probability

1. 1
EL6303 Solution to HW 3 Fall 2015
1. Let 1, 1,2,..., .( ) (1/3)k kP X k Ak    
(a) Find A so that ( )P X k representsa probabilitymass function.
(b) Find { }.E X (c) Find 2
{ }.E X
(d) Find the conditional probabilitymass function ( |1 10)P X k X   .
(Video is required.)
Solution: Let’s discuss a more general problem.
2
0
1
1
Differentiating both sides with respect to , we get1 ... .k
k
pp p p
p



    
1
2
0 (1 )
1k
k p
kp
 
 
  2
0
Differentiating both sides with respect to , we
)
e
1
t
(
g.k
k p
p
p
kp

 

2 1
3
1
1
(1 )
k
k
p
p
k p
 



 
2
2
3
1
Differentiating both sides with respect to , we
(1
e
)
g t.k
k
p p
p
pk p





2
3 1
4
1
1 4
(1 )
k
k
p p
k p
p
 

 



2. 2
1 1 2
2
1 1(1 )
1
1 (1 )k k
k kp
kp Akp A p
  
 
      
2 1 2 1
3
1 1
1 1
1(1 )
( )k k
k k
p p
pp
k p E X Ak p
  
 
 

    
2 2
3 1 3 12
4 2
1 1
1 4 1 4
( )
(1 ) (1 )
k k
k k
p p p p
k p E X Ak p
p p
  
 
   
   
 

Therefore, in our problem,
(a) 2(1 ) =4/9A p 
(b) 2 1
0 1
1
1
( ) ( ) 2k
k k
p
p
E X kP X k Ak p
  
 


     
(c)
2
2 2 3 1
2
0 1
1 4
( ) ( ) 11/2
(1 )
k
k k
p p
E X k P X k Ak p
p
  
 
 
    


(d) 10 10
1
1 1
( , 1 10) ( , 1 10)
( ) (4/9) (1/3)
( |1 10)
k
k k
P X k X P X k X
P X k k
P X k X

 
     
 
    
1177147
(4/9) (1/3)
177124
k
k


11
(4/9) (1/3) , 1,2,3,....10; 0, otherwise.
0.999870164
k
k k



3. 3
2. X is uniform in [-1, 2] and
2
Y X .
(a) Find and draw ( )X
F x . (b) Find and draw ( )Y
f y .
(c) Find and draw ( )Y
F y . (d) Find ( )E Y .
(Video is required.)
Solution:
(a)
0, 1
(1/3)( 1), 1 2
1, 2
( )X
x
x x
x
F x

    
 

(b) & (c) ( ) 0, for 0 and for 4.Yf y y y  
For 0 1, there are two roots:y  2
( )y g x x x y    
x
1 2
x
( )X
f x
1/3 1
1 2
( )X
F x

4. 4
1 2
1 2
( ) ( ) (1/ 3) (1/ 3) 1
( ) + +
( ) ( ) 32 2
X X
Y
f x f x
f y
g x g x yy y
  
   
For 1 4, there is one root :y 
( ) (1/ 3) 1
( )
( ) 62
X
Y
f x
f y
g x yy
  

1
for 0 < 1
3
1
( ) for 1 < 4
6
0 otherwise
Y
y
y
f y y
y




 




0 for < 0
2
for 0 < 1
3
( )
1
( +1) for 1 < 4
3
1 for 4
Y
y
y y
F y
y y
y


 

 
 

 
x
2y x
1 21
x
( )X
f x
1/3
1
4
1 2

5. 5
That is
1 1
( ) ( ( ) ( 1)) ( ( 1) ( 4))
3 6
Yf y u y u y u y u y
y y
      
And
2 1
( ) ( ( ) ( 1)) ( 1)( ( 1) ( 4))
3 3
YF y y u y u y y u y u y       
y
( )Y
f y
41
1
6 y
1
3 y
1/3
1/6
1/12
y
1
( )Y
F y
1
2/3
2
3
y
1
( 1)
3
y 
4

