Download Review of Asymptotics

Asymptotics
1
Example
Xi
X
V arX
¢
¡
iid ¹; ¾2 ; i = 1; 2; ::; n
1X
=
Xi
n i
1X
1X
1X
) EX = E
Xi =
EXi =
¹=¹
n i
n i
n i
"
#
1X
1 X 2 ¾2
1 X
:
= V ar
Xi = 2
V arXi = 2
¾ =
n i
n i
n i
n
»
V arX
X
2
! 0 as n ! 1
! ¹ as n ! 1:
Concepts
Probability Limits: Let Sn be a statistic whose properties depend on n. Then
(weak consistency)
plimSn = µ i¤ lim Pr [jSn ¡ µj < "] = 1 8" > 0:
n!1
We say that Sn is a weakly consistent estimator of µ.
Example (continued):
plimX n = ¹?
¯
¤
£
¤
£¯
Pr ¯X n ¡ ¹¯ < " = Pr ¡" < X n ¡ ¹ < "
" p
¢
p ¡
p #
n Xn ¡ ¹
¡ n"
n"
<
<
= Pr
¾
¾
¾
· p
p ¸
¡ n"
n"
= Pr
<Z <
¾
¾
p
where Z » N (0; 1). As n ! 1, ¾n" ! 1 for all …xed " )
· p
p ¸
¡ n"
n"
<Z <
= 1
lim Pr
n!1
¾
¾
) plimX n = ¹:
1
Alternatively, we say that, if
i
h
Pr lim jSn ¡ µj < " = 1;
n!1
then Sn is a strongly consistent estimator of µ.
Example (continued):
¯
¯
¯
¯
¯
¾
¡ ¹ ¯¯
¯X n ¡ ¹¯ = p¾ ¯ X n p
= p jZj
¯
¯
n ¾= n
n
where Z » N (0; 1) )
¾
lim p jZj
n!1
n
=
0
i
h
¯
¯
) Pr lim ¯X n ¡ ¹¯ < " = 1 8" > 0:
n!1
Thus, X n is a strongly consistent estimator of ¹.
New example:
½
X n with probability 1 ¡
Sn =
n
with probability n1
Note that
Pr [jSn ¡ ¹j < "] =
(
h
Pr jZj <
0
p i
n"
¾
1
n
:
with probability 1 ¡
with probability n1
1
n
and that, as n ! 1, n1 ! 0 and the second term disappears. ) Sn is a weakly
consistent estimator of ¹. However,
lim jSn ¡ ¹j
n!1
does not exist. So Sn is not a strongly consistent estimator of ¹. Explain why
this example is relevant.
3
Properties
plim c = c
plim cXn = cplim Xn
plim (Xn + Yn ) = plimXn + plimYn
plim (Xn Yn ) = (plimXn ) (plimYn )
plim g (X1n ; X2n ; ::; Xmn ) = g (plimX1n ; plimX2n ; ::; plimXmn )
Compare consistency and unbiasedness: If EXn ! ¹ and V arXn ! 0, then
plim Xn = ¹. The converse is not true. Examples:
2
1.
¢
¡
» iid ¹; ¾2
m
1 X
=
Xi
m i=1
Xi
Sn
for …xed m. Then
ESn
V arSn
= ¹;
¾2
=
9 0 as n ! 1
m
) plim Sn does not exist.
2.
Xi
Sn
¢
¡
» iid ¹; ¾2
" m
#¡1
1X
=
Xi
:
n i=1
Then
plimSn
But
£
¤¡1
plim X n
"
m
1X
= plim
Xi
n i=1
=
#¡1
1
:
¹
¢¡1
¡ ¢¡1 ¡
ESn = E X n
6= EX n
and, in fact, for many cases does not exist.
4
Central Limit Theorem
Let
Then
¢
¡
Xi » iid ¹; ¾ 2 ; i = 1; 2; ::; n:
¢
p ¡
n X ¡ ¹ =¾ » N (0; 1)
for a large class of distributions for Xi . The CLT generalizes in many ways,
the most important being to allow for heterogeneity in Xi .
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One more example: let U » Â2m , V » Â2n with U; V independent. Then
Z=
U=m
» Fm;n :
V=n
What happens as n ! 1? De…ne
V =
n
X
Wi
i=1
where Wi » iidÂ21
) EWi = 1; V arWi = 2
V
=1
) plim
n
U
U=m
!
) Z=
as n ! 1
V=n
m
) mZ ! U » Â2m :
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