Ch5. Homework Solution
P173 #1
Let X 1 , X 2 ,... be a sequence of independent random variables with E ( X i )  
and
Var ( X i )   i2 . Show that if n 2 i 1 i2  0 , then
n
X   in probability.
sol
VarX  Var (
0
n
1 n
1
1
xi )  2 Var ( xi )  2

n i 1
n
n
i 1
n

2
i 1
as n  
( xi , x j independen t for i  j ,
 Cov( xi , x j )  0
i  j )
由 Chebyshev’s inequality:  >0
n
P( X     ) 
VarX
2


i 1
2
2
i
n 2
 0 as n  
X n
 

P173
#2
Let X i be as in problem 1 but with E(X i )  μ i and n 1 i 1 μ i  μ.
n
Show that X  μ in probabilit y.
sol
令X i
be a Problem 1. but with EX i  μ i and μ n 
1 n
 μ i , 假定 μ n  μ
n i 1
as  
由Chebyshev' s inequality : ε  0
n
4 σ i2
Var X n
ε
P( X n  μ n  ) 
 i21 2  0 as n  
ε 2
2
n ε
( )
2
ε
 ε  0, δ  0, n 1  Ν s.t. n  n 1  P( X n  μ n  )  δ
2
 μ n  μ as n  
 n 2  Ν s.t. n  n 2  μ n  μ 

ε
2

ε

故若能証出n  n 2  X n  μ  ε   X n  μ n  , 則由n  max n 1 , n 2 
2

ε
 P( X n  μ  ε)  P( X n  μ n  )  δ
2
P

得証 X n n
 μ


ε

claim : n  n 2  X n  μ  ε   X n  μ n  
2

假定 X n  μ  ε, 則n  n 2  ε  X n  μ  X n  μ n  X n  μ  X n  μ n 
 Xn  μn 
ε
2
ε
2


ε

因此, n  n 2  X n  μ  ε   X n  μ n  
2

P173 #3
Suppose that the number of insurance claims, N, filed in a year is Poisson
distributed
with E(N)=10,000 . Use the normal approximation to the Poisson to approximate
P(N>10,200).
sol
Let N be a Poisson random var iable with mean 10000, We find
P( N  10200)by s tan dardizing :
N  10000
P( N  10200)  P(
10000
10200  10000

10000
)  1  (2)  0.0228
where  is the s tan dard normal c.d . f .
6.Using moment-generating functions, show that α   the gamma distribution with
parameters α and λ, properly standardized, tends to the standard normal
distribution.
pf﹕X～gamma(α , λ )
M x (t ) =(

 t
E(X)=


Var(X)=

2
t<λ
)2

X  E( X )
 = X  
Let Z=
=


Var ( X )
2

X
M z (t ) = e 
= e
t
t
Mx(
(
t
)= e 


 t
t
(

t


)
)
1
1
 log M z (t )    t  [log  2  log(  2  t )]
1
1
=   t   log    log(  2  t )
2
=  t 

1
3


1
t2
2
log    [ log    2 t   1   2 ( t 3 )  ]
2
2
2
3
1
=

t2
2
 2 ( t3) 
2
3
→
t2
2
M z (t ) → e
as
t2
2
 
所以 Z ～ U ( 0 ,1)
14.Answer Problem 13 under the assumption that the drunkard has some idea of
where he wants to go so that he steps north with probability 2/3 and south with
probability 1/3.
Ans：令 X i 為在第 i 分鐘內的位移(以北為正向以南為負向)
P( X i =50)=2/3
P( X i = - 50)=1/3
E( X i ) =50×(2/3)＋(- 50)×(1/3)=50/3
E( X i )  50 2  (2 / 3)  (50) 2  (1 / 3)  2500
Var( X i )= E ( X i2 )  [ E ( X i )] 2  20000 / 9
60
S 60 =  X i
i 1
令 Zn =
Sn  i
n
=
S 60  (50 / 3)  60
60  (20000 / 9)
～ N (0 , 1) by C.L.T.
所以 S 60 ～(1000 , 400000/3)
最有可能在原位置北方 1000cm
1
19.Use the Monte Carlo method with n=100and n=1000 to estimate
 cos(2X )dx .
0
Compare the estimates to the exact answer.
Ans： Iˆ( f ) 
1 1000
 cos( 2xi)
1000 i 1
1
I ( f )   c o s2(x)dx
0
Assume X i ～U(0 ,1) iid  f( xi ) iid
I ( f ) =E(f(x))

Iˆ 
1 1000
 f ( xi )
1000 i 1
1
1
1
1
1
1
f ( x)dx   c o s2(x)dx 
c o s2(x)d (2x)  s i n2(x)

2 0
2
0
0
E ( f ( x))  
0
=0
1
1
0
0
1  c o s4x
dx
2
0
1
Var ( f ( x))   ( f ( x)  0) 2 dx   c o 2s(2x)dx  
1
x 1
=  sin( 4x) =1/2
2 8
0
P( Iˆ( f )  I ( f )  )  P(  Iˆ( f )  I ( f )  )  P(

1000 / 2

Iˆ( f )  I ( f )
1000 / 2


1000 / 2
22.Using th central limit theorem to find △ such that P(∣ Iˆ (f)－I(f)∣≦ △ )=
0.05, where Iˆ (f) is the Monte Carlo estimate of
1
 cos(2X )dx
based on 1000
0
points.
Ans： X 1 , X 2 ,... X 1000 ～ U (0 ,1) 且獨立
令 Y=f(x)=cos(2πx) and Y1,Y2,……,Y1000 獨立
1 n
Iˆ( f )   Yi  Yn
n i 1
1
E (Y )  E ( f ( x))   cos( 2x)dx  I ( f )  0  E (Y )  E (Y )
0
c o s4(x)  1
1
dx 
2
2
0
1
1
E (Y )  E[( f ( x)) ]   c o s 2(x)dx  
2
2
2
0
Var(X i )=1/2
By C.L.T
Zn=
Yn  E (Yn )

Var (Yn )
Iˆ( f )  I ( f )
Var (Y ) / n

Iˆ( f )  I ( f )
1 / 2000
～ N (0 ,1)
故
P( Iˆ( f )  I ( f )  )  P(  Iˆ( f )  I ( f )  )  P(

1 / 2000

Iˆ( f )  I ( f )
1 / 2000

 ( 2000)   ( 2000)  2 ( 2000 )  1  0.05
  ( 2000)  0.525  2000  0.0 6 2 7
5   0.0 0 1 4 0 3 1 3 3
27.Prove that if a n  a
,then (1＋ a n
/ n) n  e a .
Pf：令 f n (x) (=(1+( x / n)) 2  e x  f ( x)為一連續函 數

1 / 2000
)
)
P
an 

a     0  P( a n  a )   )  0
f 在 x  a 連續     0,    0, st 0  x  a    f ( x)  f (a)  
 { f ( x)  f (a)   }  { x  a  
故   0. P( f (a n )  f (a)   )  P( a n  a   )  0
P
P
即 f (a n ) 

f ( a )  e an 

ea
 (1 
an n P
) 
 e a
n