Ch2 Homework Solution
2-8 Show that the binomial probabilities sum to 1.
X ~ B (n, p)
n n
k
  k  p
k  0 
(1  p) n  k  ( p  (1  p)) n  1
2-11 Consider the binomial distribution with n trails and probability
p of success on each trial. For what value of k is P(X=k)
maximized? This value is called the mode of the distribution.
(Hint: Consider the ratio of successive terms.)
P( X  k )
1
P ( X  k  1)
&
P( X  k )
1
P ( X  k  1)
n k
  p (1  p) n  k
n!(k  1)!(n  k  1)! p (n  k  1) p
k 


 1 －(1)

k!(n  k )!n!(1  p)
k (1  p)
 n  k 1
n  k 1

 p (1  p)
 k  1
n k
  p (1  p) n  k
n!(k  1)!(n  k  1)! p k (1  p) n  k (k  1)(1  p)
k 


 1 －(2)
k !
n  k 1
(n  k ) p
 n  k 1
k
!
(
n

k
)!
n
!
p
(
1

p
)
n  k 1

 p (1  p)
 k  1
 (1) (n  k  1) p  k  kp
k  p(n  1)
 p(n  1)  1  k  p(n  1)
(2) (k  1)(1  p)  np  kp
k  p(n  1)  1
 if p(n  1)   , k = p(n+1)-1 or
if p(n  1)   , k = [ p(n+1) ]
p(n+1)
2-15 Two terms, A and B, play a series of games. If team A has
probability 0.4 of winning each game, is it to its advantage to
play the best three out of five games or the best four out of
seven? Assume the outcomes of successive games are
independent.
(1) Three out of five games :
P(A wins )  0.43 
3!
4!
 0.43  0.6 
 0.43  0.62  0.43 (1  3x0.6  6 x0.36)
2!
2!2!
= 4.96 x 0.43
(2) Four out of seven games :
P(A w i n) 
s 0.44 
4!
5!
6!
 0.44  0.6 
 0.44  0.62 
 0.44  0.63
3!
3!2!
3!3!
 0.44 (1  4 x0.6  10 x0.36  20 x0.216)  4.528 x0.43
 “Three out of five games” is better!
2-25 The probability of being dealt a royal straight flush (ace, king,
queen, jack, and ten of the same suit) in poker is about 1.3x10-8.
Suppose that an avid poker player sees 100 hands a week, 52
weeks a year, for 20 years.
a. What is the probability that she never sees a royal straight
flush dealt?
b. What is the probability that she sees two royal straight flushes
dealt?
100 x 52 x 20 = 104000
(a) ( 1.3 x 10-8 )104000 = 0.9987
103998
104000 
 1  1.3x108
(1.3x108 ) 2  9 x107
(b) 
 2 


2-28 Let p0, p1, …, pn denote the probability mass function of the
binomial distribution with parameters n and p. Let q = 1-p.
Show that the binomial probabilities can be computed
recursively by p0 = qn and
Pk 
(n  k  1) p
Pk 1
kq
 n  k nk
  p q
Pk
(n  k  1) p
k 


Pk 1  n  k 1 n  k 1
kq

 p q
 k  1
 Pk 
(n  k  1) p
Pk 1
kq
k = 1, 2, …, n
k = 1, 2, …, n
2-29 Show that the Poisson probabilities p0, p1, …, pn can be
computed recursively by p0 = exp(-λ) and
Pk 
P( X  x) 

k
 Pk 1
k = 1, 2, …, n
 x e 
x!
Pk
k e (k  1)!




Pk 1
k!
k 1  e k
 Pk 

k
 Pk 1
2-32 For what value of k is the Poisson frequency function with
parameter λ maximized? (Hint: Consider the ratio of
consecutive terms.)
Pk
1
Pk 1


k
1
&
&
Pk
1
Pk 1
k !

1
  1  k  
 if    , k = λ or λ-1
if    , k = [λ]
2-39 The Cauchy cumulative distribution function is
1 1
 tan 1( x)
2 
F ( x) 
-∞ ＜ x ＜∞
a Show that this is a cdf.
b Find the density function.
c Find x such that P (X ＞ x) = .1.
F ( x) 
(a) (1)
1 1
 t a n1( x)
2 
-∞ ＜ x ＜∞
1 1
 tan 1 x  0
x  2 
lim
1 1
 t a n1 x  1
x  2 
1
1
(2) F ( x)  
 0, x.  F(x) 
 1  x2
lim
(3)
1 1
 tan 1 x  F ( x0 )
x  x0  2 
lim
F(x) 右連續
 0
By (1)(2)(3) F(x) is a c.d.f
(b) f ( x)  F ( x) 
(c)

1
1

 1 x2
1
a   1  x2 dx  0.1

1

1
   t a n ( x)  0.1

a
1 
(  t a n1 a )  0.1
 2
1 1
t a n1 a   (  )
2 10
a  3.078
1
,-  x  
2-49 The gamma function is a generalized factorial function.
a Show that Γ(1) = 1.
b. Show that Γ(x+1) = xΓ(x). (Hint: Use integration by parts)
c. Conclude that Γ(n) = (n-1)! for n = 1, 2, 3, …
1
2
d. Use the fact that ( )   to show that, if n is an odd integer,
n
 (n  1)!
( ) 
n 1
2
2n 1(
)!
2

