正态总体的样本均值与样本方差的分布定理

文章目录

- - 引理
  - 定理一
  - 定理二
  - 定理三
  - 定理四

引理

设总体

X

$X$

(不管服从什么分布，只要均值和方差存在)的均值为

\mu

$μ$

，方差为

2

\sigma^2

$σ^{2}$

，

1

,

X

2

,

⋯

,

X

n

X_1,X_2,\cdots,X_n

$X_{1}, X_{2}, \dots, X_{n}$

是来自

X

$X$

的一个样本，

‾

,

S

2

\overline{X},S^2

$\overline{X}, S^{2}$

分别是样本均值和样本方差，则有

(

X

‾

)

=

μ

,

D

(

X

‾

)

=

σ

2

/

n

,

E

(

S

2

)

=

σ

2

E(\overline{X})=\mu,\quad D(\overline{X})=\sigma^2/n, \quad E(S^2)=\sigma^2

$E (\overline{X}) = μ, D (\overline{X}) = σ^{2} / n, E (S^{2}) = σ^{2}$
正态分布线性可加性: 若

i

∼

N

(

μ

i

,

σ

i

2

)

,

i

=

1

,

2

,

⋯

,

n

X_i\sim N(\mu_i,\sigma_i^2),i=1,2,\cdots,n

$X_{i} \sim N (μ_{i}, σ_{i}^{2}), i = 1, 2, \dots, n$

，且他们相互独立，则他们的线性组合:

1

X

1

+

C

2

X

2

+

⋯

+

C

n

X

n

C_1X_1+C_2X_2+\cdots+C_nX_n

$C_{1} X_{1} + C_{2} X_{2} + \dots + C_{n} X_{n}$

，(

1

,

C

2

,

⋯

,

C

n

C_1,C_2,\cdots,C_n

$C_{1}, C_{2}, \dots, C_{n}$

是不全为

0

$0$

的常数)仍然服从正态分布，且有

1

X

1

+

C

2

X

2

+

⋯

+

C

n

X

n

∼

N

(

∑

i

=

1

n

C

i

μ

i

,

∑

i

=

1

n

C

i

2

σ

i

2

)

C_1X_1+C_2X_2+\cdots+C_nX_n\sim N(\sum\limits_{i=1}^nC_i\mu_i,\sum\limits_{i=1}^nC_i^2\sigma_i^2)

$C_{1} X_{1} + C_{2} X_{2} + \dots + C_{n} X_{n} \sim N (i = 1 \sum n C_{i} μ_{i}, i = 1 \sum n C_{i}^{2} σ_{i}^{2})$
n

$n$

维正态随机变量重要性质

0

1^0\quad

$1^{0}$

n

$n$

维正态随机变量

X

1

,

X

2

,

⋯

,

X

n

)

(X_1,X_2,\cdots,X_n)

$(X_{1}, X_{2}, \dots, X_{n})$

的每一个分量

i

,

i

=

1

,

2

,

⋯

,

n

X_i,i=1,2,\cdots,n

$X_{i}, i = 1, 2, \dots, n$

都是正态随机变量，反之，若

1

,

X

2

,

⋯

,

X

n

X_1,X_2,\cdots,X_n

$X_{1}, X_{2}, \dots, X_{n}$

都是正态随机变量，且相互独立，则

X

1

,

X

2

,

⋯

,

X

n

)

(X_1,X_2,\cdots,X_n)

$(X_{1}, X_{2}, \dots, X_{n})$

是

n

$n$

维正态随机变量

0

2^0\quad

$2^{0}$

n

$n$

维随机变量

X

1

,

X

2

,

⋯

,

X

n

)

(X_1,X_2,\cdots,X_n)

$(X_{1}, X_{2}, \dots, X_{n})$

服从

n

$n$

维正态分布的充要条件是

1

,

X

2

,

⋯

,

X

n

X_1,X_2,\cdots,X_n

$X_{1}, X_{2}, \dots, X_{n}$

的任意线性组合

1

X

1

+

l

2

X

2

+

⋯

+

l

n

X

n

l_1X_1+l_2X_2+\cdots+l_nX_n

$l_{1} X_{1} + l_{2} X_{2} + \dots + l_{n} X_{n}$

服从一维正态分布(其中

1

,

l

2

,

⋯

,

l

n

l_1,l_2,\cdots,l_n

$l_{1}, l_{2}, \dots, l_{n}$

不全为零)

0

3^0\quad

$3^{0}$

若

X

1

,

X

2

,

⋯

,

X

n

)

(X_1,X_2,\cdots,X_n)

