Let X and Y denote the random variables of either continuous or discrete type which have the joint probability distribution function f x,

Conditional Probability Distribution

Let X and Y denote the random variables of either continuous or discrete type which have the joint probability distribution function and marginal probability density functions of and , respectively. The conditional probability function of Y given X is

The conditional probability function of X given Y is

 , f x y   x f x f yy 

 

 

 

_{ }

, , _X

 

0 Y x X f x y f y f y x f x f x   

 

 

 

_{ }

, , _Y

 

0 X y Y f x y f x f x y f y f y   

3.7 Conditional Probability Distribution

The following properties must be satisfied: i)

ii) If X and Y are discrete random variables, then

iii) If X and Y are continuous random variables, then

(

)

0 or (

)

0

f y x



f x y



( ) 1 or ( ) 1





y x f y x f x y ( ) 1 or ( ) 1 y x f y x dy  f x y dx 





3

Example 3.19:

Using the joint probability distribution function from Example 3.9, find

a)

f(yx = 2)

b)

P(Y=1x = 2)

c)

P(Y ≥ 1│ x = 2)

X=x f(x,y) 0 1 2 Y=y 0 1/18 1/9 1/6 1 1/9 1/18 1/9 2 1/6 1/6 1/18

Solution: a) (  2)  (2, )_ ( 2) X f y f y x f x  



2    0 (2) (2, ) (2,0) (2,1) (2, 2) X y f f y f f f 3 1 18 1 9 1 6 1 _{  }    (2, )  ( , 2)  ( 2) 3 ( , 2) 1 (2) ₃ X f y f y f y x f y f : 2 , 1 , 0  y .:

5 6 1 18 1 . 3 ) 2 , 2 ( . 3 ) 2 2 ( 3 1 9 1 . 3 ) 1 , 2 ( 3 ) 2 1 ( 2 1 6 3 ) 0 , 2 ( 3 ) 2 0 (             f x f f x f f x f y=y 0 1 2 1



) 2 (y x  f 2 1 3 1 6 1

b) c) (2,1) (2,1) 1 1 ( 1 2) 3 (2,1) 3. 1 (2) 9 3 3 X f f P Y x f f        (2, 1) ( 1 2) 3. (2, 1) (2) X f y P Y x f y f       6 1 18 1 . 3 ) 2 , 2 ( 3 ) 2 2 ( 3 1 9 1 . 3 ) 1 , 2 ( 3 ) 2 1 (           f x y f f x y f

Exercise f(x,y) 0 1 2 0 1/18 4/18 6/18 1 3/18 3/18 1/18 X=x Y=y

Use the joint probability , find













i. 1 ii. 2 1 iii. 1 1 f x y f x y f x y     

Example 3.20:

If X and Y have joint probability density function

find a) b) 3 ; 0 1, 0 1 ( , ) ₄ 0 ; elsewhere xy x y f x y  _{    }     ( ) f y x 1 ( ) P Y  X

9 Solution: a) x f y x f x y f x ) , ( ) ( 



  1 0 ) 4 3 ( xy dy x f_x 4 2 3 2 4 3 2 4 3 1 0 2 x x xy y       x xy x xy x y f 2 3 4 3 4 2 3 4 3 ) (       ; 0  x  1,0  y  1 .:

b) _{P y x) f (y x)dy} 2 1 ( 1 2 1



 



 _    1 2 1 2(3 2 ) 3 3 2 3 4 3 x x dy x xy

11

Example 3.21:

The joint probability density function is given as follows:

Find

a) the marginal density of X. b) the marginal density of Y.

c) the conditional density of Y given X = x. d) the conditional density of X given Y = y.

8 ; 0 1, 0 ( , ) 0 ; elsewhere xy x y x f x y       

Solution: a) b) c) d) 3 0 2 0 4 2 8 8 ) ( ) (x g x xydy xy x f x x y y x  



   

1

0

 x



, ) 1 ( 4 2 8 8 ) ( ) ( 2 1 2 1 y y y x xydx y h y f y x y x y  



     1 0  y  , 2 3 2 4 8 ) ( ) , ( ) ( x y x xy x g y x f x y f    , 0  y 1 ) 1 ( 4 8 ) ( ) , ( ) ( ₂ y y xy y h y x f y x f    ,

0

 x



1

Exercise

Let X₁ and X₂ have the joint pdf given as



1, 2



6 2, 0 2 1 1

= 0, elsewhere.

f x x  x  x  x 

i. Find the marginal function of ii. Find f x x



2 1



.

