Ch.3 Homework Solution
3-5
(Buffon’s Neddle Problem) A needle of length L is dropped randomly on a plane ruled
with parallel lines that are a distance D apart , where D  L. Show that the probability that
the needle comes to rest crossing a line is 2L/πD.Explain how this gives a mechanical
means of estimating the value of π.
Sol:
Y :針中心點到底線之高 Y ~Uniform(0,D)
Θ:針與 X 軸正向之夾角 θ ~Uniform(0,2π)
P(針 與 平 行 線 不 相) 交

2
0
∴P(針與平行線相交)=
l
D  s i n
2
l
s i n
2

1
2l
d y d  1 
2D
D
2l
D
3-6
x2 y2
A point is chosen randomly in the interior of an ellipse: 2  2  1 .Find the marginal
a
b
densities of the x and y coordinates of the point .
Sol:
b
2b a 2
a 2  x 2 dx 
a  x 2 dx  ab

a a

a
a
2
∴
a
f
1
x2 y2
( x, y ) 
, 2  2 1
XY
ab
a
b
(  a 2  u 2 dx 
b
u
a2
u
a2  u2 
sin 1  c, a  0), f X ( x)   ab

2
2
a
a
a 2 u 2
a u
2
3-8
Let X and Y have the joint density f ( x, y ) 
6
( x  y ) 2 ,0  x  1,0  y  1
7
(a) By integrating over the appropriate regions , find
(i)P(X>Y),(ii)P(X+Y  1),(iii)P(X  ½).
Sol:
2
1
dy 
ab
f ( x, y ) 
6
( x  y ) 2 ,0  x  1,0  y  1
7
1
(a)(i) P( X  Y )  

x
0 0
12
yx
6
1
( x  y ) 2 dydx   ( x  y ) 3
dx 
0
y0
7
7
2
1
(ii) P( X  Y  1)  
0
1
1
(iii) P( X  )  
0
2
1 x

0
1
2
0

6
( x  y ) 2 dydx
7
6
( x  y ) 2 dxdy
7
3-9
Suppose that (X,Y) is uniformly distributed over the region defined by 0  y  1  x 2 and
-1  x  1
(a) Find the marginal densities of X and Y .
(b) Find the two conditional density
Sol:
1

0  y  1  x 2 and -1  x  1 ,
1
( x 2  1)dx 
4
3
, ∴ f XY ( x, y ) 
3
4
(a)
3
3
 3x 2 3
2
f X ( x)  
dy  ( x  1) 
 ,1  x  1
0
4
4
4
4
1 y 3
3
3
3
f Y ( y)  
dx 
1 y 
1 y 
1  y ,0  y  1
 1 y 4
4
4
2
 x 2 1
(b)
f X Y ( x y) 
3
4
3 1 y

1
2 1 y
, 1  x  1
2
f Y X ( y x) 
3
4
3
3
 x2 
4
4

1
 x2 1
,0  y  1
3-11
Let U 1 , U 2 and U 3 be independent random variables uniform on [0,1].Find the
probability that the roots of the quadratic U 1 x 2  U 2 x  U 3 are real .
Sol:
1
1 1
4
0 0 2 u1u3
P(U  4U 1U 3 )  
2
2

u 22
1
4
0 0
0
1
1
4 u3
1
0
2 u1u3
4
1
du 2 du1 du 3  
1
1 1
 
u 22
4 u1
0
   du du du    
3
1
2
u2
0 2
4
du 2 du1 du 3 
du 3 du1du 2 
3-12【Question】
Let f ( x , y )  c( x 2  y 2 )e  x , 0  x   ,  x  y  x
(a)Find c . (b)Find the marginal densities.
【Solution】
(a)
f ( x, y )  c ( x 2  y 2 ) e  x , 0  x  ,  x  y  x
 x
  c( x
2
0 x
 x
 y )e dydx    ce  x ( x 2  y 2 )dydx
x
2
0 x

  ce  x ( x 2 y 
0
1 3 yx
y )
dx
y  x
3

4c  x 3

e x dx
3 0
4c
 ( 4)
3
 4c  3!

