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241-460 Introduction to Queueing
Netw orks : Engineering Approach
Assoc. Prof. Thossaporn KamolphiwongCentre for Network Research (CNR)
Department of Computer Engineering, Faculty of EngineeringPrince of Songkla University, Thailand
apter u t p e o an om Variables
Email : [email protected]
Outline
Multiple of Random Variable
Events and Probabilities
Joint Probability Mass Function
Joint Cumulative Distribution Function
• Part II
Joint Pro a i ity Density Function
Marginal PMF & PDF
Conditional joint PMF & PDF
Chapter 5 : Multiple of Random Variables
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Part I
Multiple of Random Variable
Chapter 5 : Multiple of Random Variables
Random Variable
Experiment
Outcomes
Observe
Math. model Assign number
Chapter 5 : Multiple of Random Variables
Probability Random Variable
Math. model
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Multiple Random Variable
Experiment
Outcomes
ObserveInterest >one thing in
Experiment
Assign vector numberto each outcomes
Chapter 5 : Multiple of Random Variables
Random Variable Multiple RVPair of RV
Vector Random Variables
Vector random variables is a function that
ass gns a vec or o rea num ers o eac
outcome ( ) in S and several quantities
are of interest
Chapter 5 : Multiple of Random Variables
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Example
A random experiment consist of selecting a
’ .
Let = outcome of this experiment,
H ( ) = height of student in inches,
W ( ) = weight of student in ponds, and
A = a e of student in ears.
The vector ( H ( ), W ( ), A( )) is a vector random
variable
Chapter 5 : Multiple of Random Variables
Events and Probabil ities
Consider the two-dimensional variable Z = ( X,Y ).
Find the re ion of the lane corres ondin to the events
A = { X + Y < 10}
= + <
Chapter 5 : Multiple of Random Variables
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Event and P robability(2)
y
(10,0)
(0,10)
x A
(10,0)
(0,10)
xC
X+Y =10
X 2+Y 2 = 100
Chapter 5 : Multiple of Random Variables
A = { X + Y < 10} C = { X 2 + Y 2 < 100}
Probability
A2={ x1< X < x2} y
y2
A1={Y < y2}
x1 x2
A = { x1 < X < x2} {Y < y2}
A2 A1
x
A
• =
• P[ A] = P[{ X in A2} {Y in A1}]
Chapter 5 : Multiple of Random Variables
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Probability(2)
B = { x1 < X < x2} { y1 < Y < y2}
y
B
y2
y1
Chapter 5 : Multiple of Random Variables
x
P[ B] = P[{ x1 < X < x2} { y1 < Y < y2}]
x1 x2
Probability(3)
For n-dimensional random variable X = ( X 1, X 2, …, X n)
Event
A = { X 1 in A1} { X 2 in A2} … { X n in An}
The probability of events:
P[ A] = P[{ X 1 in A1}{ X 2 in A2}…{ X n in An}]
Chapter 5 : Multiple of Random Variables
P[ A] = P[ X 1 in A1,…, X n in An]
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Joint Probability Mass Function
• Probability Mass functionRandom variable X
• Joint probability mass function
Event of all outcome X () in S
PMF of random variable X
P X ( x) = P[ X = x]
Chapter 5 : Multiple of Random Variables
S X,Y = {( x,y)|P X,Y ( x,y) > 0}
P X,Y ( x,y) = P[ X = x,Y = y]
Example
• Test two integrated circuits one after the other.On each test the ossible outcomes are a(accept) and r (reject). Assume that all circuits
are acceptable with probability 0.9 and that the
outcomes of successive tests are independent.Count the number of acceptable circuits X andcount the number of successful tests Y before
you observe the first reject . (If both tests aresuccessful, let Y = 2.) Draw a tree diagram for
the experiment and find the joint PMF of
X and Y
Chapter 5 : Multiple of Random Variables
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Solution
• aa 0.81
a
0.9 a X = 2, Y = 2
0.09
0.09
0.01
• ar
• ra
• rr
.
