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

The exponential distribution has the memeoryless property. This means that at the current momnent, the distribution has no memory of past events. If wait time at your favorite restaurant is exponentially distributed with mean 30 minutes, and you’ve already waited an hour, you can still expect to wait another 30 minutes.

We can see this algebraically:

The exponential distriubtion has survial function, or the probability t > 0 \(S(x)\quad =\quad { e }^{ -\lambda t }\)

Assuming we’ve already waited x minutes, \(S(t)\quad |\quad t\quad >\quad x\), the survival function is \(S(t)\quad =\quad \frac { { e }^{ -(t+x)\lambda } }{ { e }^{ -x\lambda } }\)

This reduces to:

\(\quad { e }^{ -(t\quad +\quad x)\lambda \quad -\quad -(x\lambda ) }\)
and finally: \(\quad { e }^{ -t\lambda }\)

We can see this through simulation:

exps <- rexp(10000000)
exps <- exps * 10
mean(exps)
## [1] 9.997719
only_100s <- exps[exps > 100]
only_100s <- only_100s - 100#subtract 100 to simulate coming into contact with the system at this time
mean(only_100s)#find the mean of the expiriment
## [1] 9.981495
#the same code with the gamma distrubtion produces a different mean
gammas <- rgamma(10000000,5,.5)
mean(gammas)
## [1] 10.00003
only_100s <- gammas[gammas > 20]
only_100s <- only_100s - 20
mean(only_100s)
## [1] 2.931142