3

1
c 
1
8
x
f X ( x) 
1
 8 (x
x
2
 y 2 )e  x dy 
1 3 x
x e , x0
6

 f ( x, y )dx, y  0
y
(b) f Y ( y )   
 f ( x, y )dx, y  0

 y
1 y
 4 e (1  y ), y  0
 f Y ( y)  
 1 e y (1  y ), y  0
 4
5 1
 ln 2
36 6
5 1
 ln 2
36 6
3-15【Question】
Suppose that X and Y have the joint density function
f ( x, y )  c 1  x 2  y 2 , x 2  y 2  1
(d)Find the marginal densities of X and Y . Are X and Y independent random
variables?
【Solution】
(d)
2 1
2
2
2
 c  1  x  y dxdy  c   1  r  r  drd
x 2  y 2 1
0 0
2
 c
0
1
1
(1  r 2 ) 2 d
0
3
1
 c 
3

3
2

d
0
2c
3
1
c 
3
2
f X ( x)  
1 x 2
 1 x
3

2

2
3
1  x 2  y 2 dy
2
(let 1  x 2  a 2 )
a

a 2  y 2 dy
a
3 y2
(
2 2
a2  y2 
a2
y a
sin )
2
a a
3
(1  x 2 )
4
3
 f X ( x)  (1  x 2 ),  1  x  1
4
3
Similarly , f Y ( y )  (1  y 2 ),  1  y  1
4
 f ( x, y )  f X  f Y

 X and Y are not independent.
3-17【Question】
Let ( X , Y ) be a random point chosen uniformly on the region
R   X , Y  : X  Y  1.
(b)Find the marginal densities X and Y using your sketch. Be careful of the range
of integration.
【Solution】
(b)
1 x
f X ( x) 
1
dy  1  x ,  1  x  1
2
x 1

Similarly , f Y ( y )  1  y ,  1  y  1
3-18【Question】
Let X and Y have the joint densities function f ( x , y )  k ( x  y ), 0  y  x  1 and
0 elsewhere.
(b)Find k . (c)Find the marginal densities of X and Y .
【Solution】
(b)
f ( x, y )  k ( x  y ) , 0  y  x  1
1 x
  k ( x  y)dydx
0 0
1
yx
1
  (kxy  ky 2 )
dx
y

0
2
0
1
1
  kx 2  kx 2 dx
2
0

k x3 1
( )
2 3 0
k
1
6
k  6

x
f X ( x)   6( x  y )dy
0
1
f Y ( y )   6( x  y )dx
y
3-19【Question】
Suppose that two components have independent exponentially distributed lifetimes,
T1 and T2 , with parameters  and  , respectively. Find (a) P(T1  T2 ) and
(b) P(T1  2T2 ) .
【Solution】
(a)
f T1 ,T2 (t1 , t 2 )  e t1 e  t 2 , 0  t1  , 0  t 2  

 P (T1  T2 )    e t1  e  t 2 dt1 dt 2
0 t2

  e  t 2  e t 2 dt 2
0

  (   ) t 2 
e
0



1
 (0  1)



(b)
 
P (T1  2T2 )    e t1  e  t2 dt1 dt 2
0 2t2

  e  t2  e  2t2 dt 2
0

   e t2 (   2 ) dt 2
0



 (   2 )

2  
 (1)
3-23【Question】
Suppose that, conditional on N , X has a binomial distribution with N trials
and probability p of success, and that N is a binomial random variable with
m trials and probability r of success, Find the unconditional distribution of X .
【Solution】
X N ～ Bin ( N , p) , N ～ Bin (m, r )
m
 m
n
P( X  x)    r n (1  r ) m  n   p x (1  p ) n  x
n 0  n 
 x
m
m!n!(m  n)!