0.1r
0.1 r
0.9
0.1 r
a
X = 1, Y = 1
X = 1, Y = 0
X = 0, Y = 0
Chapter 5 : Multiple of Random Variables
P[aa] = 0.81, P[ar ] = P[ra] = 0.09, P[rr ] = 0.01
S = {aa, ar, ra, rr }
Solution(2)
Let
into the pair of RV ( X,Y )
g(aa) = (2,2), g(ar ) = (1,1)g(ra) = (1,0), g(rr ) = (0,0)
• P X,Y ( x,y) = P[ X= x, Y= y]
P X,Y (2 ,2) = P[ X= 2 , Y= 2] = P[aa] = 0.81
P X,Y (1 ,1) = 0.09, P X,Y (1 ,0) = 0.09
P X,Y (0 ,0) = 0.01
Chapter 5 : Multiple of Random Variables
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Solution(3)
x
y x
11
2,2
09.0
81.0
otherwise
y x
y x y xP Y X
0,0
0,1
0
01.0
09.0,,
P X,Y ( x,y) y = 0 y = 1 y =2
Chapter 5 : Multiple of Random Variables
x = 0 0.01 0 0
x = 1 0.09 0.09 0
x = 2 0 0 0.81
Solution(4)
y
1
2
x
.81
.09
.09.01 0
1
2
0
0.3
0.6
0.9
01
2
0.010.09
0.81
y
Chapter 5 : Multiple of Random Variables
P X,Y ( x,y)
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Joint PMF Properties
,,
X Y S x S y
Y X
0,, X Y S x S y
Y X y xP
Chapter 5 : Multiple of Random Variables
Joint PMF Theorem
For discrete random variables X and Y and any set B
in the X, Y plane, the probability of the event
B y x
Y X y xP BP
,, ,
{( X,Y ) B} is
X
P X ,Y
Chapter 5 : Multiple of Random Variables
B={ X 2 + Y 2 < 4}
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Example
• Find the probability of the event B that X , thenumber of acce table circuits e uals Y the
number of successful tests before observing thefirst failure
P X,Y ( x,y) y = 0 y = 1 y =2
x = 0 0.01 0 0
Chapter 5 : Multiple of Random Variables
x = 1 0.09 0.09 0
x = 2 0 0 0.81
Solution
X : # of acceptable circuits
observing the 1st failure
P X,Y
( x,y) y = 0 y = 1 y = 2
x = 0 0.01 0 0
x = 1 0.09 0.09 0 X
Y P X ,Y
S X,Y = {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0),
(2,1), (2,2)}
Chapter 5 : Multiple of Random Variables
.
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(Continue)
B = {( x,y)| X = Y }Y P X ,Y
B S X,Y = {(0,0), (1,1), (2,2)} X
B={( x,y) | X = Y }
Chapter 5 : Multiple of Random Variables
B = {(0,0), (1,1), (2,2)}
Solution(2)
B = {(0,0), (1,1), (2,2)} P X,Y ( x,y) y = 0 y = 1 y =2
P[ B] = ??
P B = P 0 0 + P 1 1 + P 2 2
x = 0 0.01 0 0
x = 1 0.09 0.09 0
x = 2 0 0 0.81
, , ,
= 0.01 + 0.09 + 0.81 = 0.91
Chapter 5 : Multiple of Random Variables
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Marginal PMF
• Experiment with 2 RVs, X, Y
,other ( X )
P X ( x) or PY ( y)
For discrete random variables X and Y with
joint PMF P X,Y ( x,y) ,
Chapter 5 : Multiple of Random Variables
P X ( x) = P X,Y ( x,y)
PY ( y) = P X,Y ( x,y)
ySY
xS X
Example
P X,Y ( x,y) y = 0 y = 1 y =2
x = 0 0.01 0 0
Find Marginal PMF of X and Y
x = 1 0.09 0.09 0
x = 2 0 0 0.81
Solution
Marginal PMF for X
•
Chapter 5 : Multiple of Random Variables
, ,
2
0
, , y
Y X X y xP xP
P X (0) = 0.01
P X (1) = 0.09 + 0.09 = 0.18
P X (2) = 0.81
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Solution
P X,Y ( x,y) y = 0 y = 1 y =2
x = 0 0.01 0 0
• Marginal PMF for Y P X ( x)
0.01
x = 1 0.09 0.09 0
x = 2 0 0 0.81
• PY (0) = 0.01 + 0.09
= 0.1
0
, , x
Y X Y y xP yP
PY ( y) 0.1 0.09 0.81
0.18
0.81
1
Chapter 5 : Multiple of Random Variables
• PY (1) = 0.09
• PY (2) = 0.81
Solution (2)
x
x
1
0
18.0
01.0
otherwise
x xP X
2
0
81.0
y 01.0
Chapter 5 : Multiple of Random Variables
otherwise
y
y yPY
2
0
81.0
.