$(X_{1}, X_{2}, \dots, X_{n})$

服从

n

$n$

维正态分布，设

1

,

Y

2

,

⋯

,

Y

k

Y_1,Y_2,\cdots,Y_k

$Y_{1}, Y_{2}, \dots, Y_{k}$

是

j

(

j

=

1

,

2

,

⋯

,

n

)

X_j(j=1,2,\cdots,n)

$X_{j} (j = 1, 2, \dots, n)$

的线性函数，则

Y

1

,

Y

2

,

⋯

,

Y

k

)

(Y_1,Y_2,\cdots,Y_k)

$(Y_{1}, Y_{2}, \dots, Y_{k})$

也服从多维正态分布，这一性质称为正态变量的线性变换不变性

0

4^0\quad

$4^{0}$

设

X

1

,

X

2

,

⋯

,

X

n

)

(X_1,X_2,\cdots,X_n)

$(X_{1}, X_{2}, \dots, X_{n})$

服从

n

$n$

维正态分布，则”

1

,

X

2

,

⋯

,

X

n

X_1,X_2,\cdots,X_n

$X_{1}, X_{2}, \dots, X_{n}$

相互独立”与”

1

,

X

2

,

⋯

,

X

n

X_1,X_2,\cdots,X_n

$X_{1}, X_{2}, \dots, X_{n}$

两两不相关“是等价的。

定理一

设

1

,

X

2

,

⋯

,

X

n

X_1,X_2,\cdots,X_n

$X_{1}, X_{2}, \dots, X_{n}$

是来自正态总体

(

μ

,

σ

2

)

N(\mu,\sigma^2)

$N (μ, σ^{2})$

的样本，

‾

\overline{X}

$\overline{X}$

是样本均值，则有

‾

∼

N

(

μ

,

σ

2

/

n

)

.

\overline{X}\sim N(\mu,\sigma^2/n).

$\overline{X} \sim N (μ, σ^{2} / n) .$

证明很简单，由引言1可知，

(

X

‾

)

=

μ

,

D

(

X

‾

)

=

σ

2

/

n

E(\overline{X})=\mu,\quad D(\overline{X})=\sigma^2/n

$E (\overline{X}) = μ, D (\overline{X}) = σ^{2} / n$

而

‾

=

1

n

∑

i

=

1

n

X

i

\overline{X}=\frac{1}{n}\sum\limits_{i=1}^nX_i

$\overline{X} = \frac{1}{n} i = 1 \sum n X_{i}$

，

i

X_i

$X_{i}$

服从正态分布，则根据引理2可知，

‾

∼

N

(

μ

,

σ

2

/

n

)

.

\overline{X}\sim N(\mu,\sigma^2/n).

$\overline{X} \sim N (μ, σ^{2} / n) .$

定理二

设

1

,

X

2

,

⋯

,

X

n

X_1,X_2,\cdots,X_n

$X_{1}, X_{2}, \dots, X_{n}$

是来自正态总体

(

μ

,

σ

2

)

N(\mu,\sigma^2)

$N (μ, σ^{2})$

的样本，

‾

,

S

2

\overline{X},S^2

$\overline{X}, S^{2}$

分别是样本均值和样本方差，则有

0

(

n

−

1

)

S

2

σ

2

∼

χ

2

(

n

−

1

)

1^0\quad \frac{(n-1)S^2}{\sigma^2}\sim \chi^2(n-1)

$1^{0} \frac{( n - 1 ) S ^{2}}{σ ^{2}} \sim χ^{2} (n - 1)$

0

X

‾

2^0\quad \overline{X}

$2^{0} \overline{X}$

和

2

S^2

$S^{2}$

相互独立

证明:

n

−

1

)

S

2

σ

2

=

(

n

−

1

)

σ

2

×

1

(

n

−

1

)

∑

i

=

1

n

(

X

i

−

X

‾

)

2

=

∑

i

=

1

n

(

X

i

−

X

‾

)

2

σ

2

=

∑

i

=

1

n

[

(

X

i

−

μ

)

−

(

X

‾

−

μ

)

]

2

σ

2

=

∑

i

=

1

n

(

X

i

−

μ

σ

−

X

‾

−

μ

σ

)

2

\begin{aligned} \frac{(n-1)S^2}{\sigma^2} &= \frac{(n-1)}{\sigma^2}\times \frac{1}{(n-1)}\sum\limits_{i=1}^n(X_i-\overline{X})^2 \\&= \frac{\sum\limits_{i=1}^n(X_i-\overline{X})^2}{\sigma^2}\\&=\frac{\sum\limits_{i=1}^n[(X_i-\mu)-(\overline{X}-\mu)]^2}{\sigma^2}\\&= \sum\limits_{i=1}^n\bigg(\frac{X_i-\mu}{\sigma}-\frac{\overline{X}-\mu}{\sigma}\bigg)^2\end{aligned}