(rp ) x r n  x (1  r ) m  n (1  p ) n  x
n  x n!( m  n)! x!( n  x )!( m  x )!
m
 m
(m  x)!
  ( pr ) x 
r n  x (1  p ) n  x (1  r ) m  n
n  x ( m  n)!( n  r )!
x 
let n  x  t , m  x  k
k
 m
k!
  ( pr ) x 
r t (1  p ) t (1  r ) k t
t  0 ( k  t )!t!
x 
k
 m
 k  (r  rp ) t (1  r ) k t (1  rp ) k
  ( pr ) x   
(1  rp ) t (1  rp ) k t
t 0  t 
x 
k
 m
 k  r  rp 

  ( pr ) x (1  rp ) m  x   
t  0  t  1  rp 
x 
 m
  ( pr ) x (1  rp ) m  x
x 
t
 1 r 


 1  rp 
k t
 X ~ Bin (m, rp )
3-26【Question】
Spherical particles whose radii have density function f R (r ) are dropped on a mesh
as in Problem 4. Find an expression for the density function of the particles that pass
through.
【Solution】
 w  2r 
  d  w  f R (r )dr
2
3-29【Question】
Let f ( x )  6 x 2 (1  x ) 2 , for  1  x  1.
(a) Describe an algorithm to generate random variables form this density using
the rejection method. In what proportion of the trials with the acceptance step
be taken?
【Solution】
5
 6 x 2 (1  x) 2 ,  1  x  1
32
15
 f ( x)  ,   1  x  1
4
f ( x) 
(a)
Step1:Generate T~Uniform(-1,1)
Step2:Generate U~Uniform(0,1),independent of T
f (T )
If U 
, then letX  T
15
4
o.w , go to Step1
3-34【Question】
Let N 1 and N 2 be independent random variables following Poisson distribution with
parameters  1 and  2 . Show that the distribution of N  N 1  N 2 is Poisson with
parameter 1   2 .
【Solution】
N 1 ~ Poisson (1 )
N 2 ~ Poisson ( 2 )
N  N1  N 2
M N1 (t )  e 1 ( e 1)
t
M N 2 (t )  e 2 ( e 1)
t
M N (t )  M N1 (t )  M N 2 (t )
 e ( 1  2 )( e 1)
 N ~ Poisson (1   2 )
t
3-35【Question】
For a Poisson distribution, suppose that events are independently labeled A and B with
probabilities p A  pB  1 . If the parameter of the Poisson distribution is  , show that
the number of events labeled A follows a Poisson distribution with parameter p A  .
【Solution】
N  A  B ~ Poisson ( )
A N ~ Bin ( N , p A )

p ( A  a )   p ( A  a, N  n)
n 0

n a
e   n
    p A (1  p A ) n  a 
n!
n 0  a 
(p A ) a  e 

a!


na
1  p A  na
(n  a )!
let t  n  a
(p A ) a e 

a!

1  p A  t
t 0
t!

(p A ) a  e   e (1 p A )  1  p A  

t!
a!e (1 p A )  t 0
t

e  PA  ( p A  ) a
a!
 A ~ Poisson ( p A ,  )

3-37【Question】
Let X and Y be independent standard normal random variables. Find the density of
Z=X+Y, and show that Z is normally distributed as well. (Hint: Use the technique of
completing the square to help in evaluating the integral.)
【Solution】
X～N(0,1)
Y～N(0,1)
X, Y independent
M X (t )  e
t2
2
 M Y (t )
t2
2
M X Y (t )  e  e
t2
2
∴ X+Y～N(0,2)
3-41【Question】
e
t 2 ( 2 )2
2
Let X and Y have the joint density function f ( x , y ) , and let Z  X  Y . Show
that the density function of Z is f Z ( z ) 



z 1
f ( x , ) dy .
y y
【Solution】
Z

Z  XY
X 

W

W  Y
Y  W
x x
z 1
1

w

z
J 
 w2 w  
y y
w
1
0
w z


z
1
 f Z ( z )   f ZW ( z , w)dw   f XY ( , y ) dy


y
y
3-43【Question】
Consider forming a random rectangle in two ways. Let U 1 , U 2 , and U 3 be
independent random variables uniform on [0,1]. One rectangle has sides U 1 and U 2 ,
and the other is a square with sides U 3 . Find the probability that the area of the
square is greater than the area of the other rectangle.
【Solution】
U 1 ,U 2 ,U 3 ～U(0,1) independent
1 1 1
P( U 32  U 1U 2 )= 