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Example
The number of bytes N in a message has aeometric distribution with arameter 1- and
range S N = {0,1,2 ,…}.
Suppose that messages are broken into packets of maximum length M bytes.
Let Q be then number of full packets in a messageand let R be the number of b tes left over.
Find the joint PMF PQ,R(q,r ) and the marginal PMF’s of Q and R
Chapter 5 : Multiple of Random Variables
Solution
• Joint PMF
N : messa e b tes
M : packets of maximum length
Q : number of full packets { Q = 0,1,…}
R : number of bytes left over { R = 0,1,…, M -1}
N = qM + r N bytes
Chapter 5 : Multiple of Random Variables
M bytes R bytes
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Solution(2)
PMF of Geometric RV with parameter p :
• first success on the kth attempt
P X [ X = k ] = p(1 - p)k -1 , k = 1,2,3,…
• k failures before the first success
PY [Y = k ] = p(1 - p)k , k = 0,1,2,…
(modified Geometric RV)
Chapter 5 : Multiple of Random Variables
Solution(3)
PMF of Geometric RV with parameter 1- p :
(modified Geometric RV)
P[ N = n] = pn(1- p), n = 0,1,2,…
Chapter 5 : Multiple of Random Variables
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Solution(4)
P[ N = n] = pn(1- p) N = qM+r
= P[ N = qM+r ]
= P[Q = q, R = r ] = pn(1- p)
Chapter 5 : Multiple of Random Variables
- p
P[Q = q, R = r ] = (1-p) pqM+r
Solution(5)
• Marginal PMF of Q
Q : number of full packets
R = 0 N = qM
R = 1 N = qM + 1
…
R = M – 1 N = qM + M -1
Chapter 5 : Multiple of Random Variables
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Solution(6)
P[ N = n] = P[Q = q, R = r ] = (1-p) pqM+r
Marginal PMF of Q
P[Q = q] = P[ N in {qM, qM +1 ,…, qM+( M-1)}]
1
1 M
r qM p p
Chapter 5 : Multiple of Random Variables
0r
1
0
1 M
k
k qM p p p
Solution(7)
P[Q = q]
1
1 M
r qM p p p
,...2,1,0 1
11 q p
p p p
M
qM
0r
Chapter 5 : Multiple of Random Variables
P[Q = q] = (1 - p M ) pqM
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Solution(8)
• Marginal PMF of R
R : number of bytes left over { R = 0,1,…, M -1}
Q = 1 N = r
Q = 2 N = M + r
Q = 3 N = 2 M + r
…Q = q N = qM + r
…
Chapter 5 : Multiple of Random Variables
Solution(9)
• Marginal PMF of R
P R = r
0 1q
r qM
p p
, , ,…,
1,...,2,1,0 1
M r p r
Chapter 5 : Multiple of Random Variables
1 p
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Joint CDF
• Cumulative Distribution function
F X ( x) = P[ X < x]
• Joint Cumulative Distribution
function
Chapter 5 : Multiple of Random Variables
F X,Y ( x,y) = P[ X < x, Y < y]
(Contiue)
=
( x,y)
Y
y
, , ,
Chapter 5 : Multiple of Random Variables
F X,Y ( x,y) X x
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Theorem : Joint CDF
Theorem : For any pair of random variables X, Y,
0 < F X,Y ( x,y) < 1
F X ( x) = F X,Y ( x,) = P[ X < x, Y < ]
F Y ( y) = F X,Y (, y) = P[ X < , Y < y]
F (-, y) = F ( x, -) = 0 , ,
If x < x1 and y < y1 then F X,Y ( x,y) < F X,Y ( x1 ,y1)
F X,Y (, ) = 1
Chapter 5 : Multiple of Random Variables
Summary
Multiple of Random Variable
Events and Probabilities
