Description

Consider a renewal process. Let X be the interrenewal times; and let I and R be the length of an interval interrupted at random and its remainder, respectively. The following BASIC simulation calculates the average values of X, I, and R (based on 10,000 replications of I and R, where T is a random interruption point).

100 FOR j=1 TO 10000 110 S=0 120 T = -1000*LOG(1-RND) 130 X= 140 c=c+1 150 SX=SX+X 160 S=S+X 170 IF S<T THEN 130 180 R=S-T: I=X 190 SR=SR+R: SI=SI+I 200 NEXT j 210 PRINT SX/c,SI/10000,SR/10000

a. Run the simulation for the case when X is exponentially distributed (that is, the renewal process is a Poisson process) with E(X) = 1. Fill in the following table.

E(X) E(I) E(R)
theory simulation theory simulation theory simulation

Comment on the assertion: “It is intuitively obvious that E(I) = E(X) and E(R) = E(X) / 2.”

b. Let X = Y + u, where u is a constant (to be treated as a parameter), and Y is a random variable with probability distribution: P(Y=1) = 0.9, P(Y=11) = 0.1. Run the simulation and fill in the values indicated in the table. Calculate the theoretical values of E(I) and E(R) according to the formulas (to be derived later): E(I) = E(X) + V(X) / E(X) and E(R) = E(I) / 2.

Draw the theoretical graph of E(R) versus u. On the same graph, plot the points produced by the simulation.

Comment on the assertion: “It is intuitively obvious that as the average length of the random interval X increases (that is, as u increases), the average lengths of the interrupted interval I and its remainder R will also increase.”

E(X) E(I) E(R)
u theory simulation theory simulation theory simulation
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0