r - R: How to generate a series of exponential deviates that sum to some number

Question

I am trying to generate a series of wait times for a Markov chain where the wait times are exponentially distributed numbers with rate equal to one. However, I don't know the number of transitions of the process, rather the total time spent in the process.

So, for example:

t <- rexp(100,1)
tt <- cumsum(c(0,t))

t is a vector of the successive and independent waiting times and tt is a vector of the actual transition time starting from 0.

Again, the problem is I don't know the length of t (i.e. the number of transitions), rather how much total waiting time will elapse (i.e. the floor of last entry in tt).

What is an efficient way to generate this in R?

Answer 1

If I got this right: you want to know how many transitions it'll take to fill your time interval. Since the transitions are random and unknown, there's no way to predict for a given sample. Here's how to find the answer:

tfoo<-rexp(100,1)   
max(which(cumsum(tfoo)<=10))  
[1] 10  

tfoo<-rexp(100,1)   # do another trial 
max(which(cumsum(tfoo)<=10))  
[1] 14  

Now, if you expect to need to draw some huge sample, e.g. rexp(1e10,1), then maybe you should draw in 'chunks.' Draw 1e9 samples and see if sum(tfoo) exceeds your time threshold. If so, dig thru the cumsum . If not, draw another 1e9 samples, and so on.

Answer 2

The Wikipedia entry for Poisson process has everything you need. The number of arrivals in the interval has a Poisson distribution, and once you know how many arrivals there are, the arrival times are uniformly distributed within the interval. Say, for instance, your interval is of length 15.

N <- rpois(1, lambda = 15)
arrives <- sort(runif(N, max = 15))
waits <- c(arrives[1], diff(arrives))

Here, arrives corresponds to your tt and waits corresponds to your t (by the way, it's not a good idea to name a vector t since t is reserved for the transpose function). Of course, the last entry of waits has been truncated, but you mentioned only knowing the floor of the last entry of tt, anyway. If he's really needed you could replace him with an independent exponential (bigger than waits[N]), if you like.