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Continuous Random Variables: Joint PDFs,
Conditioning, Expectation and Independence
Berlin Chen
Department of Computer Science & Information Engineering
National Taiwan Normal University
Reference:
- D. P. Bertsekas, J. N. Tsitsiklis, Introduction to Probability , Sections 3.4-3.6
Multiple Continuous Random Variables (1/2)
• Two continuous random variables X and Y associated
with a common experiment are jointly continuous and can
be described in terms of a joint PDF f X ,Y satisfying
P  X , Y   B  
–
f X ,Y
 f X ,Y  x , y dxdy
 x , y B
is a nonnegative function
 
– Normalization Probability    
f X ,Y  x , y dxdy  1
• Similarly, f X ,Y a , c  can be viewed as the “probability per
unit area” in the vicinity of a, c 
P a  X  a   , c  Y  c   
 aa   cc  f X ,Y  x , y dxdy  f X ,Y a , c    2
– Where  is a small positive number
Probability-Berlin Chen 2
Multiple Continuous Random Variables (2/2)
•
Marginal Probability
P  X  A   P  X  A and Y    ,  
  X  A  f X ,Y x , y dydx
– We have already defined that
P X  A  
•

X A
f X  x dx
We thus have the marginal PDF
f X x    f X ,Y x , y dy
Similarly
f Y  y    f X ,Y x , y dx
Probability-Berlin Chen 3
An Illustrative Example
•
Example 3.10. Two-Dimensional Uniform PDF. We are told that
the joint PDF of the random variables X and Y is a constant c
on an area S and is zero outside. Find the value of c and the
marginal PDFs of X and Y .
The correspond ing uniform joint PDF on
an area S is defined to be (cf. Example 3.9)
1

,

f X ,Y x,y    Size of area S
0,

 f X ,Y x,y  
if x,y   S
otherwise
1
for x,y   S
4
for 1  x  2
for 1  y  2
 f X  x   14 f X,Y  x,y dy
1
3
 14 dy 
4
4
for 2  x  3
 f Y  y   12 f X,Y  x,y dx
1
1
 12 dx 
4
4
for 2  y  3
 f X  x   23 f X,Y  x,y dy
1
1
 23 dy 
4
4
 f Y  y   13 f X,Y  x,y dx
1
1
 13 dx 
4
2
for 3  y  4
 f Y  y   12 f X,Y  x,y dx
1
1
 12 dx 
4
4
Probability-Berlin Chen 4
Joint CDFs
• If X and Y are two (either continuous or discrete)
random variables associated with the same experiment ,
their joint cumulative distribution function (Joint CDF) is
defined by
F X ,Y  x , y   P  X  x , Y  y 
– If X and Y further have a joint PDF f X ,Y ( X and Y are
continuous random variables) , then
F X ,Y  x , y   x y f X ,Y s , t dsdt
And
f X ,Y  x , y  
If
 F X ,Y  x , y 
2
xy
F X ,Y can be differentiated at the point  x, y 
Probability-Berlin Chen 5
An Illustrative Example
• Example 3.12. Verify that if X and Y are described by a
uniform PDF on the unit square, then the joint CDF is
given by
F X ,Y  x , y   P  X  x , Y  y   xy , for 0  x,y  1
Y
 2 FX ,Y  x, y 
xy
0,1
1,1
0,0 
1,0 
X
 1  f X ,Y  x, y , for all x, y  in the unit square
Probability-Berlin Chen 6
Expectation of a Function of Random Variables
• If X and Y are jointly continuous random variables,
and g is some function, then Z  g  X , Y  is also a
random variable (can be continuous or discrete)
– The expectation of Z can be calculated by
E Z   E g  X , Y     g  x , y  f X ,Y  x , y dxdy
– If Z is a linear function of X and Y , e.g., Z  aX  bY , then
E Z   E aX  bY   a E  X   b E Y 
• Where
a and b are scalars
We will see in Section 4.1 methods for computing the PDF of (if it has one).
Z
Probability-Berlin Chen 7
More than Two Random Variables
• The joint PDF of three random variables X , Y and Z
is defined in analogy with the case of two random
variables
P  X , Y , Z   B  
 f X ,Y , Z x , y , z dxdydz
 X ,Y , Z  B
– The corresponding marginal probabilities
f X ,Y  x , y    f X ,Y , Z x , y , z dz
f X  x     f X ,Y , Z  x , y , z dydz
• The expected value rule takes the form
E g  X , Y , Z      g x , y , z  f X ,Y , Z x , y , z dxdy dz
– If g is linear (of the form aX  bY  cZ ), then
E aX  bY  cZ   a E X   b E Y   c E Z 
Probability-Berlin Chen 8
Conditioning PDF Given an Event (1/3)
• The conditional PDF of a continuous random variable X ,
given an event A
– If A cannot be described in terms of X , the conditional PDF
is defined as a nonnegative function f X A  x  satisfying
P  X  B A   B f X
A
x dx
• Normalization property

 
fX
A
 x dx
1
Probability-Berlin Chen 9
Conditioning PDF Given an Event (2/3)
– If A can be described in terms of X ( A is a subset of the real
line with P  X  A   0 ), the conditional PDF is defined as a
nonnegative function f X A  x  satisfying
fX
 f X x 
,