$\frac{( n - 1 ) S ^{2}}{σ ^{2}} = \frac{( n - 1 )}{σ ^{2}} \times \frac{1}{( n - 1 )} i = 1 \sum n (X_{i} - \overline{X})^{2} = \frac{i = 1 \sum n ( X _{i} - X ) ^{2}}{σ ^{2}} = \frac{i = 1 \sum n [ ( X _{i} - μ ) - ( X - μ ) ] ^{2}}{σ ^{2}} = i = 1 \sum n (\frac{X _{i} - μ}{σ} - \frac{X - μ}{σ})^{2}$

为了方便，我们令

i

=

X

i

−

μ

σ

Z_i=\frac{X_i-\mu}{\sigma}

$Z_{i} = \frac{X _{i} - μ}{σ}$

，由于

i

∼

N

(

μ

,

σ

2

)

X_i\sim N(\mu,\sigma^2)

$X_{i} \sim N (μ, σ^{2})$

，因此

i

∼

N

(

0

,

1

)

Z_i\sim N(0,1)

$Z_{i} \sim N (0, 1)$

且

‾

=

X

‾

−

μ

σ

\overline{Z} = \frac{\overline{X}-\mu}{\sigma}

$\overline{Z} = \frac{X - μ}{σ}$

，则

n

−

1

)

S

2

σ

2

=

∑

i

=

1

n

(

Z

i

−

Z

‾

)

2

=

∑

i

=

1

n

(

Z

i

2

−

2

Z

i

Z

‾

+

Z

‾

2

)

=

∑

i

=

1

n

Z

i

2

−

2

Z

‾

∑

i

=

1

n

Z

i

+

∑

i

=

1

n

Z

‾

2

=

∑

i

=

1

n

Z

i

2

−

2

n

Z

‾

2

+

n

Z

‾

2

=

∑

i

=

1

n

Z

i

2

−

n

Z

‾

2

\begin{aligned} \frac{(n-1)S^2}{\sigma^2} &= \sum\limits_{i=1}^n(Z_i-\overline{Z})^2 \\&= \sum\limits_{i=1}^n(Z_i^2-2Z_i\overline{Z}+\overline{Z}^2) \\&= \sum\limits_{i=1}^nZ_i^2-2\overline{Z}\sum\limits_{i=1}^nZ_i+\sum\limits_{i=1}^n\overline{Z}^2 \\&= \sum\limits_{i=1}^nZ_i^2-2n\overline{Z}^2+n\overline{Z}^2\\&=\sum\limits_{i=1}^nZ_i^2-n\overline{Z}^2 \end{aligned}

$\frac{( n - 1 ) S ^{2}}{σ ^{2}} = i = 1 \sum n (Z_{i} - \overline{Z})^{2} = i = 1 \sum n (Z_{i}^{2} - 2 Z_{i} \overline{Z} + \overline{Z}^{2}) = i = 1 \sum n Z_{i}^{2} - 2 \overline{Z} i = 1 \sum n Z_{i} + i = 1 \sum n \overline{Z}^{2} = i = 1 \sum n Z_{i}^{2} - 2 n \overline{Z}^{2} + n \overline{Z}^{2} = i = 1 \sum n Z_{i}^{2} - n \overline{Z}^{2}$

取一个

n

$n$

阶正交矩阵

=

(

a

i

j

)

A=(a_{ij})

$A = (a_{i j})$

，其第一行元素均为

/

n

1/\sqrt{n}

$1 / n$

=

[

1

/

n

1

/

n

⋯

1

/

n

a

21

a

22

⋯

a

2

n

⋮

⋮

⋮

⋮

a

n

1

a

n

2

⋯

a

n

n

]

A = \begin{bmatrix}1/\sqrt{n} & 1/\sqrt{n} & \cdots & 1/\sqrt{n} \\ a_{21} & a_{22} & \cdots &a_{2n}\\ \vdots & \vdots & \vdots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn}\end{bmatrix}

$A = ⎣ ⎢ ⎢ ⎢ ⎡ 1 / n a_{21} ⋮ a_{n 1} 1 / n a_{22} ⋮ a_{n 2} \dots \dots ⋮ \dots 1 / n a_{2 n} ⋮ a_{n n} ⎦ ⎥ ⎥ ⎥ ⎤$

对

A

$A$

做正交变换

=

A

Z

Y=AZ

$Y = A Z$

，有

=

[

Y

1

Y

2

⋮

Y

n

]

,

Z

=

[

Z

1

Z

2

⋮

Z

n

]

Y = \begin{bmatrix}Y_1 \\Y_2\\ \vdots \\Y_n \end{bmatrix},Z = \begin{bmatrix}Z_1 \\Z_2\\ \vdots \\Z_n \end{bmatrix}

