文章目录
08.样本矩样本矩样本中心距
08.样本矩
首先说明一些记号：
设
X
1
,
X
2
,
…
,
X
n
i.i.d.
F
X_{1}, X_{2}, \ldots, X_{n}{ }^{\text {i.i.d. }} F
X1,X2,…,Xni.i.d. F
总体矩:
α
k
=
E
(
X
1
k
)
,
α
1
=
μ
\alpha_{k}=\mathrm{E}\left(X_{1}^{k}\right), \alpha_{1}=\mu
αk=E(X1k),α1=μ
总体中心矩:
μ
k
=
E
{
(
X
1
−
μ
)
k
}
,
μ
2
=
σ
2
\mu_{k}=\mathrm{E}\left\{\left(X_{1}-\mu\right)^{k}\right\}, \mu_{2}=\sigma^{2}
μk=E{(X1−μ)k},μ2=σ2
样本矩:
a
k
=
1
n
∑
i
=
1
n
X
i
k
,
a
1
=
X
ˉ
n
a_{k}=\frac{1}{n} \sum_{i=1}^{n} X_{i}^{k}, a_{1}=\bar{X}_{n}
ak=n1∑i=1nXik,a1=Xˉn
样本中心矩:
m
k
=
1
n
∑
i
=
1
n
(
X
i
−
X
ˉ
n
)
k
,
m
2
=
S
n
2
m_{k}=\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\bar{X}_{n}\right)^{k}, m_{2}=S_{n}^{2}
mk=n1∑i=1n(Xi−Xˉn)k,m2=Sn2
样本矩
样本矩可看作
n
n
n个
i
.
i
.
d
.
i.i.d.
i.i.d.随机变量的均值。在假设任意阶矩有限的条件下，由大数定律和中心极限定理可以得到如下结论：
定理 8.1:
E
(
a
k
)
=
α
k
,
Var
(
a
k
)
=
α
2
k
−
α
k
2
n
\mathrm{E}\left(a_{k}\right)=\alpha_{k}, \operatorname{Var}\left(a_{k}\right)=\frac{\alpha_{2 k}-\alpha_{k}^{2}}{n}
E(ak)=αk,Var(ak)=nα2k−αk2
a
k
→
p
α
k
a_{k} \stackrel{p}{\rightarrow} \alpha_{k}
ak→pαk
a
k
→
w
p
1
α
k
{a_{k} \stackrel{w p 1}{\rightarrow}} \alpha_{k}
ak→wp1αk
n
(
a
1
−
α
1
,
…
,
a
k
−
α
k
)
⊤
→
d
N
k
(
0
,
Σ
)
\sqrt{n}\left(a_{1}-\alpha_{1}, \ldots, a_{k}-\alpha_{k}\right)^{\top} \stackrel{d}{\rightarrow} N_{k}(\mathbf{0}, \Sigma)
n
(a1−α1,…,ak−αk)⊤→dNk(0,Σ), 其中
Σ
=
(
σ
j
1
j
2
)
k
×
k
\Sigma=\left(\sigma_{j_{1} j_{2}}\right)_{k \times k}
Σ=(σj1j2)k×k ，
σ
j
1
j
2
=
α
j
1
+
j
2
−
α
j
1
α
j
2
\sigma_{j_{1} j_{2}}=\alpha_{j_{1}+j_{2}}-\alpha_{j_{1}} \alpha_{j_{2}}
σj1j2=αj1+j2−αj1αj2
样本中心距
而对于样本中心距而言，其求和的每一项并不具有独立性。由此我们先构造另一相关统计量：
b
k
=
1
n
∑
i
=
1
n
(
X
i
−
μ
)
k
b_{k}=\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\mu\right)^{k}
bk=n1∑i=1n(Xi−μ)k，
b
k
b_k
bk为
i
.
.
i
.
d
.
i..i.d.