A x    P  X  A 
0,

if x  A
otherwise
• The conditional PDF is zero outside the
conditioning event
and for any subset
P X  B X 
fX
B
P X
A

remains the same shape as
f X except that it is scaled along
the vertical axis
 B, X  A
P X  A 
 A  B f X  x dx
P X  A 
  A  B f X A  x dx

– Normalization Property   f X
A
x dx
 A f X
A
A
 x dx
1
Probability-Berlin Chen 10
Conditioning PDF Given an Event (3/3)
• If A1 , A2 ,  , An are disjoint events with P  Ai   0 for
each i , that form a partition of the sample space, then
n
f X x    P  A i  f X
i 1
Ai
x 
– Verification of the above total probability theorem
think of as an event , X  x
B
P  X  x    P  Ai P X  x Ai 
n
i 1

x
 
n
f X t dt   P  Ai x f X
i 1
Ai
and use the total probability theorem from Chapter 1
t dt
Taking the derivative of both sides with respect to x
n
 f X x    P  Ai  f X
i 1
Ai
x 
Probability-Berlin Chen 11
Illustrative Examples (1/2)
• Example 3.13. The exponential random variable is
memoryless.
– The time T until a new light bulb burns out is exponential
distribution. John turns the light on, leave the room, and when he
returns, t time units later, find that the light bulb is still on, which
corresponds to the event A={T>t}
– Let X be the additional time until the light bulb burns out. What is
the conditional PDF of X given A ?
X  T  t , A  T  t
T is exponential
e t , t  0
f T t   
otherwise
0,
PT  t   e   t
The conditional CDF of X given A is defined by  The conditiona l PDF of X given
P X  x A  PT  t  x T  t  (where x  0)
 PT  t  x T  t  
PT  t  x 
PT  t 
e  t  x 

e  t
 e  x
PT  t  x and T  t 
PT  t 
the event A is also exponentia l
with parameter  .

Probability-Berlin Chen 12
Illustrative Examples (2/2)
•
Example 3.14. The metro train arrives at the station near your home
every quarter hour starting at 6:00 AM. You walk into the station
every morning between 7:10 and 7:30 AM, with the time in this
interval being a uniform random variable. What is the PDF of the
time you have to wait for the first train to arrive?
- The arrival time, denoted by X , is a uniform random
variable over the interval 7 : 10 to 7 : 30
- Let random varible Y model the waiting time
- Let A be a event
A  7 : 10  X  7 : 15 (You board the 7 : 15 train)
1/20
- Let B be a event
B  7 : 15  X  7 : 30 (You board the 7 : 30 train)
- Let Y be uniform conditioned on A
- Let Y be uniform conditioned on B
For 0  y  5, PY  y  
Total Probability theorem:
PY  y   P  A PY
A
 y   P B PY B  y 
1 1 3 1
1
  

4 5 4 15 10
1
3 1
1
For 5  y  15, PY  y    0  

4
4 15 20
Probability-Berlin Chen 13
Conditioning one Random Variable on Another
• Two continuous random variables X and Y have a joint
PDF. For any y with fY  y   0 , the conditional PDF of X
given that Y  y is defined by
f X Y x y  
f X ,Y  x , y 
fY  y 

– Normalization Property  
f X Y x y dx  1
• The marginal, joint and conditional PDFs are related to
each other by the following formulas
f X ,Y  x , y   f Y  y  f X Y x y ,
f X  x    f X ,Y  x , y dy .
marginalization
Probability-Berlin Chen 14
Illustrative Examples (1/2)
• Notice that the conditional PDF f X Y x y  has the same
shape as the joint PDF f X ,Y x , y  , because the
normalizing factor f Y  y  does not depend on x
f X Y x 3 . 5  
f X Y x 3 . 5  
f X ,Y  x ,3 .5 
f Y 3 .5 
f X ,Y  x , 2 .5 
f Y 2 .5 
f X Y x 3 .5  

f X ,Y  x ,1 .5 
f Y 1 .5 

1/ 4
1
1/ 4
1/ 4
 1/ 2
1/ 2

1/ 4
1
1/ 4
cf. example 3.13
Figure 3.16: Visualization of the conditional PDF f X Y x y  .
Let X , Y have a joint PDF which is uniform on the set S . For
each fixed y , we consider the joint PDF along the slice Y  y
and normalize it so that it integrates to 1
Probability-Berlin Chen 15
Illustrative Examples (2/2)
•
Example 3.15. Circular Uniform PDF. Ben throws a dart at a
circular target of radius r . We assume that he always hits the target,
and that all points of impact  x , y  are equally likely, so that the
joint PDF f X ,Y x, y  of the random variables x and y is uniform
– What is the marginal PDF f Y  y 
1