Joint Probability Mass Function
Joint Cumulative Distribution Function
Chapter 5 : Multiple of Random Variables
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Joint Probability Density Function
• Definition: The Joint PDF of the continuousrandom variable X and Y is a function f X,Y ( x,y)
x y
Y X Y X dvduvu f y xF ,, ,,
with the property
Chapter 5 : Multiple of Random Variables
Single RV X, f X ( x) measure of probability/unit length
Two RV X and Y f X,Y ( x,y) measure probability/unit area
Joint PDF
Theorem :
y x
y xF y x f
Y X
Y X
,,
,
2
,
P[ x < X < x+dx , y < Y < y + dy] = f X,Y ( x,y)dxdy
Chapter 5 : Multiple of Random Variables
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Joint PDF : Theorem
P[ x1< X < x2, y1<Y < y2] = F X,Y ( x2 ,y2) – F X,Y ( x2 ,y1)
– X,Y 1 , 2 X,Y 1 , 1
( x2 ,y2)
Y
Chapter 5 : Multiple of Random Variables
( x1 ,y1) X
Joint PDF : Theorem
• f X,Y ( x,y) > 0 for all ( x, y)
1,, dxdy y x f Y X •
• Y X dxdy y x f AP ,,
Chapter 5 : Multiple of Random Variables
A
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Solution
Case 1 completely insidethe region
Y
1
y ( x, y)
• x > 1 and y > 1
F X,Y ( x,y) = 1
Case 2 outside the region
• x < 0 or y < 0
Y
1
X 1 x
F X,Y ( x,y) = 0
Chapter 5 : Multiple of Random Variables
X 1
y
x
,
49
( x, y)
x < 0
y < 0
Solution(2)
Case 3 ( x,y) is inside the area of nonzero probability
1
Y
X ( x,y) y
f X,Y ( x, y) = 2
y x
y x
Y X Y X
dudv
dudvvu f y xF
2
,,,,
Chapter 5 : Multiple of Random Variables
X 1
y x
v
Y X dudv y xF 0
, 2, xv
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Solution(3)
y x
dudv xF 2,
Y
f X,Y ( x, y) = 2
y
dvv x0
2
22 y xy
v0
,1
X 1
X ( x,y) y
x
Chapter 5 : Multiple of Random Variables
Solution(4)
Case 4 ( x,y) above the triangle
,
1
Y
X ( x,y) y
f X,Y ( x, y) = 2
x x
vY X dudv y xF
0, 2,
x
dvv x2
Chapter 5 : Multiple of Random Variables
X 1 xv
0
2 x
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Solution(5)
Case 5 ( x,y) is the right of
triangleY
y
v
Y X dudv y xF 0
1
, 2,
• 0 < y < 1, x > 1 1
X
X ( x,y) y
x
X,Y ,
v y
dvv12
Chapter 5 : Multiple of Random Variables
0
22 y y
Solution
The resulting CDF is
1,1
1,10
10,0
10
0or0
1
2
2
0
,2
2
2
,
y x
x y
x y x
x y
y x
y y
x
y xy
y xF Y X
Chapter 5 : Multiple of Random Variables
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Marginal PDF
• Interested only in one random variable
Ignore Y and observe only X
Ignore X and observer only Y
Theorem : If X and Y are random variables with joint PDF f X,Y ( x,y),
dx y x f y f
dy y x f x f
Y X Y
Y X X
,
,
,
,
Chapter 5 : Multiple of Random Variables
Example
2
• The PDF of X and Y is
otherwise
y x x y y x f Y X
,
0,,
• Find The marginal PDFs f X ( x) and f Y ( y)
Chapter 5 : Multiple of Random Variables
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(Continue)
y x x y 1 ,114 / 5 2
-1 < x < 1
otherwiseY X
0,,
Y
1 y > x2
Chapter 5 : Multiple of Random Variables
y <
X
-1 1
0.5
0
Solution
• Find f X ( x)
1
Y X=x
0.6 )
dy y
dy y x f x f Y X X
4 / 5
,, 0.5
-1 0 1
X x2
1
4 / 52
dy y x
0.2
.
