Download Solutions to the Practice Problems for the Final Exam

Solutions to the Practice Problems for the Final Exam
1. Problem: Suppose that X and Y are independent exponential random variables with
parameter X and Y respectively.
(a) Find E (2 X  3Y ) and Var (2 X  3Y ) in terms of X and Y .
(b) Find E (2 X  3Y ) 2 in terms of X and Y .
(c) Find the correlation of X and X  Y in terms of X and Y .
(d) Assume X  3 and Y  1 . Find P( X  Y  2) .
2
Solution: (a) E (2 X  3Y )  2 E ( X )  3E (Y ) 
3

X
. Because X and Y are
Y
independent, Var (2 X  3Y )  4Var ( X )  9Var (Y ) 
4


2
X
9
Y2
.
(b)
2
 2
3 
8
12
18
E (2 X  3Y )  Var (2 X  3Y )  [ E (2 X  3Y )]  2  2  
   2 
 2
X Y  X Y  X X Y Y
2
(c) Cov( X , X  Y )  Cov( X , X )  Cov( X , Y ) 
Var ( X  Y )  Var ( X )  Var (Y )  2Cov( X , Y ) 
(X , X  Y ) 
4
2
Cov( X , X  Y )
 X  X Y

1

2
X
1

2
X
9
1

Y2
1/ X2
(1/ X ) 1/   1/ 
2
X
1
0 
2
Y
X2
and
. Thus,

Y
  Y2
2
X
.
P( X  Y  2)  1  P( X  Y  2)
(d)
 1 
2
0


2 y
0
3e3 x e  y dxdy
3 2 1 6
e  e
2
2
2. Problem: Suppose Z1 , Z 2 and Z3 are independent N (0,1) random variables. Find
(a) P(Z1  Z 2  Z3 ) ;
(b) E ( Z1Z 2 Z3 ) ;
(c) Var (Z1Z 2 Z3 ) ;
(d) P(Z1  Z3 ) ;
(e) P(max(Z1 , Z 2 )  Z3 ) (f) P( Z12  Z 22  1)
(g) P( Z1  Z 2  Z3  2) (h) P( Z1 / Z 2  1)
(i) P(3Z1  2Z 2  4Z3  1)
Solution:
(a) 1/6, by symmetry, all 3!=6 orderings of Z1 , Z 2 , Z3 are equally likely.
(b) Since Z1 , Z 2 , Z3 are independent, E (Z1Z 2 Z3 )  E (Z1 ) E (Z 2 ) E (Z3 )  0 .
(c)
Var ( Z1Z 2 Z3 )  E ( Z12 Z 2 2 Z32 )  ( E ( Z1Z 2 Z3 ))2 
E ( Z12 ) E ( Z 2 2 ) E ( Z32 )  ( E ( Z1Z 2 Z3 )) 2  1
where the second equality follows from Z1 , Z 2 , Z3 being independent.
1
(d) P( Z1  Z 3 )  P( Z1  Z 3 )  since Z1  Z3 ~ N (0, 2) is symmetric about 0.
2
(e)
P(max(Z1 , Z 2 )  Z3 )  1  P(max(Z1 , Z 2 )  Z3 )  1  P(Z3 is the largest) 
1  1/ 3  2 / 3
(f) P( Z12  Z 22  1)  e 1/ 2 (see Example 6.3b, page 282).
(g) Z1  Z 2  Z3 ~ N (0,3) so
2 
 Z  Z 2  Z3
 2 
P( Z1  Z 2  Z3  2)  P  1

  
  0.8759
3
3

 3
3
(h) P(Z1 / Z 2  1)  P(Z1  Z 2 , Z 2  0)  P(Z1  Z 2 , Z 2  0) = .
4
(i) Note that 3Z1  2Z2  4Z3 ~ N (0, 29) so
1 
 3Z  2Z 2  4Z3
 1 
P(3Z1  2Z 2  4Z3  1)  P  1

  
  0.5737
29
29 

 29 
3. Problem: Suppose that U ~ uniform(0,2 ) and independently V ~ exponential(1) .
Show that
X  2V cos U , Y  2V sin U
are independent standard normal random variables.
Solution: We use the method of Section 6.7 for finding the joint probability distribution
of functions of random variables. The transformation from (U ,V ) to ( X , Y ) is
g1 (u, v)  2v cos u, g2 (u, v)  2v sin u
and the inverse transformation is
 tan 1 ( y / x)
x>0
 1
 tan ( y / x)   x  0
u  h1 ( x, y )  
x  0, y  0,
( / 2)sign( y )
0
x  0, y  0

