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EEE376F
Signals and Systems II
H Anthony Chan
[email protected]
http://www.eng.uct.ac.za/~achan
Mqhele Dlodlo
[email protected]
http://www.uct.ac.za/depts/staff/mdlodlo/
Department of Electrical Engineering
University of Cape Town
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
Signals and Systems II
304 Page 1 May 25, 2017
Independent Random Variables
Study Bertsekas and Tsitsiklis
Do and understand exercises
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
Signals and Systems II
304 Page 2 May 25, 2017
Independent Random Variables
 The random variables X1, X2, …, Xn are
independent means that
 for any events E1, E2, …, En which involves only X1,
X2, …, Xn respectively, the probability of the joint
event, P[E], is equal to the product of the
probabilities of the individual events P[E1] P[E2] …
P[En]:
 P[X1E1, …, XnEn] = P[X1E1] … P[XnEn]
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
Signals and Systems II
304 Page 3 May 25, 2017
Independent Random Variables
Example
 Assumption: The birthday (month and day) of a
person is randomly distributed. The person’s
anniversary (month and day) is also randomly
distributed
 The event that the husband’s birthday is on
Christmas and the event that the wife’s birthday is
on Christmas are independent, so that the
probability that both birthdays are on Christmas is
1/365  1/365
 However, the event that the husband’s anniversary
is on Christmas and the event that the wife’s
anniversary is on Christmas are not independent
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
Signals and Systems II
304 Page 4 May 25, 2017
Joint Probability Mass Function
Independent Random Variables
Probability
Joint pmf at X=xj,Y=yk
Discrete: Continuous:
(xj,yk)
(x,y)
pX,Y(xj, yk) =
pX(xj) pY(yk)
FX,Y(xj, yk) =
FX(xj) FY(yk)
per unit area fX,Y(xj, yk) =
fX(xj) fY(yk)
Joint CDF for Xx,Yy
Joint pdf
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
FX,Y(x, y) =
FX,(x) FY(y)
fX,Y(x, y) = fX,(x)
fY(y)
Signals and Systems II
304 Page 5 May 25, 2017
Probability Mass Function
Independent Discrete Random Variables
 The discrete random variables X1, X2, …, Xn are
independent iff
 the joint probability mass function pX1, X2, …, Xn(x1,
x2, …, xn) is equal to the product of the marginal
probability mass functions, pX1(x1) pX2(x2) … pXn(xn),
for all x1, x2, …, xn .
 Recall that
 X1, X2, …, Xn are independent means P[X1E1, …,
XnEn] = P[X1E1] … P[XnEn]
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
Signals and Systems II
304 Page 6 May 25, 2017
Probability Mass Function
Independent Discrete Random Variables
 If X and Y are independent,
 which means P[XE1, YE2] = P[XE1] P[YE2],
 then,  xj and yk ,
 pX,Y(xj, yk)
 = P[X=xj, Y=yk] = P[X=xj] P[Y=yk]

= pX(xj) pY(yk)
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
Signals and Systems II
304 Page 7 May 25, 2017
Probability Mass Function
Independent Discrete Random Variables
 P[XE1, YE2] = P[XE1] P[YE2]
E1
E2 s11,s22
s12,s22
s13,s22
s11,s21
s12,s21
s13,s21
 then,  xj and yk ,
y2
 pX,Y(xj, yk)
 = P[X=xj,Y=yk] = P[X=xj] P[Y=yk] y1

= pX(xj) pY(yk)
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
+
+
+
x1
x2
x3
+ + ++
+ + ++
Signals and Systems II
304 Page 8 May 25, 2017
Probability Mass Function
Independent Discrete Random Variables
 If pX,Y(xj, yk) = pX(xj) pY(yk) ,  xj and yk
 then for E = (E1, E2),
 P[E]
 = xjE1 ykE2 pX,Y(xj, yk) = xjE1 ykE2 pX(xj) pY(yk)

= xjE1 pX(xj) ykE2 pY(yk)

= P[XE1] P[YE2]
 i.e., P[E] = P[XE1] P[YE2]
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
Signals and Systems II
304 Page 9 May 25, 2017
Probability Mass Function
Independent Discrete Random Variables
 pX,Y(xj, yk) = pX(xj) pY(yk) ,  xj and yk
 P[E]=pXY(x2,y2)+pXY(x3,y2) = pX(x2)pY(y2)+pX(x3)pY(y2)

= {pX(x2) + pX(x3)} pY(y2)