i..i.d.随机向量的均值，可以得到类似定理8.1的结论。以下给出引理8.2：
引理8.2：
E
(
b
k
)
=
μ
k
,
Var
(
b
k
)
=
μ
2
k
−
μ
k
2
n
\mathrm{E}\left(b_{k}\right)=\mu_{k}, \operatorname{Var}\left(b_{k}\right)=\frac{\mu_{2 k}-\mu_{k}^{2}}{n}
E(bk)=μk,Var(bk)=nμ2k−μk2
b
k
→
p
μ
k
b_{k} \stackrel{p}{\rightarrow} \mu_{k}
bk→pμk
b
k
→
w
p
1
μ
k
\quad b_{k} \stackrel{w p 1}{\rightarrow} \mu_{k}
bk→wp1μk
n
(
b
1
−
μ
1
,
…
,
b
k
−
μ
k
)
⊤
→
d
N
k
(
0
,
Σ
~
)
\sqrt{n}\left(b_{1}-\mu_{1}, \ldots, b_{k}-\mu_{k}\right)^{\top} \stackrel{d}{\rightarrow} N_{k}(\mathbf{0}, \tilde{\Sigma})
n
(b1−μ1,…,bk−μk)⊤→dNk(0,Σ~), where
Σ
~
=
(
σ
~
j
1
j
2
)
k
×
k
\tilde{\Sigma}=\left(\tilde{\sigma}_{j_{1} j_{2}}\right)_{k \times k}
Σ~=(σ~j1j2)k×k with
σ
~
j
1
j
2
=
μ
j
1
+
j
2
−
μ
j
1
μ
j
2
\tilde{\sigma}_{j_{1} j_{2}}=\mu_{j_{1}+j_{2}}-\mu_{j_{1}} \mu_{j_{2}}
σ~j1j2=μj1+j2−μj1μj2
考虑到
b
k
b_k
bk具有较好的性质，我们尝试将
b
k
b_k
bk与
m
k
m_k
mk建立联系
m
k
=
1
n
∑
i
=
1
n
(
X
i
−
X
ˉ
n
)
k
=
1
n
∑
i
=
1
n
∑
j
=
0
k
C
k
j
(
X
i
−
μ
)
j
(
μ
−
X
ˉ
n
)
k
−
j
,
m_{k}=\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\bar{X}_{n}\right)^{k}=\frac{1}{n} \sum_{i=1}^{n} \sum_{j=0}^{k} C_{k}^{j}\left(X_{i}-\mu\right)^{j}\left(\mu-\bar{X}_{n}\right)^{k-j},
mk=n1i=1∑n(Xi−Xˉn)k=n1i=1∑nj=0∑kCkj(Xi−μ)j(μ−Xˉn)k−j, i.e.,
m
k
=
∑
j
=
0
k
C
k
j
(
−
1
)
k
−
j
b
j
b
1
k
−
j
m_{k}=\sum_{j=0}^{k} C_{k}^{j}(-1)^{k-j} b_{j} b_{1}^{k-j}
mk=j=0∑kCkj(−1)k−jbjb1k−j
由此可以得到定理8.3：
定理8.3：
m
k
→
w
p
1
μ
k
m_{k} \stackrel{w p 1}{\rightarrow} \mu_{k}
mk→wp1μk
n
(
m
2
−
μ
2
,
…
,
m
k
−
μ
k
)
⊤
→
d
N
k
−
1
(
0
,
Σ
∗
)
\sqrt{n}\left(m_{2}-\mu_{2}, \ldots, m_{k}-\mu_{k}\right)^{\top} \stackrel{d}{\rightarrow} N_{k-1}\left(0, \Sigma^{*}\right)
n
(m2−μ2,…,mk−μk)⊤→dNk−1(0,Σ∗), 其中
Σ
∗
=
(
σ
j
12
∗
)
(
k
−
1
)
×
(
k
−
1
)
\Sigma^{*}=\left(\sigma_{j_{12}}^{*}\right)(k-1) \times(k-1)
Σ∗=(σj12∗)(k−1)×(k−1) ，
σ
j
j
j
∗
=
μ
j
1
+
j
2
+
2
−
μ
j
1
+
1
μ
j
2
+
1
−
(
j
1
+
1
)
μ
j
1
μ
j
2
+
2
−
(
j
2
+
1
)
μ
j
2
μ
j
1
+
2
+
(
j
1
+
1
)
(
j
2
+
1
)
μ
j
1
μ
j
2
μ
2
\sigma_{j j j}^{*}=\mu_{j_{1}+j_{2}+2}-\mu_{j_{1}+1} \mu_{j_{2}+1}-\left(j_{1}+\right.1) \mu_{j_{1}} \mu_{j_{2}+2}-\left(j_{2}+1\right) \mu_{j_{2}} \mu_{j_{1}+2}+\left(j_{1}+1\right)\left(j_{2}+1\right) \mu_{j_{1}} \mu_{j_{2}} \mu_{2}
σjjj∗=μj1+j2+2−μj1+1μj2+1−(j1+1)μj1μj2+2−(j2+1)μj2μj1+2+(j1+1)(j2+1)μj1μj2μ2
证明：
首先注意到
μ
1
=
0
\mu_1=0
μ1=0
由CMT可知：
m
k
=
∑
j
=
0
k
C
k
j
(
−
1
)
k
−
j
b
j
b
1
k
−
j
→
w
p
1
∑
j
=
0
k
C
k
j
(
−
1
)
k
−
j
μ
j
μ
1
k
−
j
=
μ
k