0-1.5 -1 -0.5 0 0.5 1
x
f X ( x
Chapter 5 : Multiple of Random Variables
58
44
18
5
22
1
4
5 x
x
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Solution
• Find f Y ( y)1
Y
2
3
)
dx y
Y X Y
4 / 5
,,
-1 0 1 X
y1/2
Y=y
-y1/2
y
ydx y 4 / 5
51
-1 -0.5 0 0.5 1 x
f Y (
Chapter 5 : Multiple of Random Variables
2
5
423
y
y y
Functions of Two Random Variable
• Use two random variable to computer a newrandom variable
• Example
X : Amplitude of signal transmitted by radio station
Y : attenuation of the signal as it travels to theantenna of a moving car
W : amplitude of the signal at the radio receiver in
W = X/Y
Chapter 5 : Multiple of Random Variables
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Theorem : For discrete random variables X
PMF of Function
and Y, the derived random variable W =
g( X,Y ) has PMF
w y xg y x
Y X W y xPwP,:,
, ,
Chapter 5 : Multiple of Random Variables
(Continue)
xPwP , P P X,Y X,Y (( x,y x,y)) w y xg y x ,:,
,
Chapter 5 : Multiple of Random Variables
w = g( x,y)
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A firm sends out two kinds of promotional facsimiles. Onekind contains only text and requires 40 seconds to
Example
.pictures that take 60 seconds per page. Faxes can be1, 2, or 3 pages long. Let random variable L : length of afax in pages. S L = {1, 2, 3}. Let the random variable T :time to send each page. ST = {40,60}. After observingmany fax transmissions, the firm derives the following
Chapter 5 : Multiple of Random Variables
P L,T (l,t ) t = 40 sec t = 60 secl = 1 page 0.15 0.1
l = 2 pages 0.3 0.2
l = 3 pages 0.15 0.1
(Continue)
• Let D = g( L,T ) = LT be the total duration in
second of a fax transmission. Find the ran e S D, the PMF P D(d ), and the expected value E [ D]
Chapter 5 : Multiple of Random Variables
64
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Solution
PMF P D(d ) P L,T (l,t ) t = 40 sec t = 60 sec
l = 1 a e 0.15 0.1
• S D = {40, 60, 80, 120, 180}
• P D(40) = P L,T (1,40) = 0.15
• P D(60) = P L,T (1,60) = 0.1
• P D(80) = P L,T (2,40) = 0.3
. .
l = 2 pages 0.3 0.2
l = 3 pages 0.15 0.1
• P D(120) = P L,T (2,60)+P L,T (3,40) = 0.35
• P D(180) = 0.1
• P D(d ) = 0;d 40, 60, 80, 120, 180
Chapter 5 : Multiple of Random Variables
(Continue)
P l t t = 40 sec t = 60 sec D (second) P D( d ) ,
l = 1 page 0.15 0.1
l = 2 pages 0.3 0.2
l = 3 pages 0.15 0.1
40 0.15
60 0.1
80 0.3
120 0.35
Chapter 5 : Multiple of Random Variables
180 0.1
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Solution (cont.)
