Homework - Reliable Intuition?
Consider a renewal process. Let X be the inter renewal 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
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
|
|
|
|
|
|
|