$Y = ⎣ ⎢ ⎢ ⎢ ⎡ Y_{1} Y_{2} ⋮ Y_{n} ⎦ ⎥ ⎥ ⎥ ⎤, Z = ⎣ ⎢ ⎢ ⎢ ⎡ Z_{1} Z_{2} ⋮ Z_{n} ⎦ ⎥ ⎥ ⎥ ⎤$

由于

i

=

∑

j

=

1

n

a

i

j

Z

j

,

i

=

1

,

2

,

⋯

,

n

Y_i=\sum\limits_{j=1}^na_{ij}Z_j,\quad i=1,2,\cdots ,n

$Y_{i} = j = 1 \sum n a_{i j} Z_{j}, i = 1, 2, \dots, n$

，因此

i

Y_i

$Y_{i}$

仍服从正态分布，由

i

∼

N

(

0

,

1

)

Z_i \sim N(0,1)

$Z_{i} \sim N (0, 1)$

可知

(

Y

i

)

=

E

(

∑

j

=

1

n

a

i

j

Z

j

)

=

∑

j

=

1

n

a

i

j

E

(

Z

j

)

=

0

E(Y_i) = E(\sum\limits_{j=1}^na_{ij}Z_j) = \sum\limits_{j=1}^na_{ij}E(Z_j) = 0

$E (Y_{i}) = E (j = 1 \sum n a_{i j} Z_{j}) = j = 1 \sum n a_{i j} E (Z_{j}) = 0$

(

Y

i

)

=

D

(

∑

j

=

1

n

a

i

j

Z

j

)

=

∑

j

=

1

n

a

i

j

2

D

(

Z

j

)

=

∑

j

=

1

n

a

i

j

2

=

1

D(Y_i) = D(\sum\limits_{j=1}^na_{ij}Z_j) = \sum\limits_{j=1}^na_{ij}^2D(Z_j) = \sum\limits_{j=1}^na_{ij}^2=1

$D (Y_{i}) = D (j = 1 \sum n a_{i j} Z_{j}) = j = 1 \sum n a_{i j}^{2} D (Z_{j}) = j = 1 \sum n a_{i j}^{2} = 1$

又由

o

v

(

Z

i

,

Z

j

)

=

δ

i

j

=

{

0

,

i

≠

j

1

,

i

=

j

,

i

,

j

=

1

,

2

,

⋯

,

n

Cov(Z_i,Z_j) = \delta_{ij}=\begin{cases}0,\quad i\neq j \\1,\quad i=j\end{cases} \quad ,i,j=1,2,\cdots,n

$C o v (Z_{i}, Z_{j}) = δ_{i j} = {0, i \neq = j 1, i = j, i, j = 1, 2, \dots, n$

o

v

(

Y

i

,

Y

k

)

=

C

o

v

(

∑

j

=

1

n

a

i

j

Z

j

,

∑

l

=

1

n

a

k

l

Z

l

)

=

∑

j

=

1

n

∑

l

=

1

n

a

i

j

a

k

l

C

o

v

(

Z

j

,

Z

l

)

=

∑

j

=

1

n

a

i

j

a

k

j

=

{

0

,

i

≠

j

1

,

i

=

j

(

正

交

矩

阵

性

质

，

各

行

均

是

单

位

向

量

且

两

两

正

交

)

\begin{aligned}Cov(Y_i,Y_k) &= Cov(\sum\limits_{j=1}^na_{ij}Z_j,\sum\limits_{l=1}^na_{kl}Z_l)\\&=\sum\limits_{j=1}^n\sum\limits_{l=1}^n a_{ij}a_{kl}Cov(Z_j,Z_l) \\&=\sum\limits_{j=1}^na_{ij}a_{kj} \\&=\begin{cases}0,\quad i\neq j \\1,\quad i=j\end{cases}(正交矩阵性质，各行均是单位向量且两两正交)\end{aligned}

$C o v (Y_{i}, Y_{k}) = C o v (j = 1 \sum n a_{i j} Z_{j}, l = 1 \sum n a_{k l} Z_{l}) = j = 1 \sum n l = 1 \sum n a_{i j} a_{k l} C o v (Z_{j}, Z_{l}) = j = 1 \sum n a_{i j} a_{k j} = {0, i \neq = j 1, i = j (正交矩阵性质，各行均是单位向量且两两正交)$

由此可知

1

,

Y

2

,

⋯

,

Y

n

Y_1,Y_2,\cdots,Y_n

$Y_{1}, Y_{2}, \dots, Y_{n}$

两两互不相关。又由于

n

$n$

维随机变量

Y

1

,

Y

2

,

⋯

,

Y

n

)

(Y_1,Y_2,\cdots,Y_n)