0 0
u1u 2
1 1
=
 1
0 0
1
=  (u 2 
0
1
=  (1 
0
du 3 du 2 du1
u1 u 2 du 2 du1
2 u1 u
3
3
2
2
1
) du1
0
2
u1 )du1
3
3
1
4
=( u1  u12 )
9
0
4 5
=1- =
9 9
3-45【Question】
A point is generated on a unit disk in the following way: The radius, R , is uniform
on [0,1] , and the angle  is uniform on [0,2 ] and is independent of R .
(a) Find the joint density of X = R cos  and Y = R sin  .
(b) Find the marginal densities of X and Y .
【Solution】
(a)
R ～U [0,1]
 ～U [0,2 ]
2
2

 X  R cos  R  X  Y


1 Y
Y  R sin 
  tan
X

r
x
J 

x
r
x
y
 r

x
y

y
r
y

1

1
x2  y2
y
1
1
f XY ( x, y )  f R ( x 2  y 2 , tan 1 )  J 

, x2  y2  1
2
2
x
2
x y
(b)
f X ( x)  
1 x 2
 1 x 2
f XY ( x, y )dy
1 x 2

2
2

2
1 x2
[ln( y  x 2  y 2 )]
2
0

1


x2  y2
0
ln(
1
dy
1 x2 1
),  1  x  1
x


1
2
2


du

ln(
u

a

u
)

c
,
a

0


2
2
a

u


1 y2 1
),  1  y  1
Similarly, f Y ( y )  ln(

y
1
3-53【Question】
Let X and Y be jointly continuous random variables.
(b) Develop an expression for the joint density of XY and Y X .
【Solution】
(b)


Z
Z
Z  XY

X 

W or  W

Y 
W  X
Y  ZW
 ZW


X
J  Z
Y
Z
 Z

Z


or  W
or  W
 ZW
 ZW


X
1
1
W 
 2
Y
Z W W  Y X , Z  XY
w
W
X X
Z W
Y Y
f ZW ( z , w)  2
1
w


z
z
z
z
, zw )  f XY (
, zw )  f XY (
, zw )  f XY (
, zw )
 f XY (
w
w
w
w


3-54【Question】
Find the joint density of X  Y and X Y , where X and Y are independent
exponential random variables with parameter λ. Show that X  Y and X Y
are independent.
【Solution】
X ～exp(λ)
Y ～exp(λ)
X , Y independent
f XY ( x, y)  2 e  ( x  y )
WV

X 
W  X  Y



1V

Let 
X
V  Y
Y  W

1V
1
1 v
J 
v
1 v
w
w
(1  v) 2

w
(1  v) 2
2
(1  v)
f W V ( w, v)  2 e w 
w
1
 2 e w  w 
2
(1  v)
(1  v) 2
∴ X  Y , X Y are independent.
3-56【Question】
Each component of the following system (Figure 3.17) has an independent
exponentially distributed lifetime with parameter λ. Find the cdf and the density
of the system’s lifetime.
●Figure 3.17
T
T
T
1
T
2
T
T
a
b
【Solution】
Ti ～exp(λ), i=1,…,6
T
3
T
4
∴ P( Ti  t )=1- e  t
5
6
T
c
P( T  t )=P( Ta  t )P( Tb  t )P( Tc  t )
= 1  P(T1  t ) P(T2  t ) 1  P(T3  t ) P(T4  t ) 1  P(T5  t ) P(T6  t )