The Jacobian of the transformation is
2v ( sin u ) (2v)-1/2 cos u
J (u, v) 
 1
2v cos u
(2v)-1/2 sin u
v  h2 ( x, y ) 
x2  y 2
2
Thus, using formula (7.1) on page 300,
f X ,Y ( x, y )  fU ,V (h1 ( x, y ), h2 ( x, y )) | J (u, v) |1
 x2  y 2 
1

exp  

2
2 

This is the joint density of two independent standard normal random variables.
4. Problem: I toss a coin which lands heads with probability p  1/ 3 . Let N H be the
number of tosses till I get a head, N HH the number of tosses till I get two heads in a row
and N HHH the number of tosses till I get three heads in a row. Find
(a) E ( N H ) ;
(b) E ( N HH ) ;
(c) E ( N HHH ) ;
(d) Generalize to find the expected number of tosses to obtain m heads in a row.
Solution:
(a) Note that N H ~ Geometric( p) so E ( N H ) 
1
 3.
p
(b) Condition on whether the first toss was a head or a tail. Let Y be the number of heads
in the first toss. So Y  1 with probability p and 0 with probability 1  p . Then
E ( N HH )  E ( E ( N HH | Y ))  pE ( N HH | Y  1)  (1  p) E ( N HH | Y  0) .
Note that E ( N HH | Y  0)  1  E ( N HH ) and E ( N HH | Y  1)  2 p  (2  E( N HH ))(1  p) .
Let q  1  p and x  E ( N HH ) . Then it follows that
x  p(2 p  (2  x)(1  p))  (1  p)(1  x)  q  qx  2 p 2  2 pq  xpq
and hence E ( N HH ) 
q  2 pq  2 p 2 1  p 1 1
 2   2  3  9  12 .
1  q  pq
p
p p
(c) Let NT be the number of tosses to obtain the first tail. Then N HHH  3 if NT  4 and
N HHH  k  N where the random variable N has the same distribution as N HHH if
NT  k , k  1, 2,3 . Set x  E ( N HHH ) . Then
x  3 p3  (3  x) p 2 q  (2  x) pq  (1  x)q
 q  2 pq  3 p 2 q  3 p3  ( p  pq  p 2 q) x
and so x 
q  2 pq  3 p 2 q  q  3 p3 1  p  p 2 1 1
1

  2  3  3  9  27  39 .
2
3
1  q  pq  p q
p
p p
p
(d) x  k 1 (k  x) p k 1q  mp m and so
m

x


kp k 1q  mp m
k 1
m
1   k 1 p k 1q

 p m1
pm
m
1   k 1 p k 1   k 1 p k
m
1 p 
kp k 1   k 1 kp k  mp m
k 1
m
m
1 1


p p2

m

1
 3  32 
pm
 3m 
3m1  3
2
5. Problem: Let X 1 , X 2 , be a sequence of iid continuous random variables. Let N  2
be such that
X1  X 2   X N 1  X N .
That is, N is the point at which the sequence stops decreasing. Show that E ( N )  e .
Proof:
P( X 1  X 2 
 X N 1  X N )  P( X1  X 2   X N 1 ) P( X N  X N 1 | X1  X 2   X N 1 )
1
We have P( X1  X 2   X N 1 ) 
because X1  X 2   X N 1 is one of
( N  1)!
( N  1)! equally likely orderings of X 1 , , X N 1 . We have
1 N 1
P( X N  X N 1 | X 1  X 2   X N 1 )  1  P( X N is minimum of {X 1 , , X N })  1  
N
N
1
N 1
1
Thus, P( X1  X 2   X N 1  X N ) 
and

( N  1)! N
( N  2)! N



1
1
1
E(N )   N

  e .
( N  2)! N N  2 ( N  2)! i 0 i !
N 2
6. Problem: Let X and Y have the joint density
f ( x, y)  cx( y  x)e y , 0  x  y   .
(a) Find the constant c
(b) Find the conditional distribution of X given Y  y .
(c) Find the conditional distribution of Y given X  x .
(d) Find E ( X | Y  y ) and E (Y | X  x) .
Solution:
(a) 1  



 
f ( x, y )dxdy  

0
y

y
0
cx( y  x)e  y dxdy 
(b) Note that fY ( y )   x( y  x)e  y dx 
0
f X ( x | Y  y) 
c  3 y
y e dy  c , thus c  1 .
6 0
1 3 y
y e . Hence,
6
f ( x, y )
 6 x( y  x) y 3 , 0  x  y ,
f X ( x)