$(Y_{1}, Y_{2}, \dots, Y_{n})$

是由

n

$n$

维随机变量

X

1

,

X

2

，

⋯

,

X

n

)

(X_1,X_2，\cdots,X_n)

$(X_{1}, X_{2} ， \dots, X_{n})$

经线性变换得到，因此

Y

1

,

Y

2

,

⋯

,

Y

n

)

(Y_1,Y_2,\cdots,Y_n)

$(Y_{1}, Y_{2}, \dots, Y_{n})$

也是

n

$n$

维正态随机变量，由

引理3性质4

可知，

1

,

Y

2

,

⋯

,

Y

n

Y_1,Y_2,\cdots,Y_n

$Y_{1}, Y_{2}, \dots, Y_{n}$

两两互不相关也即是

1

,

Y

2

,

⋯

,

Y

n

Y_1,Y_2,\cdots,Y_n

$Y_{1}, Y_{2}, \dots, Y_{n}$

互相独立。前面已经计算出

(

Y

i

)

=

0

,

D

(

Y

i

)

=

1

E(Y_i)=0,D(Y_i)=1

$E (Y_{i}) = 0, D (Y_{i}) = 1$

因此

i

∼

N

(

0

,

1

)

,

i

=

1

,

2

,

⋯

,

n

.

Y_i\sim N(0,1),i=1,2,\cdots,n.

$Y_{i} \sim N (0, 1), i = 1, 2, \dots, n .$

1

=

∑

j

=

1

n

a

1

j

Z

j

=

∑

j

=

1

n

1

n

Z

j

=

1

n

∗

n

Z

‾

=

n

Z

‾

\begin{aligned}Y_1&=\sum\limits_{j=1}^na_{1j}Z_j \\&= \sum\limits_{j=1}^n\frac{1}{\sqrt{n}}Z_j\\&=\frac{1}{\sqrt{n}}*n\overline{Z}\\&=\sqrt{n}\overline{Z}\end{aligned}

$Y_{1} = j = 1 \sum n a_{1 j} Z_{j} = j = 1 \sum n \frac{1}{n} Z_{j} = \frac{1}{n} * n \overline{Z} = n \overline{Z}$

i

=

1

n

Y

i

2

=

Y

T

Y

=

(

A

Z

)

T

(

A

Z

)

=

Z

T

A

T

A

Z

=

Z

T

Z

=

∑

i

=

1

n

Z

i

2

\begin{aligned}\sum\limits_{i=1}^nY_i^2&=Y^TY=(AZ)^T(AZ)\\&=Z^TA^TAZ = Z^TZ = \sum\limits_{i=1}^nZ_i^2\end{aligned}

$i = 1 \sum n Y_{i}^{2} = Y^{T} Y = (A Z)^{T} (A Z) = Z^{T} A^{T} A Z = Z^{T} Z = i = 1 \sum n Z_{i}^{2}$

此时有

n

−

1

)

S

2

σ

2

=

∑

i

=

1

n

Z

i

2

−

n

Z

‾

2

=

∑

i

=

1

n

Y

i

2

−

Y

1

2

=

∑

i

=

2

n

Y

i

2

\begin{aligned} \frac{(n-1)S^2}{\sigma^2} &=\sum\limits_{i=1}^nZ_i^2-n\overline{Z}^2 \\&=\sum\limits_{i=1}^nY_i^2-Y_1^2\\&=\sum\limits_{i=2}^nY_i^2\end{aligned}

$\frac{( n - 1 ) S ^{2}}{σ ^{2}} = i = 1 \sum n Z_{i}^{2} - n \overline{Z}^{2} = i = 1 \sum n Y_{i}^{2} - Y_{1}^{2} = i = 2 \sum n Y_{i}^{2}$

由于

2

,

Y

3

,

⋯

,

Y

n

Y_2,Y_3,\cdots,Y_n

$Y_{2}, Y_{3}, \dots, Y_{n}$

相互独立，且

i

∼

N

(

0

,

1

)

Y_i\sim N(0,1)

$Y_{i} \sim N (0, 1)$

，因此

n

−

1

)

S

2

σ

2

=

∑

i

=

2

n

Y

i

2

∼

χ

2

(

n

−

1

)

.

\frac{(n-1)S^2}{\sigma^2}=\sum\limits_{i=2}^nY_i^2\sim\chi^2(n-1).

