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Chapter 6 parameter estimation. § 6.1 Point estimation. Set the distribution function of population X is F(x), which contains the unknown parameters ,we call any function of a sample point estimator. 1, Moment estimate. - PowerPoint PPT Presentation
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Chapter 6 parameter estimation
§ 6.1 Point estimation
Set the distribution function of population X is F(x), which contains the unknown parameters ,we call any function of a sample point estimator.
1( ,..., )nX X
1, Moment estimate
If population distribution contains k unknown parameters,method of moments estimator are found by equating the first k sample moments to the corresponding k population moments,and solving the resulting system of simultaneous equations.
Example 6.1 suppose are iid ,We have
according to the method of moments,we yield the moments estimator of is .
1 2( , ,..., )nX X X ( )P
1 1, ,X
X
Example 6.2 Suppose are iid ,
We have ,hence we must solve the equation
Solving for yields the moments estimator of is
1 2( , ,..., )nX X X 2( , )N 2 2 2
1 2 1 21
1, , ,
n
ii
X Xn
2 2 2
1
,
1,
n
ii
X
Xn
2, 2,
2 2 2
1
,
1 1( ) ,
n
ii
X
nS X X
n n
Example 6.3 suppose are iid,and the
common pdf is
， is a known positive constant ， we have
Let ,we yield the moment estimator of is .
1 2( , ,..., )nX X X1 1
(1 )1, ,
( ; )
0,
c x x cf x
其它,10 c
1
1 1(1 )
( ; )
1
,1
c
EX xf x dx
xc x dx
c
X 1
c
X
2, Maximum likelihood estimate
Consider a random sample from a distribution having pdf or probability distribution , are unknown parameters,The joint pdf of is ,this may be regard as a function of ,when so regarded,it is called the likelihood function L of the random sample,and we write
1 2( , ,..., )nX X X
( ; )x 1( ,..., )m
1 2( , ,..., )nX X X
1
( ; )n
ii
x
11
( ,..., ; ) ( ; )n
n ii
L x x x
Suppose that we can find a nontrivial function of ,say ,when is replaced by ,the likelihood function L is a maximum,then the statistic will be called a maximum likelihood estimator of ,and will be denoted by the symbol .
1 2( , ,..., )nx x x
1 2( , ,..., )nu x x x
1 2( , ,..., )nu x x x
1 2( , ,..., )nu X X X
1 2( , ,..., )nu X X X
The function can be maximized by setting the first partial derivation of ,with respect to ,equal to zero,that is to say
and solving the resulting equation for ,which i
s the maximum likelihood estimator of .
L
ln L
1
ln0,
ln0.
m
L
L
Example 6.4 Let denote a random sample
from a distribution that is ,we shall find , the maximum likelihood estimator of .
1 2( , ,..., )nX X X
( )P ̂
Example 6.5 Let denote a random sample
from a distribution that is are unknown parameters, we shall find , the maximum likelihood estimators of .
1 2( , ,..., )nX X X
2( , ),N
2,
2ˆ ˆ, 2,
Example 6.6 If the probability distribution of is
( ) is unknown parameter,we observed the following eight values of ,
3, 1, 3, 0, 3, 1, 2, 3, Find the moment estimate and the
maximum likelihood estimate of .
X
2 2
0 1 2 3
2 (1 ) 1 2
10
2
X
Example 6.7 Let denote a random
sample from a distribution that is , Find the maximum likelihood estimators of .
1 2( , ,..., )nX X X
[ , ]U a b,a b
3, Methods of evaluating estimators
3.1 Unbiased estimator If an estimator satisfies
that ,the estimator is called unbiased.
1( ,..., )nX X
1( ,..., )nE X X
Example 6.8 Let denote a random
sample from , ,Prove that and are the unbiased estimator of respectly.
1 2( , ,..., )nX X XX 2,EX DX
X 2 2
1
1( )
1
n
ii
S X Xn
2,
3.2 Efficient estimator Let and are both the unbiased estimator of
,if , We say is more efficient than .If the number
of sample is fixed,and the variance of is less than or equal to the variance of every other unbiased estimator of ,we say is the efficient estimator of .
1
2
1 2D D
1
2
n
Example 6.9 Consider a random sample from a
distribution having pdf
,Prove that and are both the unbiased estimators of ,and when , is more efficient than .
1 2( , ,..., )nX X X
1, 0,
( )0 ,
xe xp x
others,
1min( , , )nZ X X X
1n XnZ
nZ
3.3 Uniform estimator
If for any ,we have , We call is the uniform
estimator of .