1. X

3.8 Transformation Technique

Theorem:

Let X and Y be continuous random variables having joint density function f(x,y). Let us define U =



₁(X,Y), V=



₂(X,Y)

where X=



₁(U,V), Y=



₂(U,V). Then the joint density function of U and V is given by g(u,v) where

1 2 ( , ) ( , ) or ( , ) ( , ) ( , ) [( ( , ), ( , )] ( , ) g u v dudv f x y dxdy x y g u v f x y f u v u v J u v       

15 The Jacobian determinant atau Jacobian, J is given as follows;

( ,

)

( , )

x

x

x y

_u

_v

J

y

y

u v

u

v

 



_{ }





 



 

Example 3.22:

The joint probability density function of X and Y is given as follows: Let and find 1 ; 0 1, 0 1 ( , ) 0 ; elsewhere y x f x y       

U

 

X

Y

V  X Y, ( , ). f u v

17

Solution:

Substitute (2) into (1) and substitute (1) into (2)

Y

X

U





V



X



Y

Y

U

X





…..(1)

Y



X



V

_…..(2) 2 1 , 2 1 2 2 ) (             dv dx du dx V U X V U X V X U X V X U X 2 1 , 2 1 2 2 ) (              dv dy du dy V U Y V U Y V Y U Y V Y U Y

2 1 ) 2 1 )( 2 1 ( ) 2 1 )( 2 1 ( 2 1 2 1 2 1 2 1       J

2

1

2

1 





J

2 1 2 1 . 1 ) , ( ) , (u v  f x y J   g 1 ; 0 2, 0 2 ( , ) ₂ 0; elsewhere. U V U V g u v           

19

Example 3.23:

are standard normal distributions functions. X₁ dan X₂ are independent. If and

find 1 ~ (0,1) and 2~ (0,1) X N X N 1 1 2 Y  X  X 1 2 2 X Y X  f y y( ,1 2)

Solution:

X₁~N(0,1) X₂~N(0,1)

X₁ and X₂ are independent.

) ( 2 1 2 1 2 1 2 2 2 1

2

1

)

(

).

(

)

,

(

x

x

f

x

f

x

e

x x

f





 



.:

2 2 1 2 2 2 1 2 2 1 2 2 2 2 1 ) 1 ( , 1 1 1 : . Y Y dY dX Y dY dX Y Y X X X Y Y          2 2 1 2 1 2 2 1 1 2 1 2 1 2 1 . 2 1 ) 1 ( , 1 1 : . 1 Y Y dY dX Y Y dY dX Y Y Y X Y Y Y X         Therefore, 2 2 1 2 2 2 1 2 2 ) 1 ( 1 1 ) 1 ( 1 Y Y Y Y Y Y Y J      

23 2 ) 2 1 ( 1 3 ) 2 1 ( ) 2 1 ( 1 3 ) 2 1 ( 1 3 ) 2 1 ( 2 1 Y Y Y Y Y Y Y Y Y Y             J y x f Y Y g( ₁, ₂)  ( , )





2 2 2 1 1 1 1 2 2 2 2 ₂ 1 1 exp ; , 2 2 1 1 ₁ Y Y Y Y Y Y Y Y _Y   _{   }   _{ }  _ _{  } _{ }             _{   }  _  _{ }  

Example 3.24:

The joint probability density function is given as follows

; 0

4, 1

5

( , )

₉₆

0 ; otherwise



_{ }

_{ }



 



xy

x

y

f x y

Find the joint probability density function g(u,v) if U = X + 2Y and V=X.

25 Solution: Substitute (2) into (1). ) 1 ...( 2 y x U   V  x...(2) 2 1 , 2 1 2 2 2          dV dy dU dy V U y V U y y V U

1

,

0







dV

dx

dU

dx

x

V

2 1 2 1 2 1 1 0      dV dy dU dy dV dx dU dx J 2 1 2 1    J

J

y

x

f

V

U

g

(

,

)



(

,

)

.: ( ) ; 2 10, 0 4 ( , ) 384 . 0 ; elsewhere V U V U V V g u v   _{    }     

27

Example 3.25:

The joint probability density function of X₁ and X₂ is given as follows:

Let and , find

1 2 1 2 1 2

; 0

, 0

( ,

)

0 ; elsewhere

x x

e

x

x

f x x

 



  



 

 



1 1 1 2 X Y X X   Y2  X1  X2 f y y( ,1 2).

Solution: Solve: and As a result: and Therefore: 1 1 2 y  x x 2 1 1 2 x y x x   1 1 2 x  y y 2  2(1 1) x y y





2 1 2 1 1 2 2 2 1 1 2 2 2 1 1 1           y y J y y y y y y y y y y y y 2 2 ; 0 1 , 0 2 ( , )          y e y y y g y y

29

Quiz 3.8

The joint density function is given by

If and , find 1 ( ) 2 1 ; 0, 0 ( , ) ₈ 0 ; elsewhere x y xe x y f x y          y U x 

v



x

_{f u v}

_{( , )}