Expected value E [ D]
DSd
D d dP D E
= 40(0.15)+60(0.1)+80(0.3)+120(0.35)+180(.1)
= 96 secP L,T (l,t ) t = 40 sec t = 60 sec
Chapter 5 : Multiple of Random Variables
l = 1 page 0.15 0.1
l = 2 pages 0.3 0.2
l = 3 pages 0.15 0.1
Theorem : For continuous random
PDF of Function
,
W = g( X,Y ) is
w y xg
Y X W dxdy y x f wW PwF ,
, ,
Chapter 5 : Multiple of Random Variables
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(Continue)
Y X W dxdy y x f wW PwF , ,
P P X,Y X,Y (( x,y x,y))
w y xg ,
Chapter 5 : Multiple of Random Variables
w = g( x,y)
Example
Y x
• X and Y have the joint PDF
otherwise y x f Y X
,
0,,
Y
Y=wX
• Find the PDF of W = Y / X
Chapter 5 : Multiple of Random Variables
X
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36
Solution
• F W (w) = P[W w]
= w = w
Y
Y=wX Y< wX
Chapter 5 : Multiple of Random Variables
X
= w
Solution
For w < 0, F W (w) = 0,
w > 0, CDF is
0 0
dxdyewX Y P y x Y
Y=wX Y< wX
0 0
dxdye
wx
y x
Chapter 5 : Multiple of Random Variables
0 0
dxdyee
wx
y x X
= w
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37
(Continue)
dxdyeewX Y P
wx
y x 0 0
w
dxee wx x
1
10
Chapter 5 : Multiple of Random Variables
(Continue)
0
01 w
w
wX Y PwF W
00
wwdF w w
Chapter 5 : Multiple of Random Variables
02 ww
dw
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38
Theorem : For variable X and Y , the expectedvalue of W = g( X,Y ) is
Expected Value of Function
Discrete:
X Y S x S y
Y X y xP y xgW E ,, ,
Continuous :
Chapter 5 : Multiple of Random Variables
dxdy y xP y xgW E Y X ,,
,
Example
P L,T (l,t ) t = 40
sec
t = 60
sec
• Compute E [ D] directlyfrom P L,T (l,t )
l = 1 page 0.15 0.1
l = 2 pages 0.3 0.2
l = 3 pages 0.15 0.1
• g(l,t ) = lt
3
L T Sl St
T L t lPt lg D E ,, ,
Chapter 5 : Multiple of Random Variables
T L
l t
,,
1 60,40
1.060315.04032.06023.04021.060115.0401
sec 96 D E
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39
Theorem:
Expected Value of Function
E [g1( X,Y ) +…+ gn( X,Y )] = E [g1( X,Y )] +…+ E [gn( X,Y )]
E [ X+Y ] = E [ X ] + E [Y ]
Chapter 5 : Multiple of Random Variables
Conditioning by an Event
Definition : For discrete random variable X and , ,
conditional joint PMF of X and Y given B is
P X,Y | B( x,y) = P[ X = x, Y=y| B]
Chapter 5 : Multiple of Random Variables
EventPair of RV
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40
Conditional joint PMF & PDF
• Know joint of PMF orPDF
• Given Event B
• Want to know condition joint Probability when
P P X,Y X,Y (( x,y x,y))
EventEventBB
probability that ( x,y) B]
Chapter 5 : Multiple of Random Variables
Conditional joint PMF & PDF
• Theorem : Condition Joint PMF of
the X,Y plane with P[ B] > 0 ,
otherwise
B y x BP
y xP
y xP
Y X
BY X
,
0
,
,
,
|,
• Theorem: Conditional Joint PDF of
P( x, y)
P[ B]
Chapter 5 : Multiple of Random Variables
,
otherwise
B y x BP
y x f
y x f
Y X
BY X
,
0
,
,
,
|,
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41
1/16
y
P X,Y ( x,y)
Example
• Random variables X any Y
have the oint PMF P x
1/8
1/8
1/12
1/12
1/12
1/16
1/16
1/16
1
2
3
1 2 3 4
1/4
x
,
as shown. Let B denote theevent X + Y < 4. Find theconditional PMF of X and Y