= P[XE1] P[YE2]
E1
E2 s11,s22 s12,s22 s13,s22 s13,s22
s11,s21 s12,s21 s13,s21 s13,s21
y2
y1
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
+
+
+
+
+
+
+
x1
+
x2
+
x3
+
+
+
+
x3
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
Signals and Systems II
304 Page 10 May 25, 2017
Example
Marginal probability mass function
 Marginal pmf: pX(xj) = k pX,Y(xj, yk)
 pX(0) = (1+7)/64 = 1/8
 pX(1) = (3+21)/64 = 3/8
 pX(2) = (3+21)/64 = 3/8
 pX(3) = (1+7)/64 = 1/8
y
1 +7/64 +21/64 +21/64 +7/64
0 +1/64 +3/64 +3/64 +1/64x
0
1
2
3
 pY(0) = (1+3+3+1)/64 = 1/8
 pY(1) = (7+21+21+7)/64 = 7/8
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
Signals and Systems II
304 Page 11 May 25, 2017
Example
Joint probability mass function
 Joint pmf pX,Y(x, y) = pX(x) pY(y)
 pX,Y(0, 0) = 1/64 = pX(0) pY(0)
 pX,Y(1, 0) = 3/64 = pX(1) pY(0)
 pX,Y(2, 0) = 3/64 = pX(2) pY(0)
 pX,Y(3, 0) = 1/64 = pX(3) pY(0)
y
7/64
21/64
21/64
7/64
7/8 1 +
+
+
+
1/64
1/8 0 +
0
1/8
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
3/64
+
1
3/8
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
3/64
+
2
3/8
1/64
+ x
3
1/8
Signals and Systems II
304 Page 12 May 25, 2017
Example
Joint probability mass function
 Joint pmf pX,Y(x, y) = pX(x) pY(y)
 pX,Y(0, 1) = 7/64 = pX(0) pY(1)
 pX,Y(1, 1) = 21/64 = pX(1) pY(1)
 pX,Y(2, 1) = 21/64 = pX(2) pY(1)
 pX,Y(3, 1) = 7/64 = pX(3) pY(1)
y
7/64
21/64
21/64
7/64
7/8 1 +
+
+
+
1/64
1/8 0 +
0
1/8
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
3/64
+
1
3/8
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
3/64
+
2
3/8
1/64
+ x
3
1/8
Signals and Systems II
304 Page 13 May 25, 2017
Joint Cumulative Distribution Function
Independent Random Variables
Probability
Joint pmf at X=xj,Y=yk
Discrete: Continuous:
(xj,yk)
(x,y)
pX,Y(xj, yk) =
pX(xj) pY(yk)
FX,Y(xj, yk) =
FX(xj) FY(yk)
per unit area fX,Y(xj, yk) =
fX(xj) fY(yk)
Joint CDF for Xx,Yy
Joint pdf
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
FX,Y(x, y) =
FX,(x) FY(y)
fX,Y(x, y) = fX,(x)
fY(y)
Signals and Systems II
304 Page 14 May 25, 2017
Cumulative Distribution Function
 The random variables X1, X2, …, Xn are
independent iff the joint Cumulative Distribution
Function FX1,X2,…,Xn(x1, x2, …, xn) is equal to the
product of the marginal Cumulative Distribution
Functions, FX1(x1) FX2(x2) … FXn(xn), for all x1, x2, …,
xn .
 X, Y are independent iff
 FX,Y(x, y) = FX(x) FY(y),  x and y
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
y
(x, y)
x
Signals and Systems II
304 Page 15 May 25, 2017
Joint probability density function
Independent Random Variables
Probability
Joint pmf at X=xj,Y=yk
Discrete: Continuous:
(xj,yk)
(x,y)
pX,Y(xj, yk) =
pX(xj) pY(yk)
FX,Y(xj, yk) =
FX(xj) FY(yk)
per unit area fX,Y(xj, yk) =
fX(xj) fY(yk)
Joint CDF for Xx,Yy
Joint pdf
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
FX,Y(x, y) =
FX,(x) FY(y)
fX,Y(x, y) = fX,(x)
fY(y)
Signals and Systems II
304 Page 16 May 25, 2017
Probability density function
Independent Random Variables
 The random variables X1, X2, …, Xn are
independent iff the joint probability density function
fX1,X2,…,Xn(x1, x2, …, xn) is equal to the product of the
marginal probability density functions, fX1(x1)
fX2(x2) … fXn(xn), for all x1, x2, …, xn .
 X, Y are independent iff
 fX,Y(x, y) = fX(x) fY(y),  x and y
y
(x, y)
x
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
Signals and Systems II
304 Page 17 May 25, 2017
Conditional Probability
for Multiple Random Variables
Study Bertsekas and Tsitsiklis
Do and understand exercises
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
Signals and Systems II
304 Page 18 May 25, 2017
Conditional probability
Review
 Define
P( A  B)
 P(A|B) 
P( B)
provided P(B)  0
 Then:
 P(AB) = P(B) P(A|B) = P(A) P(B|A)
A
B
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
Signals and Systems II
304 Page 19 May 25, 2017
Conditional probability
Review
 Independent: P(AB) = P(A)P(B)
P( A  B)
 P(A|B) 
P( B)


P( A) P( B)
=
P( B)
P ( A)
= P(A) =
P( S )
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
BC
B
AB
A
AC
A
AC
BC
B
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
Signals and Systems II
304 Page 20 May 25, 2017
Conditional probability
Review
 Not Independent
P( A  B)
 P(A|B) 
P( B)
BC
B
AB
A
 P(A) =
P ( A)
P( S )
 where P(S) = 1
BC
B
AB
A
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
AC
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
AC
Signals and Systems II
304 Page 21 May 25, 2017
Conditional Probability
 Conditional probability
P[ X  E1,Y  yk ]
 P[XE1|Y=yk] =
for P[Y = yk] > 0
P[Y  yk ]

= 0 for P[Y = yk] = 0
 If X and Y are independent

What I have is only borrowed from God so
that I may serve others. H Anthony Chan
=
P[ X  E1 ]P[Y  yk ]
= P[XE1]
P[Y  yk ]
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
Signals and Systems II
304 Page 22 May 25, 2017
Conditional Probability
 Vector Random Variables: (X,Y): S  (SX,SY )
E1
y2
y1
E2 s11,s22
s12,s22;s14,s22
s13,s22
s11,s21
s12,s21;s14,s21
s13,s21
+
+
+ +
+ +
++
++
x1
x2
x3
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
+
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
Signals and Systems II
304 Page 23 May 25, 2017
EEE376F
Signals and Systems II
Whatever you did for one of these least
ones, you did for me. (Matthew 25:45)
© 2005
What I have is only borrowed from God so
that I may serve others. H Anthony Chan
Whatever you did for one of these least ones, you did for me.
(Matthew 25:45)
Signals and Systems II
304 Page 24 May 25, 2017
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