$\frac{( n - 1 ) S ^{2}}{σ ^{2}} = i = 2 \sum n Y_{i}^{2} \sim χ^{2} (n - 1) .$

其次，

‾

=

σ

Z

‾

+

μ

=

σ

Y

1

n

+

μ

\overline{X} = \sigma\overline{Z}+\mu = \frac{\sigma Y_1}{\sqrt{n}}+\mu

$\overline{X} = σ \overline{Z} + μ = \frac{σ Y _{1}}{n} + μ$

仅跟

1

Y_1

$Y_{1}$

有关，而

2

=

σ

2

n

−

1

∑

i

=

2

n

Y

i

2

S^2=\frac{\sigma^2}{n-1}\sum\limits_{i=2}^nY_i^2

$S^{2} = \frac{σ ^{2}}{n - 1} i = 2 \sum n Y_{i}^{2}$

仅依赖于

2

,

Y

3

,

⋯

,

Y

n

Y_2,Y_3,\cdots,Y_n

$Y_{2}, Y_{3}, \dots, Y_{n}$

，又因为

1

,

Y

2

,

⋯

,

Y

n

Y_1,Y_2,\cdots,Y_n

$Y_{1}, Y_{2}, \dots, Y_{n}$

相互独立，因此有

‾

\overline{X}

$\overline{X}$

和

2

S^2

$S^{2}$

相互独立

定理三

设

1

,

X

2

,

⋯

,

X

n

X_1,X_2,\cdots,X_n

$X_{1}, X_{2}, \dots, X_{n}$

是来自正态总体

(

μ

,

σ

2

)

N(\mu,\sigma^2)

$N (μ, σ^{2})$

的样本，

‾

,

S

2

\overline{X},S^2

$\overline{X}, S^{2}$

分别是样本均值和样本方差，则有

‾

−

μ

S

/

n

∼

t

(

n

−

1

)

\frac{\overline{X}-\mu}{S/\sqrt{n}}\sim t(n-1)

$\frac{X - μ}{S / n} \sim t (n - 1)$

证明：

由定理一可知，

‾

∼

N

(

μ

,

σ

2

/

n

)

.

\overline{X}\sim N(\mu,\sigma^2/n).

$\overline{X} \sim N (μ, σ^{2} / n) .$

进行标准化之后有

‾

−

μ

σ

/

n

∼

N

(

0

,

1

)

.

\frac{\overline{X}-\mu}{\sigma /\sqrt{n}}\sim N(0,1).

$\frac{X - μ}{σ / n} \sim N (0, 1) .$

由定理二可知，

n

−

1

)

S

2

σ

2

∼

χ

2

(

n

−

1

)

\frac{(n-1)S^2}{\sigma^2}\sim \chi^2(n-1)

$\frac{( n - 1 ) S ^{2}}{σ ^{2}} \sim χ^{2} (n - 1)$

根据

t

$t$

分布定义有

‾

−

μ

σ

/

n

(

n

−

1

)

S

2

σ

2

(

n

−

1

)

=

X

‾

−

μ

S

/

n

∼

t

(

n

−

1

)

\begin{aligned} \frac{\frac{\overline{X}-\mu}{\sigma/ \sqrt{n}}}{\sqrt{\frac{(n-1)S^2}{\sigma^2(n-1)}}} = \frac{\overline{X}-\mu}{S/\sqrt{n}} \sim t(n-1)\end{aligned}

$\frac{\frac{X - μ}{σ / n}}{\frac{( n - 1 ) S ^{2}}{σ ^{2} ( n - 1 )}} = \frac{X - μ}{S / n} \sim t (n - 1)$

证明完毕

定理四

设

1

,

X

2

,

⋯

,

X

n

1

X_1,X_2,\cdots,X_{n_1}

$X_{1}, X_{2}, \dots, X_{n_{1}}$

与

1

,

Y

2

,

⋯

,

Y

n

1

Y_1,Y_2,\cdots,Y_{n_1}

$Y_{1}, Y_{2}, \dots, Y_{n_{1}}$

是来自正态总体

(

μ

1

,

σ

1

2

)

N(\mu_1,\sigma_1^2)

$N (μ_{1}, σ_{1}^{2})$

和

(

μ

2

,

σ

2

2

)

N(\mu_2,\sigma_2^2)

$N (μ_{2}, σ_{2}^{2})$

的样本，且这两个样本相互独立. 设

‾

=

1

n

1

∑

i

=

1

n

1

X

i

,

Y

‾

=

1

n

2

∑

i

=

1

n

2

Y

i

\overline{X}=\frac{1}{n_1}\sum\limits_{i=1}^{n_1}X_i,\overline{Y}=\frac{1}{n_2}\sum\limits_{i=1}^{n_2}Y_i

$\overline{X} = \frac{1}{n _{1}} i = 1 \sum n_{1} X_{i}, \overline{Y} = \frac{1}{n _{2}} i = 1 \sum n_{2} Y_{i}$

分别是这两个样本的样本均值；

1

2

=

1

n

1

−

1

∑

i

=

1

n

1

(

X

i

−

X

‾

)

2

,

S

2

2

=

1

n

2

−

1

∑

i

=

1

n

2

(

Y

i

−

Y

‾

)