Gamma function (t )   e  y y t 1dy
0




(a) (1)   e y dy   e y 0  1
0
(b) ( x  1) 
 y x
0 e
y dy




=  e y y x 0   (e y ) xyx 1dy
0

= x   e  y y x 1dy
0
= x   (x)
(c) (n)  (n  1)(n  1)
= (n  1)( n  2)(n  2)
= …
= (n  1)( n  2)3  2  1  (1)
= (n  1)!
1
(d) ( )   , n is an odd integer.
2
n
 (n  1)!
Claim: ( ) 
n 1
2
2n 1(
)!
2
1
Pf: (1) n = 1, ( )   .
2
(2) 設 n = k (k is an odd integer.)
k
 (k  1)!
( ) 
成立.
k 1
2
2k 1(
)!
2
(3) n = k + 2.
k 1
by(b) k
k2
k
k
k    (k  1)!
(
)  (  1) 
( ) 
x 2
k

1
k 1
2
2
2 2
2  2k 1  (
)!
2
2
=
  (k  1)!
k 1
)!
2
By (1)(2)(3) 由數學歸納法知,原命題成立.
2k 1  (
2-51 Show that the normal density integrates to 1. (Hint: First
make a change of variables to reduce the integral to that for
the standard normal.) The problem is then to show that


 x2
e 2 dx 
2 . Square both sides and reexpress the problem as
that of showing


 x 2 
 y2
 
 

  e 2 dx   e 2 dy  = 2
 
  




Finally, write the product of integrals as a double integral and
change to the polar coordinates.
claim:


i.e. claim:
1
2
 x2
e 2 dx
 x2

e 2 dx 


1
2
Sol :
 x2

Let  
e 2 dx



 x2 



 2    e 2 dx 
 



2


 x 2 
 y2
 
 

2
2
dx   e
dy 
=  e
 
  




=
Let X  r cos
Y  r sin 
 dxdy  rdrd
=
( x 2  y 2 )
 
2
e
dxdy
 
 
 2
0 0
r 2
e 2 rd dr
r 2

r  e 2 dr
0
= 2 

 r 2 


=  2 e 2 


0
= 2
 x2


e 2 dx 



1
 2

2-57
X ~ N (  ,  2 ) and
2
 x2
e 2 dx
1
Y  aX  b , where a<0 , show that Y ~ N (a  b, a 2 2 )
Sol:
X ~ N ( , 2 )
Y  aX  b, a  0
 X=
( x   ) 2
f X ( x) 
1
2 
e
2 2
Y b
a
,
J 
1
a
(
1
f Y ( y) 
2 
e
y b
 )
a
2 2
2
 ( y ( b  a )) 2
1
1
=
a
2  a
e
2 a 2 2
 Y ~ N (a  b, a 2 2 )
2-58
U .
If U is uniform on [0,1] , find the density of
Sol:
U ~uniform[0,1]
fU (u)  1 , Y= U
P(Y  y)  P( U )  y)  P(U  y 2 )  y 2
 f Y ( y) (y)=2y , 0  y  1
2-60
Find the density function of Y= e Z , where Z ~ N (  ,  2 ) .This is called the
lognormal density, since log Y is normally distributed.
Sol:
Z~N(μ,σ²)
Let Y= e Z , Z= ln Y
, J 
1
y
1
1
∴ f Y ( y )  f Z ( l ny ) (y)=
e
y
2 
=
 ( l yn  ) 2
1
2 y
0  x   ,      ,  0
2 2
exp( (
1
y
ln x   2
) )
2 2
2-63
Suppose that  follows a uniform distribution on the interval [ 
density of tan  .
Sol:
 ~U[ 
f  ( ) =
 
, ]
2 2
1

,
θ [ 
 
, ]
2 2
 
, ].Find the cdf and
2 2
tan 1 y
1


Y= tan  , P(Y  y)=P(tan   y)=P(   tan 1 y )   
2
d 
1

tan 1 y 
1
,
2
  x
2-67
The Weibull cumulative distribution function is F ( x)  1  e
( x )

, x ≧0,α>0,β>0
(a) Find the density function
(b) Show that is W follows a Weibull distribution , then X= (W )  follows an

exponential distribution .
(c) How could Weibull random variable be generated from a uniform random
number generator ?
Sol:
X ~ Weibull , x  0,α>0,β>0
(a) F (x) =, x  0,α>0,β>0
f ( x )  e
x
( ) 

( ) 
x
1

  ( )  1   x  1e  , x  0,α>0,β>0
x



(b) X= (W )  , X ~ Weibull ( ,  )

 1
J  x

1
1
W   (x)

,
1

f X ( x)   (x  )  1 e

1
(c)
F
x  
(
)

1
( z)  z 
 F W (w)  1  e
?
Z
  1
x
 e  x ~ exp(1)

let z= 1  e
w
( ) 

1-Z= e
w
( ) 

w
( ) 

w
ln( 1  z )  ( )   w 
又 if Z~ uniform[0,1] then 1-Z~uniform[0,1]
∴we can firstly generate U ~uniform(0,1)

let W=
then X ~ Weibull ( ,  )
(  ln U ) 


( ln( 1  z )) 