0 lim (| | ) 1nnP
n̂
§6.2 Interval estimation
An interval estimator of a real-valued parameter is any pair of function and of a sample that satisfy
for all sample x.if the sample x is observed,the inference
is made.The random interval is called an interval estimator;If ,we call the confidence coefficient of is .we often set =0.01,0.05,0.1 etc.
2 1( ,..., )nX X 1 1( ,..., )nX X
1 1( ,..., )nX X
2 1( ,..., )nX X
1 1 2 1( ,..., ) ( ,..., )n nX X X X
1 2( , )
1 1 2 1( ( ,..., ) ( ,..., )) 1n nP X X X X
1 2( , )
1
1, Interval estimator for mean in normal population
1.1 Let denote a random sample from a distribution that is is known, is unknown parameter,
Set
,We have By the definition of we have ,
,
,the interval estimator of with the confidence coefficient is .
1 2( , ,..., )nX X X2( , ),N 2
/
XU
n
(0,1),U N /2u
/2(| | ) 1P U u
/2(| | ) 1/
XP u
n
/2 /2( . / . / ) 1P X u n X u n
1 /2 /2( . / , . / )X u n X u n
Example 6.10 Let a random sample of size 10 from the
normal distribution They are18.3, 17.5, 18.1, 17.7, 17.9, 18.5, 18,
18.1,17.8,17.9,Determine a 95 per cent confidence
interval for .
2( ,0.2 ),X N
1.2 Let denote a random sample from a distribution that is and are unknown parameters, Set
, We have ,so
, Hence the interval estimator of with the
confidence coefficient is .
1 2( , ,..., )nX X X2( , ),N
2
/
XT
S n
~ ( 1)T t n
/2(| | ( 1)) 1P T t n
1
/2 /2( ( 1) / , ( 1) / )X t n S n X t n S n
Example 6.11 Let a random sample of size 9 from the
normal distribution ,They are 0.497,0.506 ， 0.518 ， 0.524 ， 0.488 ， 0.510 ，
0.510 ， 0.515 ， 0.512 ， Determine a 99 per cent confidence
interval for .
),( 2N
2, Interval estimator for variance in normal population
Let denote a random sample from a distribution that is and are unknown parameters,
Set , We have ,so
,
,So the interval estimator of with the confidence coefficient is
.
1 2( , ,..., )nX X X2( , ),N 2
22 2
2 21
1 ( 1)( )
n
ii
n SX X
2 2~ ( 1)n 2 2 21 /2 /2( ( 1) ( 1)) 1P n n
2 2
21 12 2/2 1 /2
( ) ( )1
( 1) ( 1)
n n
i ii i
X X X XP
n n
21 2 2
1 12 2/2 1 /2
( ) ( ),
( 1) ( 1)
n n
i ii i
X X X X
n n
Example 6.12 Let a random sample of size 9 from the
normal distribution ,They are 600, 612, 598, 583, 609, 607, 592, 588, 593, Determine a 95 per cent confidence interval
for .
),( 2N
2
3, Interval estimator for difference of means and ratio of variance in two normal populations
Let denote a random sample of size from a distribution that is denote a random sample of size from a distribution that is where , are unknown parameters, the two random samples are independent, What is the interval estimator of and with the confidence coefficient ?
11 2( , ,..., )nX X X1n
21 1( , ),N
21 2( , ,..., )nY Y Y
2n2
2 2( , ),N 21 1,
22 2,
1 2 2 21 2/
1
3.1 Difference of means If ,but is unknown,set
, So ,hence
, So the interval estimator of with the confidence coefficient is
.
2 2 21 2 2
1 2
1 2
( ) ( )
1 1T
X Y
Sn n
1 2~ ( 2)T t n n
/2 1 2(| | ( 2)) 1P T t n n
1 2 1
/ 2 1 21 2
/ 2 1 21 2
1 1( ( 2) ,
1 1( 2) )
X Y t n n Sn n
X Y t n n Sn n
3.2 ratio of variance Set
, so .hence
So the interval estimator of with the confidence coefficient is
2 21 22 21 2
SF
S
1 2( 1, 1)F F n n
1 2 1 2 2 1 2( ( 1, 1) ( 1, 1) 1P F n n F F n n
1 2 2 2 21 2 1 2
2 1 2 1 2 1 2
( , )( 1, 1) ( 1, 1)
S S S S
F n n F n n
2 21 2/
Example 6.13 Let a random sample of size 4 from the
normal distribution ,They are 0.143 0.142 0.143 0.137Another random sample of size 5 from the normal distribution ,They are0.140 0.142 0.136 0.138 0.140The two random samples are independent,determine the interval estimator of and with the confidence coefficient 0.95.
21( , )N
22( , )N
1 2 2 21 2/