, if  x , y  is in the circle

f X ,Y  x , y    area of the circle
 0,
otherwise
 1
2
2
2
 2, x y r
  πr
0,
otherwise

fY  y  



 
f X,Y  x,y dx  x 2  y 2  r 2
1
πr 2
2
2
x 2  y 2  r 2 1dx 
1
πr 2
r  y , if y  r
2

fX
1
πr
2
Y x y  
f X,Y  x,y
fY y 
dx
r2  y2
 r2 y

1dx
2
2
πr
(Notice here that PDF f Y  y  is not uniform)

1
πr 2
2
πr 2
r2  y2
1

2
r
2
 y
For each value y , f X
2
Y
,
x y 
if x
2
 y2  r2
is uniform
Probability-Berlin Chen 16
Conditional Expectation Given an Event
• The conditional expectation of a continuous random
variable X , given an event A ( P  A   0 ), is defined by
E X A    xf
X A
 x dx
– The conditional expectation of a function
form
Eg  X  A   g  x  f X
– Total Expectation Theorem

n
EX    P  Ai E X Ai
and
i 1
A
g  X  also has the
x dx

Eg  X    P  Ai Eg  X  Ai 
n
i 1
A1 , A2 ,  , An
• Where
are disjoint events with P  Ai   0 for
each i , that form a partition of the sample space
Probability-Berlin Chen 17
An Illustrative Example
•
Example 3.17. Mean and Variance of a Piecewise Constant PDF.
Suppose that the random variable X has the piecewise constant
PDF
if 0  x  1,
1 / 3,

f X  x    2 / 3,
0,

if 1  x  2 ,
otherwise.
Define event A1  X lies in the first interval [0,1] 
event A2  X lies in the second interval [1,2] 
 P  A1   01 1 / 3 dx 1 / 3, P  A2   12 2 / 3 dx  2 / 3
fX
 f X x 
 1, 0  x  1



P
X
A

x




1
A1
0,
otherwise

Recall that the mean and second moment of
a uniform random variable over an interval


[ a, b ] is a  b / 2 and a 2  ab  b 2 / 3

E X A   3 / 2 , E X
 E X A1   1 / 2 , E X
2

A1  1 / 3
2
2

A2  7 / 3
fX
 f X x 
 1, 1  x  2



P
X
A

x




2
A2
0,
otherwise

 E  X   P  A1 E X A1   P  A2 E X A2 
 1 / 3 1 / 2  2 / 3  3 / 2  7 / 6
   P  A E X
E X
2
1
2


A1  P  A2 E X
2
A2

 1 / 3  1 / 3  2 / 3  7 / 3  15 / 9
 var  X   15 / 9  7 / 6 2  11 / 36
Probability-Berlin Chen 18
Conditional Expectation Given a Random Variable
• The properties of unconditional expectation carry though,
with the obvious modifications, to conditional expectation
E X Y  y    xf X Y x y dx
E g  X Y  y    g  x  f X Y x y dx
E g  X , Y Y  y    g  x , y  f X Y x y dx
Probability-Berlin Chen 19
Total Probability/Expectation Theorems
• Total Probability Theorem
– For any event A and a continuous random variable Y
P A 

 
P  A Y  y  f Y  y dy
• Total Expectation Theorem
– For any continuous random variables X and Y
E  X    E X Y  y  f Y  y dy





E g X    E g  X Y  y  f Y  y dy
E g  X , Y    E g  X , Y Y  y  f Y  y dy
Probability-Berlin Chen 20
Independence
• Two continuous random variables X and Y are
independent if
f X ,Y  x , y   f X  x  f Y  y , for all x,y
– Since that
f X ,Y  x , y   f Y  y  f X Y x y   f X  x  f Y
X
y x 
• We therefore have
f X Y x y   f X  x , for all x and all y with fY  y   0
• Or
fY X  y x   fY  y , for all y and all x with f X  x   0
Probability-Berlin Chen 21
More Factors about Independence (1/2)
• If two continuous random variables X and Y are
independent, then
– Any two events of the forms X  Aand Y  B  are
independent
P  X  A , Y  B   x A  y B f X ,Y x , y dydx
 x A  y B f X x  f Y  y dydx

 x A f X x dx   y B f Y  y dy
 P  X  A P Y  B 

– It also implies that
F X ,Y x , y   P  X  x , Y  y   P  X  x P Y  y   F X x FY x 
– The converse statement is also true (See the end-of-chapter
problem 28)
Probability-Berlin Chen 22
More Factors about Independence (2/2)
• If two continuous random variables X and Y are
independent, then
–
–
E  XY   E  X E Y 
var  X  Y   var  X   var Y 
– The random variables g X  and hY  are independent for any
functions g and h
• Therefore,
E g  X h Y   E g  X E h Y 
Probability-Berlin Chen 23
Recall: the Discrete Bayes’ Rule
• Let A1, A2 ,, An be disjoint events that form a partition of
the sample space, and assume that P  Ai   0, for all i .
Then, for any event B such that P B   0 we have


