Question

I´m trying to simulate death over 7 years with the cumulative probability as follows:

tab <- data.frame(id=1:1000,char=rnorm(1000,7,4))

cum.prob <- c(0.05,0.07,0.08,0.09,0.1,0.11,0.12)

How can I sample from tab$id without replacement in a vectorized fashion according to the cumulative probability in cum.prob ? The ids sampled from yr 1 can necessarily not be sampled again in yr 2. Hence the lapply(cum.prob,function(x) sample(tab$id,x*1000)) will not work. Is it possible to vectorize this?

//M

Answer 1

Here’s one way: First get the probability of a given individual’s dying in a given year as probYrDeath, i.e. probYrDeath[i] = Prob( individual dies in year i ), where i=1,2,...,7.

probYrDeath <- c(diff(c(0,cum.prob)).

Now generate a random sample of 1000 “Death Years”, with replacement, from the sequence 1:8, according to the probabilities in probYrDeath, augmented by the probability of not dying by year 7:

set.seed(1) ## for reproducibility
tab$DeathYr <- sample( 8, 1000, replace = TRUE, 
                       prob = c(probYrDeath, 1-sum(probYrDeath)))

We interpret “‘DeathYr = 8′” as “not dying within 7 years”, and extract the subset of tab where DeathYr != 8:

tab_sample <- subset(tab, DeathYr != 8 )

You can verify that the cumulative proportions of deaths in each year approximate the values in cum.prob:

> cumsum(table(tab_sample$DeathYr)/1000)
    1     2     3     4     5     6     7 
0.045 0.071 0.080 0.094 0.105 0.115 0.124