given B
Chapter 5 : Multiple of Random VariablesChapter 5 : Multiple of Random Variables
(Continue)
y P X,Y ( x,y) B = {( x,y)| X + Y < 4}
1/8
1/8
1/12
1/12
1/12
1/16
1/16
1/16
1
2
3
41/16
1/4
x
= , , , , , , ,
P[ B] = P X,Y (1,1) + P X,Y (2,1) +
P X,Y (2,2) + P X,Y (3,1)= ¼ + 1/8 + 1/8 + 1/12
= 7/12
Chapter 5 : Multiple of Random Variables
X+Y = 4
X+Y < 4
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42
Solution
y xP
y xPY X
BY X
,,
,
|,
• P X,Y|B(1,1) = (¼)/(7/12) = 3/7
• P X,Y|B(2,1) = (1/8)/(7/12) = 3/14
• P X,Y|B(3,1) = (1/12)/(7/12) = 1/7 1
2
3
4
3/14
y
P X,Y|B( x,y)
3/7 3/14 1/7
• P X,Y|B(2,2) = (1/8)/(7/12) = 3/14
Chapter 5 : Multiple of Random Variables
1 2 3 4 x
• Theorem : For random variable X and Y and an
event B o nonzero robabilit the conditional
Conditional Expected Value
expected value of W = g( X,Y ) given B is
Discrete :
X Y S x S y
BY X y xP y xg BW E ,,| |,
Continuous:
Chapter 5 : Multiple of Random Variables
dxdy y x f y xg BY X ,, |,
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43
Example
Find the conditional expected value of W = X + Y
given the event B = { X + Y < 4}
B = {(1,1), (2,1), (3,1), (2,2)}
P X,Y|B( x,y) = {3/7, 3/14, 1/7, 3/14}
1/4
1/8
1/8
1/12
1/12
1/12
1/16
1/16
1/16
1
2
3
41/16
y P X,Y ( x,y)
BY X y xP y x BW E |, ,|
14
41
14
34
7
14
14
33
7
32
Chapter 5 : Multiple of Random Variables
1 2 3 4 x
,
Conditioning by Random Variable
• Outcome of experiment ( x,y) B
Derive new robabilit model for ex eriment conditioning by event
• Change B = { X = x} or B = {Y = y} RVKnowledge of Y derive new probability model
of X conditioning by RV
Con itioning y RV
Conditional PMF of X given Y
Conditional PDF of X given Y
Chapter 5 : Multiple of Random Variables
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44
Condition PMF
• Definition : For any event Y = y such that PY ( y) >
0 the condition PMF o X iven Y = is
P X|Y ( x|y) = P[ X=x|Y=y]
• Theorem : For random variables X and Y with joint
PMF P X,Y ( x,y) and x and y such that P X ( x) > 0 and
Y , X,Y , X|Y Y Y|X X
Chapter 5 : Multiple of Random Variables
• Definition : For y such that f Y ( y) > 0 , the conditional
PDF of X given Y {Y = y} is
Condition PDF
• Theorem : f X,Y ( x,y) = f X|Y ( x|y) f Y ( y) = f Y|X ( y|x) f X ( x)
y f
y x f y x f
Y
Y X
Y X
,|
,
|
x f
y x f x y f
X
Y X
X Y
,|
,
|
Chapter 5 : Multiple of Random Variables
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45
Example
Random variable X and Y
have the oint PMF1/16
y
P X,Y ( x,y), as shown. Findthe condition PMF of Y
given X = x for each x S x
1/8
1/8
1/12
1/12
1/12
1/16
1/16
1/16
1
2
3
1 2 3 4
1/4
x y xP xP
Y X ,|
,
Chapter 5 : Multiple of Random Variables
xP X
(Continue)
X = 1 P( X = 1) = ¼
1/16
y
x
x
2,
1,
8 / 18 / 1
4 / 1
1/8
1/8
1/12
1/12
1/12
1/16
1/16
1/16
12
3
1 2 3 4
1/4
x
Chapter 5 : Multiple of Random Variables
otherwise
x
X
,
4,
,
0
16 / 116 / 116 / 116 / 1
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46
(Continue)
1/16
y
x
x
2,
1,
4 / 1
4 / 1
1/8
1/8
1/12
1/12
1/12
1/16
1/16
1/16
1
2
3
1 2 3 4
1/4
x
otherwise
x
x xP X
,
4,
3,
0
4 / 1
4 / 1
