (runif(1)<newprob/oldprob) {
        vals[i] <- newval
    } else vals[i] <- vals[i-1]
   oldprob <- newprob
}

Notes:

1. completely untested
2. efficiency depends on candidate distribution (i.e. value of cand.sd). For maximum efficiency, tune cand.sd to an acceptance rate of 25-40%
3. results will be autocorrelated ... (although I guess you could always sample() the results to scramble them, or thin)
4. may need to discard a "burn-in", if your starting value is weird

The classical approach to this problem is rejection sampling (see e.g. Press et al Numerical Recipes)

Answer 2

Use cumulative distribution function http://en.wikipedia.org/wiki/Cumulative_distribution_function

Then just use its inverse. Check here for better picture http://en.wikipedia.org/wiki/Normal_distribution

That mean: pick random number from [0,1] and set as CDF, then check Value

It is also called quantile function.

Answer 3

This is a comment but I don't have enough reputation to drop a comment to Ben Bolker's answer.

I am new to Metropolis, but IMHO this code is wrong because:

a) the newval is drawn from a normal distribution whereas in other codes it is drawn from a uniform distribution; this value must be drawn from the range covered by the random number. For example, for a gaussian distribution this should be something like runif(1, -5, +5).

b) the prob value must be updated only if acceptance.

Hope this help and hope that someone with reputation could correct this answer (especially mine if I am wrong).

# the distribution 
ddist <- dnorm
# number of random number
n <- 100000
# the center of the range is taken as init
init <- 0
# the following should go into a function
vals <- numeric(n)
vals[1] <- init 
oldprob <- 0
for (i in 2:n) {
  newval <- runif(1, -5, +5)
  newprob <- ddist(newval)
  if (runif(1) < newprob/oldprob) {
    vals[i] <- newval
    oldprob <- newprob
  } else vals[i] <- vals[i-1]
}
# Final view
hist(vals, breaks = 100)
# and comparison
hist(rnorm(length(vals)), breaks = 100)

Answer 4

Here is a (slow) implementation of the inverse cdf method when you are only given a density.

den<-dnorm #replace with your own density

#calculates the cdf by numerical integration
cdf<-function(x) integrate(den,-Inf,x)[[1]]

#inverts the cdf
inverse.cdf<-function(x,cdf,starting.value=0){
 lower.found<-FALSE
 lower<-starting.value
 while(!lower.found){
  if(cdf(lower)>=(x-.000001))
   lower<-lower-(lower-starting.value)^2-1
  else
   lower.found<-TRUE
 }
 upper.found<-FALSE
 upper<-starting.value
 while(!upper.found){
  if(cdf(upper)<=(x+.000001))
   upper<-upper+(upper-starting.value)^2+1
  else
   upper.found<-TRUE
 }
 uniroot(function(y) cdf(y)-x,c(lower,upper))$root
}

#generates 1000 random variables of distribution 'den'
vars<-apply(matrix(runif(1000)),1,function(x) inverse.cdf(x,cdf))
hist(vars)