Bivariate Distributions
Wen-Guey Tzeng
Computer Science Department
National Chiao Tung University
Bivariate distribution
Motivation: multiple factors affect the result
• The relation of annual incomes and years of
education
• The age distribution for married couples
• Investment return between stocks and bonds
• Life span of incomes, races, areas, smoking, …
2
Example
• Xavier and Yvette are two real estate agents. Let X
and Y denote the number of houses that Xavier
and Yvette will sell next week, respectively.
3
0.42
p(x,y)
0.21
0
1
2
0
0.12
0.42
0.06
1
0.12
0.06
0.03
2
0.07
0.02
0.01
0.12
0.06
0.06
0.07
0.02
0.01
Y
X=0
y=0
X
0.03
y=1
y=2
X=1
X=2
4
Joint probability mass function (joint pmf)
• Let X and Y be discrete random variables defined
on the same sample space S
• X’s value set is A and Y’s value set is B
• The joint pmf of X and Y is
𝑝 𝑥, 𝑦 = 𝑃 𝑋 = 𝑥, 𝑌 = 𝑦
where
𝑝 𝑥, 𝑦 = 1
𝑥∈𝐴
𝑦∈𝐵
5
6
Marginal pmf
• X’s marginal probability mass function is
𝑝𝑋 𝑥 = 𝑦∈𝐵 𝑝 𝑥, 𝑦
• Y’s marginal probability mass function is
𝑝𝑌 𝑦 = 𝑥∈𝐴 𝑝 𝑥, 𝑦
7
Joint cumulative distribution function (Joint cdf)
8
• A small college has 90 male and 30 female
professors. An ad-hoc committee of 5 is selected at
random to write the vision and mission of the
college.
• Let X and Y be the number of men and women on
this committee, respectively.
(a) Find the joint pmf of X and Y.
(b) Find pX and pY, the marginal pmf’s of X and Y.
9
Sol:
  90  30 
   
x
y

   
(a ) p ( x, y )   120 

 
  5 

0
if x, y  {0, 1, 2, 3, 4, 5}, x  y  5
otherwise.
 90  30 
 

x  5  x 

(b) p X ( x) 
, x  {0, 1, 2, 3, 4, 5},
120




 5 
 90  30 

 
5  y  y 
pY ( y )  
, y  {0, 1, 2, 3, 4, 5}.
120 


 5 
• Roll a balanced die and let the outcome be X.
• Then toss a fair coin X times and let Y denote the
number of tails.
• What is joint probability function?
• What are pX and pY?
11
Sol:
y
pX(x)
0
1
2
3
4
5
6
1
1/12
1/12
0
0
0
0
0
1/6
2
1/24
2/24
1/24
0
0
0
0
1/6
3
1/48
3/48
3/48
1/48
0
0
0
1/6
4
1/96
4/96
6/96
4/96
1/96
0
0
1/6
5
1/192
5/192
10/192 10/192
5/192
1/192
0
1/6
6
1/384
6/384
15/384 20/384
15/384
6/384
1/384
1/6
63/384
120/384
99/384 64/384
29/384
8/384
1/384
x
pY(y)
12
Joint probability density function (joint pdf)
• Continuous random variables X and Y, defined on
the same sample space, have a joint pdf f(x,y) on
R2, if for any region S in the xy-plane
P(( X , Y )  S )   f ( x, y )dxdy.
S
where
 f ( x, y)dxdy  1
R2
Properties
(a) P(X  a, Y  b)  
a
a

b
b
f ( x, y )dxdy  0
(b) P(a  X  b, c  Y  d)  
b
a
(c) P(X  A, Y  B)  

A B

d
c
f ( x, y )dxdy
f ( x, y )dxdy
16
Marginal pdf
• X’s and Y’s marginal probability density functions are

f X ( x)   f ( x, y )dy


fY ( y )   f ( x, y )dx

Joint (cumulative) distribution function
• X and Y’s joint distribution function is
F (t , u )  P( X  t , Y  u )  -   t, u  .
• X’s marginal distribution function is
FX (t )  P( X  t )  P( X  t , Y  )  F (t , )
Similarly FY (u )  P(Y  u )  F (, u )
Suppose that the joint density function of X and Y is f ( x, y )
F ( x, y )  P ( X  x, Y  y )  
y

x
 
f (t , u )dtdu.
Assuming that the partial derivatives of F exist, we get
2
f ( x, y ) 
F ( x, y ).
xy

x


Moreover, FX ( x)  F ( x,)   (  f (t, u )dt )du
x

x



  (  f (t, u )du )dt   f X (t )dt ,
y
and similarly, FY (y)   fY (u )du.

