Chapter 2 Question 9

Suppose that we have a sequence of occurrences. We assume that the time X between occurrences is exponentially distributed with λ= 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.

library(plyr)

x<-function() {
  RSTD<-sqrt(rpois(10, 0.1)) 
  print(RSTD)
  return (mean(RSTD))
}

# repet s function 100 time to generate a series of STD 
RSTD<-do.call(rbind, rlply(100, x)) 
##  [1] 0 0 0 1 0 0 0 0 0 0
##  [1] 1 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 1
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 1 0 0 0 0 0 0 0
##  [1] 0 1 0 0 0 1 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 1 0 1 1 0 0 0
##  [1] 0 1 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 1 0 0 0 0 0
##  [1] 0 1 0 0 0 1 1 0 0 0
##  [1] 0.000000 0.000000 0.000000 0.000000 1.414214 0.000000 0.000000
##  [8] 0.000000 0.000000 0.000000
##  [1] 0 0 0 0 0 1 0 1 0 1
##  [1] 0 0 0 1 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 1 0 0 0 0 0 0 1 0 0
##  [1] 1 0 0 0 0 0 0 0 1 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 1 0 0 0 0 0
##  [1] 0 0 0 0 0 1 0 0 0 0
##  [1] 0 0 0 0 0 0 1 0 0 0
##  [1] 0 0 0 0 1 0 0 1 0 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 1 0 1 0 0 0 0 1 0 0
##  [1] 0 0 0 0 1 0 1 0 0 1
##  [1] 0 0 0 0 0 0 1 0 1 0
##  [1] 0 0 1 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 1 0 1 0 0 0 0 0
##  [1] 0 1 1 0 0 0 0 0 0 0
##  [1] 0 1 0 0 0 0 0 0 0 1
##  [1] 1 0 0 0 0 0 1 0 0 0
##  [1] 0 0 0 1 0 0 0 0 1 1
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 1 0
##  [1] 0 0 0 1 0 0 0 0 0 0
##  [1] 0 0 1 0 1 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 1 0 0 0 0 0 0 0 0
##  [1] 0 1 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 1 0 0 0
##  [1] 0 0 0 1 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 1
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 1 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 1 0 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 1
##  [1] 1 1 0 0 0 1 0 0 0 0
##  [1] 0 1 0 0 0 0 0 0 0 0
##  [1] 0 0 1 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 1 0 0 0 0
##  [1] 0 0 0 0 1 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 1 0 0 0 0 1 0 0 1
##  [1] 1 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 1 0 0 1 0 0
##  [1] 0 0 0 0 0 1 0 0 0 1
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 1 0 0 0 0 0 0 1 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 1 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 1 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 1
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 1 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 1 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 1 0 0 1
##  [1] 0 0 1 0 0 0 0 0 0 0
##  [1] 1.414214 0.000000 1.000000 0.000000 0.000000 0.000000 0.000000
##  [8] 0.000000 0.000000 0.000000
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 1
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 1 0 0 0 0 0 0 0 0 0
##  [1] 0 0 1 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 1 1 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 1 0
##  [1] 0 0 0 0 1 0 0 0 0 0
##  [1] 0 0 0 1 0 1 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 1 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 0 0 0
##  [1] 1 0 0 0 0 0 0 0 0 0
##  [1] 0 0 1 0 0 0 0 0 0 0
##  [1] 0 0 0 0 0 0 0 1 0 0
##  [1] 0 0 0 0 0 0 0 0 0 0
# mean deviates (from 10 mins)
rSTD<-mean(RSTD[,1]) 
# 95% confidence interval 
c(10-1.96*rSTD/sqrt(100), 10+1.96*rSTD/sqrt(100)) 
## [1]  9.980238 10.019762
hist(RSTD[,1],main="100 Random Standart Deviation ") #