= 1  e 2t , t  0
3


FT (t )  1  e 2t , t  0
3

f T (t )  3 1  e 2t


2
 2  e 2t

 6e 2t 1  e 2t , t  0
2
3-62【Question】
Let X 1 , X 2 ,..., X n be independent continuous random variables each with
cumulative distribution function F . Show that the joint cdf of X (1 ) and X ( n ) is
F ( x , y )  F n ( y )  F ( y )  F ( x ) , x  y .
n
【Solution】
F ( x, y ) = P( X (1)  x, X ( n )  y ) = P( X ( n )  y )  P( X (1)  x, X ( n )  y )
P( X ( n )  y ) = P( X 1  y, X 2  y,..., X n  y) = F ( y )
n
n 2
Let X (1) =V, X (n ) =U, fUV (u, v)  n(n  1)  f (u) f (v)F (u)  F (v)
 P( X (1)  x, X ( n )  y ) = P( x  V  U  y )
=
y
x
 n(n  1) f (u) f (v)F (u)F (v)
u
n2
x
=   nf (u )F (u )  F (v)
y
n 1
x
dvdu
u
du
x
=  nf (u)F (u)  F ( x) du
y
n 1
x
= F (u )  F ( x)
n
y
n
= F ( y )  F ( x)
x
 F ( x, y )  F ( y )  F ( y )  F ( x) , x  y
n
n
3-63【Question】
If X 1 ,..., X n are independent random variables, each with the density function f ,
show that the joint density of X (1) ,..., X ( n ) is
n! f ( x1 ) f ( x 2 )  f ( x n ), x1  x 2    x n
【Solution】
n! f ( x )  f X ( xn )
f X (1) ,..., X ( n ) ( x1 ,..., xn )   X 1
otherwise
0
   x1    xn  
The n! naturally comes into this formula because, for any set of values, x1 ,..., xn ,
there are n! equally likely assignments for those values to X 1 ,..., X n which all
yield the same values for the order statistics.
3-67【Question】
Find the density of U ( k )  U ( k 1) if the U i , i  1,..., n are independent uniform
random variables. This is the density of the spacing between adjacent points chosen
uniformly in the interval [0,1].
【Solution】
U i ～Uniform(0,1)
f U ( k )U ( k 1) (u k , u k 1 ) 
n!
 1  1  (u k 1 ) k 2 (1  u k ) nk
(k  2)!(n  k )!
 X  U ( k )  U ( k 1)
Let 
Y  U ( k )
U ( k )  Y

U ( k 1)  Y  X
0< U ( k 1) < U (k ) <1, 0< y  x < y <1, 0< x < y <1
J 
0 1
1
1 1
1
f X ( x)  
x
n!
yx
dy
( y  x) k  2 (1  y ) n  k dy, Let
z
 dz
(k  2)!(n  k )!
1 x
1 x
n!
k 2
nk


1  x z   1  z 1  x   1  x dz
0 ( k  2)! ( n  k )!
1


1
n!
1  x n1  0 z k 2  1  z nk dz
(k  2)!(n  k )!
n!1  x 
k  1n  k  1


(k  2)!(n  k )!
n 
n 1
k  2!n  k!  n1  x n1 , 0  x  1
n!1  x 

n  1!
(k  2)!(n  k )!
n 1

Note: X～Beta ( ,  )
f X ( x) 
(   )  1
x (1  x)  1 , 0  x  1
( )(  )
3-69【Question】
If T1 and T2 are independent exponential  random variables, find the
density function of R  T( 2 )  T(1) .
【Solution】
Ti ~ exp(  ) i  1,2
T1 ,T2 indep .
f Ti (t )  e t , t  0
Let V  T(1) , U  T( 2 )
f U ,V (u , v) 
2!
(e u )(e v )( F (u )) 0 ( F (u )  F (v)) 0 ( F (v)) 0
0!0!
 22 e   (u  v ) , 0  v  u
0 1
R  U  V
V  V

J 
1

1 1
V  V
U  R  V
f R,V (r, v)  22 e  ( r vv) 1

f R (r )   22 e 2v e r dv
0

= e r  2e 2v dv
0
= e  r
 R  T( 2)  T(1) ~ exp(  )
0 v  r v