And FX' ( x)  f X ( x), FY' ( y )  fY ( y )
• The joint density function of random variables X and
Y is given by
xy 2
f ( x, y )  
0
0  x  y 1
otherwise.
(a) Determine the value of 
(b) Find the marginal density function of X and Y
22
1
1
0
x
(a)  (  xy 2 dy ) dx  1 
1 3 1
1   (  y dy )xdx   [ y ] x xdx
0
x
0 3
1 1
1 3

   (  x ) xdx 
0 3
3
10
   10
1
10
2
(b) f X ( x )   10 xy dy 
x (1  x 3 ), 0  x  1
x
3
1
1
fY ( y ) 
2

y
0
1
10 xy 2 dx  5 y 4 , 0  y  1
23
For   0, let
1  e  ( x  y )
F ( x, y )  
0
if x  0, y  0
otherwise.
Determine if F is the joint distribution function of 2
random variables X and Y.
24
If F is the joint density of X and Y then
2
F ( x, y ) is the joint density function of X and Y.
xy
But
3  ( x  y )

 e

F ( x, y )  
xy
0
2
if x  0, y  0
otherwise.

Since
F ( x, y )  0, it cannot be a joint density function.
xy
Therefore, F is not a joint distribution function.
2
25
• A circle of radius 1 is inscribed in a square with sides
of length 2.
• A point is selected at random from the square
• What is the probability that the point is within the
circle.
y
x
26
c
f ( x, y )  
0
where




 
if 0  x  2, 0  y  2
otherwise,
f ( x, y )dxdy  0,
implying c  1/4.
Now let R be the region inside the circle.
1

dxdy  .

4
4
R
27
• Let S be a subset of the plane with area A(S).
• A point is randomly selected from S
• For any subset R of S with area A(R), the probability
that R contains the point is A(R)/A(S).
28
• A man invites his fiancee to a fine hotel for a Sunday
brunch.
• They decide to meet in the lobby of the hotel
between 11:30 A.M. and 12 noon.
• They arrive at random times during this period.
• What is the probability that they will meet within 10
minutes?
29
• Sol:
S  {( x, y ) : 0  x  30, 0  y  30}
R  {( x, y )  S : | x  y |  10}
So the desired probability is 500/900  5/9
30
30
30
Theorem. Let f(x,y) be the density function of X and Y.
If h is a function of two variables from R2 to R, then
Z=h(X,Y) is a random variable with the expected value
E (Z )  



 
h( x, y ) f ( x, y )dxdy.
31
Corollary For X and Y, E(X+Y) = E(X) + E(Y)
Proof:
E( X  Y )  


 

 


 
( x  y ) f ( x, y )dxdy
xf ( x, y )dxdy  



 
yf ( x, y )dxdy
 EX  EY .
32
• Let X and Y have joint density function
3 2

 ( x  y 2 ) if 0  x  1, 0  y  1
f ( x, y )   2

otherwise.
0
Find E(X2+Y2)
33
Sol:
E( X  Y )  
2
2



 
( x 2  y 2 ) f ( x, y )dxdy
3 2
2 2
   ( x  y ) dxdy
0 0 2
3 1 1 4
14
2 2
4
   ( x  2 x y  y )dxdy  .
2 0 0
15
1 1
34
Independent random variables
Definition. Two random variables are independent if,
for A, B, in R, the events {XA} and {YB} are
independent
P( X  A, Y  B)  P( X  A) P(Y  B)
• Using the axioms of probability, we can prove that
X and Y are independent iff for any real numbers a
and b,
P( X  a, Y  b)  P( X  a ) P(Y  b)
Theorem. Let X and Y be two random variables defined
on the same sample space.
If F is the joint distribution function of X and Y, then X
and Y are independent
∀ reals 𝑡, 𝑢 𝐹 𝑡, 𝑢 = 𝐹𝑋 𝑡 𝐹𝑌 (𝑢)
Theorem. Let X and Y be two discrete random variables
defined on the same sample space.
If p(x, y) is the joint probability function of X and Y, then
X and Y are independent iff
∀ reals 𝑥, 𝑦 𝑝 𝑥, 𝑦 = 𝑝𝑋 𝑥 𝑝𝑌 (𝑦)
Theorem. Let X and Y be two continuous random
variables defined on the same sample space.
If f(x, y) is the joint density function of X and Y, then X
and Y are independent iff
∀ reals 𝑥, 𝑦 𝑓 𝑥, 𝑦 = 𝑓𝑋 𝑥 𝑓𝑌 (𝑦)
• Store A and B, which belong to the same owner, are
located in 2 different towns.
• If the density function of the weekly profit of each store,
in thousands of dollars, is given by
 x / 4 if 1  x  3
f ( x)  
otherwise,
0
• The profit of one store is independent of the other.
• What is the probability that next week one store makes at
least $500 more than the other store?
Sol:
Let X and Y denote next week' s profits of A and B, respectively.
The desired probability is
P( X  Y  1/2)  P( Y  X  1/2)  2P( X  Y  1/2).
 x / 4 if 1  x  3
f X ( x)  
otherwise,
0
 y / 4 if 1  y  3
and fY ( y )  
otherwise.
0
 xy / 16 if 1  x  3, 1  y  3
So f ( x, y )  
otherwise.
0
2 P( X  Y  1/ 2)
 2 P(( X,Y )  {( x, y ) : 3 / 2  x  3, 1  y  x  1 / 2})
3
 2 ( 
3/ 2
x 1 / 2
1
xy
1 3 xy 2 x 1/ 2
dy )dx   [
]1 dx
16
8 3/ 2 2
1 3
  x[( x  1 / 2) 2  1]dx
16 3 / 2
1 3 3
3
549
2
  ( x  x  x)dx 
 0.54
16 3 / 2
4
1024
• Prove that 2 random variables X and Y with the
following joint density function are not
independent.
8 xy 0  x  y  1
f ( x, y )  
otherwise.
0
Sol:
1
f X ( x)   8 xydy  4 x(1  x 2 ), 0  x  1,
x
y
fY ( y )   8 xydx  4 y 3 , 0  y  1.
0
Now since f ( x, y )  f X ( x) fY ( y ), X and Y are dependent.
This is expected because of the range 0  x  y  1, which
is not a Cartesian product of 1 - dim regions.
• Theorem. Let X and Y be independent random variables
and g: RR and h: RR be real-valued functions; then
g(X) and h(Y) are also independent random variables.
• Theorem. Let X and Y be independent random variables
and g: RR and h: RR be real-valued functions; then
E[g(X)h(Y)]=E[g(X)]E[h(Y)]
By the above theorem, if X and Y are independent,
then E(XY)=E(X)E(Y)
But, the converse is not necessarily true.
• Let X be a r. v. with p(-1)=p(0)=p(1) =1/3.
• Letting Y=X2, we have
•
•
•
•
E(X)=(-1+0+1)/3=0,
E(Y)=E(X2)=((-1)2+02+(1)2)/3=2/3,
E(XY)=E(X3)=((-1)3+03+(1)3)/3=0.
Thus E(XY)=E(X)E(Y)
• Clearly, X and Y are dependent.
Conditional distributions
• Definition. Let X and Y be two discrete random
variables. The conditional probability function of X
given that Y=y is
p X |Y ( x | y )  P( X  x | Y  y )
P ( X  x, Y  y )