P Ai B 



P  Ai P B Ai
P B 


P  Ai P B Ai

Multiplication rule
Total probability theorem

 nk 1 P  Ak P B Ak



P  Ai P B Ai



P  A1 P B A1    P  An P B An

Probability-Berlin Chen 24
Inference and the Continuous Bayes’ Rule
• As we have a model of an underlying but unobserved
phenomenon, represented by a random variable X with
PDF f X , and we make a noisy measurement Y , which
is modeled in terms of a conditional PDF f Y X . Once the
experimental value of Y is measured, what information
does this provide on the unknown value of X ?
X
f X x 
Y
Measurement
fY
f X Y x y  
X
y x 
Inference
f X Y x y 
y x 
 
fY  y 
  f X t  f Y X  y t dt
f X ,Y  x , y 
f X x  fY
X
Note that
f X fY
X
 f X ,Y  f Y f X
Y
Probability-Berlin Chen 25
Inference and the Continuous Bayes’ Rule (2/2)
Inference about a Discrete Random Variable
• If the unobserved phenomenon is inherently discrete
– Let N is a discrete random variable of the form N  n that
represents the different discrete probabilities for the unobserved
phenomenon of interest, and p N be the PMF of N
P N  n Y  y   P N  n y  Y  y   
P N  n P  y  Y  y   N  n 

P y  Y  y   


p N n  f Y
N
 y n 
f Y  y 
y n 
i  f  y i 
p N n  f Y
p
i
N
N
Total probability theorem
Y N
Probability-Berlin Chen 26
Illustrative Examples (1/2)
•
Example 3.19. A lightbulb produced by the General Illumination
Company is known to have an exponentially distributed lifetime Y .
However, the company has been experiencing quality control
problems. On any given day, the parameter    of the PDF of Y
is actually a random variable, uniformly distributed in the interval
1, 3 / 2  .
– If we test a lightbulb and record its lifetime ( Y  y ), what can
we say about the underlying parameter  ?
f Y   y     e   y , y  0,   0
Conditioned on    , Y has a exponential distribution
with parameter 
 2, for 1    3 / 2
f     
 0, otherwise
f  Y  y  
f    f Y   y  
3/ 2
1
f  t  f Y

 y t dt

2 e y
3/ 2
 ty
1 2te dt
,
for 1  λ  3 / 2
Probability-Berlin Chen 27
Illustrative Examples (2/2)
•
Example 3.20. Signal Detection. A binary signal S is transmitted,
and we are given that P S  1  p and P S   1  1  p .
– The received signal is Y  S  N , where N is a normal noise
with zero mean and unit variance , independent of S .
S  1 , as a function of the observed value
– What is the probability that
y of Y ?
fY
S y s  
Conditioned on S
2
1
e   y  s  / 2 , for s  1 and -1, and -  y  
2 
s,Y
P S  1 Y  y  
has a normal distribution with mean
p S 1 f Y
S
 y 1
fY  y 

p S 1 f Y
p S 1 f Y
S
S
s
and unit variance
 y 1
 y 1  p S  1 fY S  y  1
2
1
e   y 1 / 2
2
2
2
1
1
e   y 1 / 2  1  p 
e   y 1 / 2
2
2
p

p

e


 y 2 1 / 2
e  y
y
2

1 / 2
 pe  e

 pe y
  1  p e  y
 y 2 1 / 2

pe y
pe y  1  p e  y
Probability-Berlin Chen 28
Inference Based on a Discrete Random Variable
• The earlier formula expressing P A Y  y  in terms of
f Y A  y  can be turned around to yield
fY
A
y  

P A fY
A
y  
  P  A  f Y
f Y  y P A Y  y 
P A
?
f Y  y P A Y  y 

 
f Y t P A Y  t dt
f Y  y P A Y  y 



y
dy


A

f Y  y P A Y  y dy
 P  A    f Y  y P A Y  y dy ( normalizat ion property :  f Y
A
 y dy  1)
Probability-Berlin Chen 29
Recitation
• SECTION 3.4 Joint PDFs of Multiple Random Variables
– Problems 15, 16
• SECTION 3.5 Conditioning
– Problems 18, 20, 23, 24
• SECTION 3.6 The Continuous Bayes’ Rule
– Problems 34, 35
Probability-Berlin Chen 30
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