University of Illinois Spring 2008
ECE 313: Problem Set 11
Many Random Variables
This Problem Set contains six problems
Due: Wednesday April 16 at the beginning of class.
Reading: Ross Chapter 6
Noncredit Exercises: Chapter 6: Problems 1, 9-15, 20-23
Reminder: Hour Exam II on Monday April 14, 7:00 p.m. – 8:00 p.m
Room 165 Everitt Laboratory
1. The joint pmf pX,Y(u, v) of Xand Yis shown in the table below.
u→
v↓0 1 3 5
4 0 1/12 1/6 1/12
3 1/6 1/12 0 1/12
-1 1/12 1/6 1/12 0
(a) Find the marginal pmfs pX(u) and pY(v).
(b) Are Xand Yindependent random variables?
(c) Find P{X ≤ Y} and P{X +Y ≤ 4}.
(d) Find pX |Y (u|3), E[X |Y = 3] and var(X |Y = 3).
2. Packets arriving at a router are addressed either to Server A or to Server B. Each
packet address (A or B) may be regarded as an independent trial of an experiment
whose outcomes are A and B with probabilities pand q= 1 −prespectively. The
router sends each packet to the appropriate server.
We model the packet arrivals as a Poisson process with arrival rate λ. Let X=N(0, T ]
denote the number of packets arriving at the router during (0,T ] and Y=NA(0, T ]
the number of packets that are routed to Server A. Then, all the above implies that
given that X=n, the conditional pmf pY|X (m|n) of Yis binomial with parameters
(n, p). Of course, Xis a Poisson random variable with parameter λT.
(a) Show that the unconditional pmf of Yis Poisson with parameter (λp)T.
(b) Explain briefly why the unconditional pmf of Z=X − Y is Poisson with pa-
rameter (λq)T. Note that Z=NB(0, T ] counts the number of packets that are
addressed to Server B.
The packet streams routed to the two servers are in fact Poisson processes with
arrival rates λp and λq respectively. The division of a Poisson stream into two
Poisson substreams is called Poisson splitting. Poisson splitting also occurs if
the packets are jobs and it is the router that chooses to randomly allot the jobs
to Server A or B with probabilities pand q. However, Poisson splitting does not
occur if the incoming jobs are simply routed alternately to Servers A and B: the
probability that Server A receives kjobs in (0,T ] is P{X = 2k−1}+P{X = 2k}
which does not match up with a Poisson probability mass function.
