This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 6.262 Discrete Stochastic Processes, Spring 2010 Problem Set 6 Solutions due: Friday, March 19, 2010 Problem 1 (Exercise 3.6) Let Y ( t ) denote the residual life at time t , that is Y ( t ) = S N ( t )+1- t , and let the inter-renewal times be distributed as f X ( x ) = e- x for x 0. As shown in the notes, the long-term time-averaged residual life is given by lim t R =0 Y ( )d t = E( X 2 ) 2E( X ) = 2 / 2 2 / = 1 w.p.1 . Recall that for a Poisson process with arrival rate , given any time t 0, the expected time until the next arrival is given by 1 / . In this case, the time average matches the ensemble average. To find the time-average second moment of Y ( t ), consider the function R ( t ) = Y ( t ) 2 and note that R ( t ) is a valid reward function since its value is uniquely determined by the position of t within the current inter-renewal interval. Since R ( t ) 0 for all t and E( X ) < , by Theorem 3.4 in the notes, it follows that lim t R =0 R ( )d t = E( R n ) E( X ) w.p.1 . Supposing that the inter-renewal interval containing t has length x , the total accumulated reward becomes R n = R x y 2 d y = x 3 / 3. It follows that E( R n ) = E X (E R n ( R n | X = x )) = E( X 3 ) / 3, where the subscripts indicate the random variable with respect to which the expectations are computed. It follows that lim t R =0 R ( )d t = E( X 3 ) 3E( X ) = 3! / 3 3 / = 2 2 w.p.1 . which matches once again the ensemble average of Y ( t ) 2 for a Poisson process, computed as E( Y ( t ) 2 ) = E( X 2 ) = 2 2 , where the first equality follows by the memoryless property of the exponential random variable. Letting instead f X ( x ) = 3 / ( x + 1) 4 for x 0, the first three moments of X are the following: E( X ) = Z 3 x (1 + x ) 4 d x = 1 2 E( X 2 ) = Z 3 x 2 (1 + x ) 4 d x = 1 E ( X 3 ) = Z 3 x 3 (1 + x ) 4 d x = Following the same reasoning as previously, the first two time-averaged moments are given by lim t R =0 Y ( )d t = E( X 2 ) 2E( X ) = 1 2 / 2 = 1 w.p.1 1 and lim t R =0 Y 2 ( )d t = E( X 3 ) 3E( X ) = w.p.1 . Note that the the time-average residual life of this process is less than that of the Poisson process with rate = 1. However, the squared residual life is infinite, while that of any Poisson process is finite. Problem 2 (Exercise 3.7) a) To show that S 1 ,S 2 ,... form a renewal process, it suffices to show that the putative inter- arrival times X 1 ,X 2 ,... , where X i = S i- S i- 1 , are IID. First note that X 1 = S 1 = Y 1 + U 1 , where Y 1 is the first service time and U 1 is the time from when the first customer completes service and the following customer arrives. Since customer arrivals form a Poisson process, it follows that U 1 is simply an inter-arrival time of the Poisson process, that is f U i ( u ) = e- u for u 0. (Why? Use the stationary increments property combined with the fact0....
View Full Document
This note was uploaded on 10/21/2010 for the course EE 5581 taught by Professor Moon,j during the Spring '08 term at Minnesota.
- Spring '08