6. 6
(d) 1 1 1 11 4 1 4
0 1 0 13 6 3 6
(1/ ) (1/ ){ } y y dy y y dy ydy ydyE Y      
1 41 2 1 2 2 13/2 3/2
13 3 6 3 9 90
| | (8 1) 1y y    
Check: 0
(1 ( )) ... 1{ } F y dyE Y 
   . You check the details.
3. ( 1)
( ) ( 1)x
X
f x Ae u x 
  . And 3
Y X . Then
(1) Find A so that f(x) is a valid probabilitydensity function.
(2) Find and draw ( )X
F x .
(3) Find ( )Y
f y .
(4) Find 3
{ }YE .
Solution: (1) Clearly, A=1.
(2) ( 1)
( ) ( 1)x
X
f x e u x 
   ( 1)
( ) (1 ) ( 1)x
X
F x e u x 
  
(3) 1/3
2/3 2/3
1/3 1/3
1 1( 1) ( 1)
1 1
3 3
( ) ( ) ( )
| '( )|
( )X y y
y y
Y y y
f y e u e u
g
f x
x
   
   
1
( )X
F x
1
x

7. 7
because for 1/3
1 11, ( ) ( )y yy u u   .
Or, think this way:
1/3
1/3 3
1 1
1/3 3
1, for 1 1, for 1 1, for 1
( ) ( )
0, for 10, for 1 0, for 1
y y
y y y
u u
yy y
 
         
    
(4) 3
{ } { } 2YE E X  .
4. A fine needleof length a is droppedat randomon a board covered with
parallel lines distance l apart where l a as in the figure.
(1) Find the probabilitythat the needleintersects any of the lines.
(2) Find the probabilitythat the needlevertically intersects any of the lines.
(3) Find the probabilitythat the needleintersects a line with an angle between
needleand line is within 0 ～30 or 150 ～180 .
l
a

8. 8
Solution:
(1) Let X be the distance between the central point of the needle and the
nearest line, and  be the angle between the needleand the line counter-
clockwise.
Clearly, 0 / 2, 0 .X l      In order to intersect a line, we must have
( / 2)sin .X a 

/2a
2/l
x


/2a
( / 2)sina 
a
 X
l

9. 9
0
( / 2)sin 2
(intersection) .
( / 2)
a d a
P
l l

 
 
 

(2) P(the needlevertically intersects any of the lines)=0.
(3) P(The needleintersects a line with an angle between needleand line is
within 0 ～30 or 150 ～180 )
/6
0 2 /6
( / 2)sin ( / 2)sin
(intersection) (2 3).
( / 2)
a d a d a
P
l l
 

   
 

   
 
5. X ～ 2
( , ), .X
N n n Y e  
(1) Find ( )Y
f y . (2) Find E{Y }. (3) Find the second moment of Y.
Solution:
(1)
( )
( )
| '( )|
X
Y
f x
f y
g x
 
2(ln )
22
2
1
2
y n
ne
y n


 


(2) { } { } ( )xXE Y E e e f x dx

 
2( )
22
2
1
2
x n
x ne e dx
n


 




  ------- (*)
Let 2
2
2
2
n
n
x n z x z n

     

10. 10
(*)
2 2( 2 ) 2
2
1
( 2 )
2
n nz ze e e d n z n
n
   
 


 
2 2
2 ( 2 )
2
1
2
2
nn zz
e n e e dz
n
 
 



 
2 221 ( )nn z z
e e dz




 
2 2
( )
2 21
n
n
n
e e e
 





 
(3) Similarly, 22 2{ } { } ( )xXE Y E e e f x dx

 
2
2 ( )n
e  
 .
6. Randomly pick a number from ( j , 1 1
2 2
j , 1 1
2 2
j , 1 1
2 2
j ).
Find the probabilitythat the picked number equals j .
Solution: 1 90 1 /2) 1(90 45j      1 1
2 2
j  and
1 270 1 /2) 1(270 135j      1 1
2 2
j 
Or:
1 1
2 2 2
90
4590 1 1
2 2
1 1 1( )
j jj jj e e e
         and
1 1
1352 2 2
027
270 1 1
2 2
1 1 1( )
j jj jj e e e       
Therefore, the probabilitythat the picked number equals j is ½.