X = 1, Y = 1
1
1,11|1
,
|
X
Y X
X Y P
PP
Chapter 5 : Multiple of Random Variables
141
41
1
1,11|1
,
| X
Y X
X Y P
PP
(Continue)
1/16
y
x
x
2,
1,
4 / 1
4 / 1
2
1
41
81
2
1,22|1
,
| X
Y X
X Y P
PP
1/8
1/8
1/12
1/12
1/12
1/16
1/16
1/16
12
3
1 2 3 4
1/4
x
otherwise
x
x xP X
,
4,
3,
0
4 / 1
4 / 1
Chapter 5 : Multiple of Random Variables
2
1
41
81
2
2,22|2
,
| X
Y X
X Y P
PP
otherwise
y yP X Y
2,1
0
2 / 12||
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47
(Continue)
From Theorem
y xP Y X ,,
otherwise
y yP X Y
2,1
0
2 / 12||
Then
otherwise
y yP X Y
1
0
11|
|
xP X
X Y |
Chapter 5 : Multiple of Random Variables
otherwise
y yP X Y
4,3,2,1
0
4 / 14||
otherwise
y yP
X Y
3,2,1
0
3 / 13||
Example
A server cluster has two servers labeled A and B.Incoming jobs are independently routed by thefront end equipment (called server switch) toserver A with probability p and to server B with
probability (1- p). The number of jobs, X , arrivingper unit time is Poisson distributed withparameter .
of jobs, Y , received by server A, per unit time.
Chapter 5 : Multiple of Random Variables
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48
Solution
A PY [k ] = ?? p
erverSwitch
B
Poisson()
n o s
1- pServer cluster Server cluster
Let X = # of jobs arrive per unit time at servercluster
Y = # of jobs arrive per unit time at server A
Chapter 5 : Multiple of Random Variables
(Continue)
Y X Y k nPk P ,,4
X
X
X is Poisson Distribution
X
Sn
X X Y Y nPnk Pk P ||
1
2
3
1 2 3 4Y
!n
enP
n
X
Chapter 5 : Multiple of Random Variables
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49
(Continue)
• Jobs occur as a sequence of n independent
Bernoulli trials
k nk
X Y p pk
nnk P
1||
• Condition probability that [Y=k ] given [ X=n] is
Chapter 5 : Multiple of Random Variables
Binomial PMF
(Continue)
X X Y Y nPnk Pk P ||
k n
k nk
k n
nk nk
Y
k n
p
k
e p
n
e p p
k
nk P
!
1
!
!1
X Sn
Chapter 5 : Multiple of Random Variables
p
k
Y ek
e pk P
1
!
e (1- p)
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50
• Theorem : Discrete
Conditional Expected Value of a
Function
• Theorem : Continuous
X S x
Y X ,, |
X S x
Y X y x xP yY X E || |
Chapter 5 : Multiple of Random Variables
dx y xP y xg yY Y X g E Y X |,|, |
dx y x xP yY X E Y X
|| |
Independent Random Variables
Definition : Random variables X and Y are independent
i and onl i
Discrete : P X,Y ( x,y) = P X ( x)PY ( y)
Continuous : f X,Y ( x,y) = f X (x)f Y (y)
Chapter 5 : Multiple of Random Variables
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Example
• Are Q and R independent?
r
M
q M M p p
p p pr RPqQP
1
11
r RP
p pr Mq
1
r P ,
Therefore Q and R are independent
Chapter 5 : Multiple of Random Variables
,
References
1. Alberto Leon-Garcia, Probability and RandomProcesses for Electrical En ineerin 3rd Ed. Addision-Wesley Publishing, 2008
2. Roy D. Yates, David J. Goodman, Probabilityand Stochastic Processes: A FriendlyIntroduction for Electrical and ComputerEngineering, 2nd, John Wiley & Sons, Inc, 2005
3. Jay L. Devore, Probability and Statistics forEngineering and the Sciences, 3rd edition,Brooks/Cole Publishing Company, USA, 1991.