library(plyr)
## Warning: package 'plyr' was built under R version 3.3.3

Page 72 Question 9

Suppose that we have a sequence of occurrences. We assume that the time X between occurrences is exponentially distributed with \(lambda\) = 1/10, so on the average, there is one occurrence every 10 minutes (see Example 2.17). You come upon this system at time 100, and wait until the next occurrence. Make a conjecture concerning how long, on the average, you will have to wait. Write a program to see if your conjecture is right

s<-function() {
  #rpois generates random standar deviates(from 10 mins) based given lambda.
  Randomstd<-sqrt(rpois(10, 0.1)) 
  print(Randomstd)
  return (mean(Randomstd))
}

# repet s function 100 time to generate a series of STD 
randomSTD<-do.call(rbind, rlply(100, s)) 
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##  [8] 0.000000 1.414214 0.000000
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#print(dd[,1]) 
rSTD<-mean(randomSTD[,1]) # average deviates(from 10 mins) 
c(10-1.96*rSTD/sqrt(100), 10+1.96*rSTD/sqrt(100)) # 95% confident of waitting time 
## [1]  9.981629 10.018371
hist(randomSTD[,1],main="100 Random Standar Deviates (from 10 mins) ") #