2

S_1^2=\frac{1}{n_1-1}\sum\limits_{i=1}^{n_1}(X_i-\overline{X})^2,S_2^2=\frac{1}{n_2-1}\sum\limits_{i=1}^{n_2}(Y_i-\overline{Y})^2

$S_{1}^{2} = \frac{1}{n _{1} - 1} i = 1 \sum n_{1} (X_{i} - \overline{X})^{2}, S_{2}^{2} = \frac{1}{n _{2} - 1} i = 1 \sum n_{2} (Y_{i} - \overline{Y})^{2}$

分别是两个样本的样本方差，则有

0

S

1

2

/

S

2

2

σ

1

2

/

σ

2

2

∼

F

(

n

1

−

1

,

n

2

−

1

)

1^0\quad \frac{S_1^2/S_2^2}{\sigma_1^2/\sigma_2^2}\sim F(n_1-1,n_2-1)

$1^{0} \frac{S _{1}^{2} / S _{2}^{2}}{σ _{1}^{2} / σ _{2}^{2}} \sim F (n_{1} - 1, n_{2} - 1)$

0

2^0\quad

$2^{0}$

当

1

2

=

σ

2

2

=

σ

2

\sigma_1^2=\sigma_2^2=\sigma^2

$σ_{1}^{2} = σ_{2}^{2} = σ^{2}$

时，

X

‾

−

Y

‾

)

−

(

μ

1

−

μ

2

)

S

w

1

n

1

+

1

n

2

∼

t

(

n

1

+

n

2

−

2

)

\frac{(\overline{X}-\overline{Y})-(\mu_1-\mu_2)}{S_w\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\sim t(n_1+n_2-2)

$\frac{( X - Y ) - ( μ _{1} - μ _{2} )}{S _{w} \frac{1}{n _{1}} + \frac{1}{n _{2}}} \sim t (n_{1} + n_{2} - 2)$

其中

w

2

=

(

n

1

−

1

)

S

1

2

+

(

n

2

−

1

)

S

2

2

n

1

+

n

2

−

2

,

S

w

=

S

w

2

S_w^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}, S_w=\sqrt{S_w^2}

$S_{w}^{2} = \frac{( n _{1} - 1 ) S _{1}^{2} + ( n _{2} - 1 ) S _{2}^{2}}{n _{1} + n _{2} - 2}, S_{w} = S_{w}^{2}$

证明:

0

1^0 \quad

$1^{0}$

由定理二可知

n

1

−

1

)

S

1

2

σ

1

2

∼

χ

2

(

n

1

−

1

)

(

1

)

(

n

2

−

1

)

S

2

2

σ

2

2

∼

χ

2

(

n

2

−

1

)

(

2

)

\frac{(n_1-1)S_1^2}{\sigma_1^2}\sim \chi^2(n_1-1) \quad(1) \\\frac{(n_2-1)S_2^2}{\sigma_2^2}\sim \chi^2(n_2-1)\quad(2)

$\frac{( n _{1} - 1 ) S _{1}^{2}}{σ _{1}^{2}} \sim χ^{2} (n_{1} - 1) (1) \frac{( n _{2} - 1 ) S _{2}^{2}}{σ _{2}^{2}} \sim χ^{2} (n_{2} - 1) (2)$

此时有

n

1

−

1

)

S

1

2

(

n

1

−

1

)

σ

1

2

/

(

n

2

−

1

)

S

2

2

(

n

2

−

1

)

σ

2

2

∼

F

(

n

1

−

1

,

n

2

−

1

)

\frac{(n_1-1)S_1^2}{(n_1-1)\sigma_1^2}\bigg/\frac{(n_2-1)S_2^2}{(n_2-1)\sigma_2^2} \sim F(n_1-1,n_2-1)

$\frac{( n _{1} - 1 ) S _{1}^{2}}{( n _{1} - 1 ) σ _{1}^{2}} / \frac{( n _{2} - 1 ) S _{2}^{2}}{( n _{2} - 1 ) σ _{2}^{2}} \sim F (n_{1} - 1, n_{2} - 1)$

即

1

2

/

S

2

2

σ

1

2

/

σ

2

2

∼

F

(

n

1

−

1

,

n

2

−

1

)

\frac{S_1^2/S_2^2}{\sigma_1^2/\sigma_2^2}\sim F(n_1-1,n_2-1)

$\frac{S _{1}^{2} / S _{2}^{2}}{σ _{1}^{2} / σ _{2}^{2}} \sim F (n_{1} - 1, n_{2} - 1)$

0

2^0\quad

$2^{0}$

由式

1

)

(1)

$(1)$

和

2

)

(2)