11. 11
7. Show that ( ) ( | ) ( ) ( | )[1 ( )]P A P A x F x P A x F x    X X
Solution: ( ) ( ( ) ( ))P A P A x A x     X X
( ( )) ( ( ))P A x P A x     X X
( | ) ( ) ( | ) ( )P A x P x P A x P x     X X X X
( | ) ( ) ( | )[1 ( )]P A x F x P A x F x    X X
8. Show that
( | ) ( )
( | ) .
( )
x
x
P A x F x
F x A
P A


X
Solution:
(( ) )
( | ) ( | )
( )
x
P x A
F x A P x A
P A
 
  
X
X
( | ) ( )( | ) ( )
.
( ) ( )
xP A x F xP A x P x
P A P A
 
 
XX X
9. Find 2( ) and ( ) if 4 3 ( ) 2 ( ).y y x
Y X xF y f y and f x e u x  
(In this solution, we are going to apply some properties of impulse functions:
a) ( ) ( )
d
u x x
dx
 . b) ( ) ( ) (The delta function is an even function.)x x  
c) For any function ( ),h x we have ( ) ( ) ( ) ( ),h x x a h a x a    which is known
as the sampling propertyof the impulse function.)

12. 12
Solution: ( ) ( )x
x x
F x f d 


2
2 ( )x
e u d

 


 0
2
2 ( ) ( )x
e u d u x

 


0
2
2 ( )x
e d u x



 0
2
| ( )
x
e u x


2
(1 ) ( )
x
e u x

 
3
( ) ( ) ( 4 3 ) ( )
4y
Y X X
y
F y P y P y P

       
3 3
( ) 1 ( )
4 4
X X
y y
P P
 
    
3
1 ( )
4x
y
F

   3
4
2
1 (1 ) ( )| y
x
x
e u x 


  
3
2( )
4 3
4
1 (1 ) ( )
y
y
e u

 

  
3
2( )
4
(3 )1 (1 )
y
ye u

 
  
3
2
(3 )1 (1 )
y
ye u

  
3
2
(3 ) (3 )1
y
y yu e u

   
3
2
( 3) (3 )
y
y yu e u

  
3
2
1
for y 3
for y 3
y
e








3
4
3
( )x
f x
x
( )x
F x2
y
x

13. 13
3
2
( 3) (3 )( ) ( ) ( )
y
d d
y y
y ydy dy
f y F y u e u

   
3 3
2 2
( 3) (3 ) (3 )( )
y y
d d
y y y
dy dy
e u e u
 
    
3 3
2 2
( 3) (1/ 2) (3 ) (3 )
y y
y y ye u e 
 
    
3
2
( 3) (1/ 2) (3 ) (3 )
y
y y ye u 

    
3
2
( 3) (1/ 2) (3 ) ( 3)
y
y y ye u 

    
3
2
(1/ 2) (3 )
y
ye u


x
( )y
f y
1/2
3
y
( )y
F y 1
3

14. 14
10. Show that if 2
,Y X then 0
( )( )
1 (0) 2
( | )y
X
f yu y x
F yx
f y 

 .
Solution:
0 0)( | ) ( | ( | 0)y y
X X
d d
f y F y P Y y X
dy dy
    
( , 0) 1
( , 0)
( 0) 1 ( 0)
d P Y y X d
P Y y X
dy P X P X dy
 
   
  
1 1
1 (0) 1 (0)
2( , 0) ( , 0)
F F
x x
d d
P Y y X P X y X
dy dy 
     
1 1
1 (0) 1 (0)
( , 0) (0 )
F F
x x
d d
P y X y X P X y
dy dy 
       
11 1
(1/ 2)
1 (0) 1 (0)
( ( ) (0)) ( )x x xF F yx x
d
F y F f y
dy 
  
( )( )
1 (0) 2
( is always nonnegative.)
f yu y x
F yx
y