P(Y  y )
p ( x, y )

,
pY ( y )
where x  A, y  B, and p Y ( y )  0.
Note that
p
xA
X |Y
( x | y)  
xA
p ( x, y )
1.
pY ( y )
Hence for any fixed y  B, p X |Y ( x | y ) is itself a probability
function with the set of possibe values A.
Similar to p X |Y ( x | y ), the conditional distribution function of X,
given that Y  y is defined as follows :
FX |Y ( x | y )  P( X  x | Y  y )   P( X  t | Y  y )   p X |Y (t | y ).
tx
tx
• Definition. Let X and Y be two continuous random
variables with joint density f(x, y). The conditional
density function of X given that Y=y is:
f ( x, y )
f X |Y ( x | y ) 
,
fY ( y )
provided that f Y ( y )  0.
The conditional distribution function of X,
given that Y  y is defined as follows :
x
FX |Y ( x | y )  P( X  x | Y  y )   f X |Y (t | y )dt.

d
Therefore,
FX |Y ( x | y )  f X |Y ( x | y )
dt
• Let X and Y be continuous random variables with
joint density function
 3 2
2
( x  y ) if 0  x  1, 0  y  1
f ( x, y )   2
0
otherwise.
Find f X |Y ( x | y )
Sol :
fY ( y )  


3 2
3 2 1
2
f ( x, y )dx   ( x  y )dx  y  .
0 2
2
2
1
Thus
3 / 2( x  y ) 3( x  y )
f X |Y ( x | y ) 

2
(3 / 2) y  1 / 2
3y2 1
for 0  x  1 and 0  y  1.
Everywhere else, f X |Y ( x | y )  0.
2
2
2
2
• Definition. The conditional expectation of the
random variable X given that Y=y is
(discrete)
E ( X | Y  y )   xpX |Y ( x | y ), where pY ( y )  0.
xA
(continuous)

E ( X | Y  y )   x f X |Y ( x | y )dx, where fY ( y )  0.

• Calculate the expected number of aces in a
randomly selected poker hand that is found to have
exactly 2 jacks.
Sol:
Let X be the number Aces in a hand.
Let Y be the number of Jacks in a hand.
3
p ( x,2)
E ( X | Y  2)   xp X |Y ( x | 2)   x
pY ( 2)
x A
x 0
 4  4  44 
  

 2  x  3  x 
 52 
 4  44 
 
 

3
5
x  3  x 


x
x
 0.25.
 4  48 
 48 
x 0
  
 
 2  3 
 3
 52 
 
 5
• Let X and Y be continuous random variables with
joint density function
e  y
f ( x, y )  
0
Find E(X | Y  2).
if y  0, 0  x  1
elsewhere.
Sol :
1
f ( x ,2 )
e 2
E ( X | Y  2)   x f X |Y ( x | 2)dx   x
dx   x
dx.

0
0
fY ( 2)
fY ( 2)

1
1
1
0
0
But fY (2)   f ( x,2)dx   e 2dx  e 2 ; therefore,
e 2
1
E ( X | Y  2)   x 2 dx  .
0
e
2
1

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