$(2)$

，以及

2

\chi^2

$χ^{2}$

分布的可加性可知

n

1

−

1

)

S

1

2

σ

1

2

+

(

n

2

−

1

)

S

2

2

σ

2

2

服

从

χ

2

(

n

1

+

n

2

−

2

)

\frac{(n_1-1)S_1^2}{\sigma_1^2}+\frac{(n_2-1)S_2^2}{\sigma_2^2} 服从\chi^2(n_1+n_2-2)

$\frac{( n _{1} - 1 ) S _{1}^{2}}{σ _{1}^{2}} + \frac{( n _{2} - 1 ) S _{2}^{2}}{σ _{2}^{2}} 服从 χ^{2} (n_{1} + n_{2} - 2)$

由

1

2

=

σ

2

2

=

σ

2

\sigma_1^2=\sigma_2^2=\sigma^2

$σ_{1}^{2} = σ_{2}^{2} = σ^{2}$

可知，

n

1

−

1

)

S

1

2

+

(

n

2

−

1

)

S

2

2

σ

2

∼

χ

2

(

n

1

+

n

2

−

2

)

(

3

)

\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{\sigma^2} \sim \chi^2(n_1+n_2-2)\quad (3)

$\frac{( n _{1} - 1 ) S _{1}^{2} + ( n _{2} - 1 ) S _{2}^{2}}{σ ^{2}} \sim χ^{2} (n_{1} + n_{2} - 2) (3)$

由定理一可知

‾

∼

N

(

μ

1

,

σ

2

/

n

)

,

Y

‾

∼

N

(

μ

2

,

σ

2

/

n

)

.

\overline{X}\sim N(\mu_1,\sigma^2/n),\overline{Y}\sim N(\mu_2,\sigma^2/n).

$\overline{X} \sim N (μ_{1}, σ^{2} / n), \overline{Y} \sim N (μ_{2}, σ^{2} / n) .$

因此

‾

−

Y

‾

∼

N

(

μ

1

−

μ

2

,

σ

2

/

n

1

+

σ

2

/

n

2

)

\overline{X}-\overline{Y}\sim N(\mu_1-\mu_2,\sigma^2/n_1+\sigma^2/n_2)

$\overline{X} - \overline{Y} \sim N (μ_{1} - μ_{2}, σ^{2} / n_{1} + σ^{2} / n_{2})$

，对其进行标准化有

X

‾

−

Y

‾

)

−

(

μ

1

−

μ

2

)

σ

2

/

n

1

+

σ

2

/

n

2

∼

N

(

0

,

1

)

(

4

)

\frac{(\overline{X}-\overline{Y})-(\mu_1-\mu_2)}{\sqrt{\sigma^2/n_1+\sigma^2/n_2}} \sim N(0,1)\quad (4)

$\frac{( X - Y ) - ( μ _{1} - μ _{2} )}{σ ^{2} / n _{1} + σ ^{2} / n _{2}} \sim N (0, 1) (4)$

由式

3

)

、

(

4

)

(3)、(4)

$(3) 、 (4)$

可知

X

‾

−

Y

‾

)

−

(

μ

1

−

μ

2

)

σ

2

/

n

1

+

σ

2

/

n

2

/

(

n

1

−

1

)

S

1

2

+

(

n

2

−

1

)

S

2

2

σ

2

(

n

1

+

n

2

−

2

)

∼

t

(

n

1

+

n

2

−

2

)

\frac{(\overline{X}-\overline{Y})-(\mu_1-\mu_2)}{\sqrt{\sigma^2/n_1+\sigma^2/n_2}}\bigg/\sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{\sigma^2(n_1+n_2-2)}} \sim t(n_1+n_2-2)

$\frac{( X - Y ) - ( μ _{1} - μ _{2} )}{σ ^{2} / n _{1} + σ ^{2} / n _{2}} / \frac{( n _{1} - 1 ) S _{1}^{2} + ( n _{2} - 1 ) S _{2}^{2}}{σ ^{2} ( n _{1} + n _{2} - 2 )} \sim t (n_{1} + n_{2} - 2)$

即

X

‾

−

Y

‾

)

−

(

μ

1

−

μ

2

)

S

w

1

n

1

+

1

n

2

∼

t

(

n

1

+

n

2

−

2

)

\frac{(\overline{X}-\overline{Y})-(\mu_1-\mu_2)}{S_w\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\sim t(n_1+n_2-2)

$\frac{( X - Y ) - ( μ _{1} - μ _{2} )}{S _{w} \frac{1}{n _{1}} + \frac{1}{n _{2}}} \sim t (n_{1} + n_{2} - 2)$

证明完毕

原文链接：https://blog.csdn.net/SpiritedAway1106/article/details/108608994

文章目录

引理

定理一

定理二

定理三

定理四

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