(c) f X ( x)   x( y  x)e y dy  xe x . So
x
fY ( y | X  x ) 
f ( x, y )
 ( y  x )e  ( y  x ) , x  y   .
f X ( x)
y
(d) E ( X | Y  y )   xf X ( x | Y  y )dx   6 x 2 ( y  x) y 3dx 
0
1
y and
2

E (Y | X  x)   yfY ( y | X  x)dy   y( y  x)e( y  x ) dy  x  2 .
x
7. Problem: An urn contains 10 red balls, 8 blue balls and 12 white balls. From this urn,
12 balls are randomly withdrawn. Let X be the number of red, and Y the number of blue
balls that are withdrawn. Find Cov( X , Y )
(a) by defining appropriate indicator random variables X i , Yi such that X  i 1 X i and
10
Y   j 1 Y j .
8
(b) by conditioning (on either X or Y) to determine E ( XY ) .
Solution:
(a) Number the red balls and the blue balls and let X i  1 if the ith red ball is selected and
let X i  0 otherwise. Similarly, let Y j  1 if the jth blue ball is selected and let
Y j  0 otherwise. Cov(i 1 X i , j 1Y j )  i 1  j 1 Cov( X i , Y j ) . Now,
10
8
10
8
E ( X i )  E (Y j )  12 / 30  2 / 5 and
 28   30 
E ( X iY j )  P(red ball i and blue ball j are selected)=   /    22 /145 . Thus,
10  12 
2

96
 22  2  

.
Cov( X , Y )  80 
    
145
145  5  


(b) We shall calculate E ( XY ) by conditioning on X. Note that given X, there are
12  X additional balls to be selected from among 8 blue and 12 white balls. Hence,
E ( XY | X )  XE (Y | X )  X (12  X )(8 / 20) .
Now since X is a hypergeometric random variable, it follows that E ( X )  12(10 / 30)  4
and E ( X 2 )  Var ( X )  ( E ( X ))2  12(18)(1/ 3)(2 / 3) / 29  42  512 / 29 . As
2
E (Y )  12(8 / 30)  16 / 5 , we obtain E ( XY )  (48  512 / 29)  352 / 29 and
5
Cov( X , Y )  352 / 29  4(16 / 5)  96 /145 .
8. Problem: Let X  2U  V and Y  3U  2V where U and V are independent standard
normal random variables.
(a) Find P( X  2Y  1) .
(b) Find the conditional distribution of U given X  x .
(c) Find the conditional distribution of Y given X  x .
Solution:
(a)
P( X  2Y  1)  P(2U  V  6U  4V  1)  P(4U  3V  1)
4
3
1
 P( U  V  )  (0.2)  0.5793
5
5
5
(b) Let X '  U  2V . Then X ' ~ N (0,5) and Cov( X , X ')  0 . So X and X ' are
uncorrelated and hence independent (since X and X ' are jointly bivariate normal).
2
1
2
1
2 1
U  X  X ' . Given X  x , U  x  X ' ~ N ( x, ) .
5
5
5
5
5 5
(c) Similar to part (b), Y  3U  2V 
Y
8
1
X  X ' . Given X  2U  V  x ,
5
5
8
1
8 1
x  X ' ~ N ( x, ) .
5
5
5 5
9. Problem: Let U1 ,U 2 ,
For x  (0,1) , let
be a sequence of independent uniform (0,1) random variables.
n
N ( x)  min{n :  U i  x} .
i 1
Find E[ N ( x )] by first showing by induction on n that for 0  x  1 and all n  0 ,
P( N ( x)  n  1)  x n / n!.
Solution: We first show by induction on n that for 0  x  1 and all n  0 ,
P( N ( x)  n  1)  x n / n!. The result is true when n  0 , so assume that
P( N ( x)  n)  x n1 /(n  1)!
Now,
1
P( N ( x)  n  1)   P{N ( x)  n  1| U1  y}dy
0
x
  P{N ( x  y )  n}dy
0
x
  P{N (u )  n}du
0
x
  u n 1 /(n  1)!du by the induction hypothesis
0
 xn / n!
which completes the proof. Then, using the result in Theoretical Exercises 5.2,



E[ N ( x)]  n0 P{N ( x)  n} n0 P{N ( x)  n  1} n0 x n / n!  e x .