m_{k}=\sum_{j=0}^{k} C_{k}^{j}(-1)^{k-j} b_{j} b_{1}^{k-j}\stackrel{w p 1}{\rightarrow}\sum_{j=0}^{k} C_{k}^{j}(-1)^{k-j} \mu_{j} \mu_{1}^{k-j}=\mu_k
mk=∑j=0kCkj(−1)k−jbjb1k−j→wp1∑j=0kCkj(−1)k−jμjμ1k−j=μk,则（1）成立
将
m
k
m_k
mk改写为
m
k
=
∑
j
=
0
k
C
k
j
(
−
1
)
k
−
j
b
j
b
1
k
−
j
=
b
k
−
k
b
k
−
1
b
1
+
∑
j
=
0
k
−
2
C
k
j
(
−
1
)
k
−
j
b
j
b
1
k
−
j
\begin{aligned} m_{k} &=\sum_{j=0}^{k} C_{k}^{j}(-1)^{k-j} b_{j} b_{1}^{k-j} \\ &=b_{k}-k b_{k-1} b_{1}+\sum_{j=0}^{k-2} C_{k}^{j}(-1)^{k-j} b_{j} b_{1}^{k-j} \end{aligned}
mk=j=0∑kCkj(−1)k−jbjb1k−j=bk−kbk−1b1+j=0∑k−2Ckj(−1)k−jbjb1k−j 有
n
(
m
k
−
μ
k
)
=
n
(
b
k
−
μ
k
)
−
k
b
k
−
1
n
b
1
+
n
b
1
∑
j
=
0
k
−
2
C
k
j
(
−
1
)
k
−
j
b
j
b
1
k
−
j
−
1
=
n
(
b
k
−
μ
k
)
−
k
μ
k
−
1
n
b
1
+
o
p
(
1
)
=
n
(
b
k
−
μ
k
−
k
μ
k
−
1
b
1
)
+
o
p
(
1
)
\begin{aligned} &\sqrt{n}\left(m_{k}-\mu_{k}\right) \\ &\quad=\sqrt{n}\left(b_{k}-\mu_{k}\right)-k b_{k-1} \sqrt{n} b_{1}+\sqrt{n} b_{1} \sum_{j=0}^{k-2} C_{k}^{j}(-1)^{k-j} b_{j} b_{1}^{k-j-1} \\ &\quad=\sqrt{n}\left(b_{k}-\mu_{k}\right)-k \mu_{k-1} \sqrt{n} b_{1}+o_{p}(1) \\ &\quad=\sqrt{n}\left(b_{k}-\mu_{k}-k \mu_{k-1} b_{1}\right)+o_{p}(1) \end{aligned}
n
(mk−μk)=n
(bk−μk)−kbk−1n
b1+n
b1j=0∑k−2Ckj(−1)k−jbjb1k−j−1=n
(bk−μk)−kμk−1n
b1+op(1)=n
(bk−μk−kμk−1b1)+op(1) 则
n
(
m
2
−
μ
2
,
…
,
m
k
−
μ
k
)
⊤
=
n
(
b
2
−
μ
2
−
2
μ
1
b
1
,
…
,
b
k
−
μ
k
−
k
μ
k
−
1
b
1
)
⊤
=
n
−
1
/
2
∑
i
=
1
n
(
Z
i
2
,
…
,
Z
i
k
)
⊤
\begin{aligned} &\sqrt{n}\left(m_{2}-\mu_{2}, \ldots, m_{k}-\mu_{k}\right)^{\top} \\ &\quad=\sqrt{n}\left(b_{2}-\mu_{2}-2 \mu_{1} b_{1}, \ldots, b_{k}-\mu_{k}-k \mu_{k-1} b_{1}\right)^{\top} \\ &\quad=n^{-1 / 2} \sum_{i=1}^{n}\left(Z_{i 2}, \ldots, Z_{i k}\right)^{\top} \end{aligned}
n
(m2−μ2,…,mk−μk)⊤=n
(b2−μ2−2μ1b1,…,bk−μk−kμk−1b1)⊤=n−1/2i=1∑n(Zi2,…,Zik)⊤ 其中
Z
i
j
=
(
X
i
−
μ
)
j
−
μ
j
−
j
μ
j
−
1
(
X
i
−
μ
)
Z_{i j}=\left(X_{i}-\mu\right)^{j}-\mu_{j}-j \mu_{j-1}\left(X_{i}-\mu\right)
Zij=(Xi−μ)j−μj−jμj−1(Xi−μ) ,
j
=
2
,
…
,
k
j=2, \ldots, k
j=2,…,k
计算可得：
E
(
Z
i
j
)
=
0
\mathrm{E}\left(Z_{i j}\right)=0
E(Zij)=0
Cov
(
Z
i
j
1
,
Z
i
j
2
)
=
μ
j
1
+
j
2
−
μ
j
1
μ
j
2
−
j
1
μ
j
1
−
1
μ
j
2
+
1
−
j
2
μ
j
2
−
1
μ
j
1
+
1
+
j
1
j
2
μ
j
1
−
1
μ
j
2
−
1
μ
2
\operatorname{Cov}\left(Z_{i j_{1}}, Z_{i j_{2}}\right)=\mu_{j_{1}+j_{2}}-\mu_{j_{1}} \mu_{j_{2}}-j_{1} \mu_{j_{1}-1} \mu_{j_{2}+1}-j_{2} \mu_{j_{2}-1} \mu_{j_{1}+1}+j_{1} j_{2} \mu_{j_{1}-1} \mu_{j_{2}-1} \mu_{2}
Cov(Zij1,Zij2)=μj1+j2−μj1μj2−j1μj1−1μj2+1−j2μj2−1μj1+1+j1j2μj1−1μj2−1μ2
综上则(2)成立，证毕。