數理統計基礎 統計量

坐上蝸牛去地球發表於2020-11-01

目錄

https://blog.csdn.net/weixin_45792450/article/details/109314584

統計量由來與定義

如前所述，在進行統計推斷時，構造樣本的適當函式是關鍵，一個好的表示式，可以更方便研究總體的未知分佈及相關性質。

為此，我們給出下列定義.

定義：設 ( X 1 , X 2 , . . . , X n ) ({X_1},{X_2},...,{X_n}) (X1​,X2​,...,Xn​)為總體 X {X} X的一個樣本，稱此樣本的任一不含總體分佈未知引數的函式為該樣本的統計量。

舉例說明：設總體 X {X} X服從正態分佈， E ( X ) = 5 {E(X)=5} E(X)=5, D ( X ) = σ 2 D(X) = {\sigma ^2} D(X)=σ2, σ {\sigma} σ未知。 ( X 1 , X 2 , . . . , X n ) ({X_1},{X_2},...,{X_n}) (X1​,X2​,...,Xn​)為總體 X {X} X的一個樣本，令：
X ˉ = X 1 + X 2 + . . . + X n n \bar X = {{{X_1} + {X_2} + ... + {X_n}} \over n} Xˉ=nX1​+X2​+...+Xn​​
U = X 1 + X 2 + . . . + X n n σ U = {{{X_1} + {X_2} + ... + {X_n}} \over {n\sigma }} U=nσX1​+X2​+...+Xn​​
那麼 X ˉ \bar X Xˉ為該樣本的統計量，而 U {U} U不是，因為其表示式中含有了 X {X} X的未知引數 σ \sigma σ

常用的統計量

設 ( X 1 , X 2 , . . . , X n ) ({X_1},{X_2},...,{X_n}) (X1​,X2​,...,Xn​)為總體 X {X} X的一個樣本

樣本均值

稱樣本的算術平均值為樣本均值，記為 X ˉ \bar X Xˉ，即
X ˉ = X 1 + X 2 + . . . + X n n \bar X = {{{X_1} + {X_2} + ... + {X_n}} \over n} Xˉ=nX1​+X2​+...+Xn​​

樣本方差

樣本方差是用來描述樣本中諸分量與樣本均值的均方差異的，它有兩種定義方式。

未修正樣本方差

S 0 2 = 1 n ∑ i = 1 n ( X i − X ˉ ) 2 S_0^2 = {1 \over n}\sum\limits_{i = 1}^n {{{\left( {{X_i} - \bar X} \right)}^2}} S02​=n1​i=1∑n​(Xi​−Xˉ)2

修正樣本方差

S 0 2 = 1 n − 1 ∑ i = 1 n ( X i − X ˉ ) 2 S_0^2 = {1 \over {n - 1}}\sum\limits_{i = 1}^n {{{\left( {{X_i} - \bar X} \right)}^2}} S02​=n−11​i=1∑n​(Xi​−Xˉ)2

由於存在數學關係 E ( S 2 ) = σ 2 E({S^2}) = {\sigma ^2} E(S2)=σ2(後面證明)，修正樣本方差具有更好的統計性質，故樣本方差均採用修正樣本方差。

樣本標準差

樣本標準差定義為樣本方差的算術平方根

S = 1 n − 1 ∑ i = 1 n ( X i − X ˉ ) 2 S = \sqrt {{1 \over {n - 1}}\sum\limits_{i = 1}^n {{{\left( {{X_i} - \bar X} \right)}^2}} } S=n−11​i=1∑n​(Xi​−Xˉ)2 ​

樣本原點矩

k階原點矩為

A k = 1 n ∑ i = 1 n X i k {A_k} = {1 \over n}\sum\limits_{i = 1}^n {X_i^k} Ak​=n1​i=1∑n​Xik​

樣本中心距

k階中心距為

B k = 1 n ∑ i = 1 n ( X i − X ˉ ) k {B_k} = {1 \over n}\sum\limits_{i = 1}^n {{{\left( {{X_i} - \bar X} \right)}^k}} Bk​=n1​i=1∑n​(Xi​−Xˉ)k

樣本均值與方差性質討論

均值性質

如果總體 X {X} X具有數學期望$E(X) = \mu $，則

E ( X ˉ ) = E ( X ) = σ E(\bar X) = E(X) = \sigma E(Xˉ)=E(X)=σ

結論顯然，證明從略。

方差性質

如果總體 X {X} X具有方差 D ( X ) = σ 2 D(X) = {\sigma ^2} D(X)=σ2，則

D ( X ˉ ) = 1 n D ( X ) = σ 2 n D(\bar X) = {1 \over n}D(X) = {{{\sigma ^2}} \over n} D(Xˉ)=n1​D(X)=nσ2​
E ( S 2 ) = D ( X ) = σ 2 E({S^2}) = D(X) = {\sigma ^2} E(S2)=D(X)=σ2

證明：

對於 D ( X ˉ ) = 1 n D ( X ) = σ 2 n D(\bar X) = {1 \over n}D(X) = {{{\sigma ^2}} \over n} D(Xˉ)=n1​D(X)=nσ2​：

D ( X ˉ ) = D ( X 1 + X 2 + . . . + X n n ) D(\bar X) = D({{{X_1} + {X_2} + ... + {X_n}} \over n}) D(Xˉ)=D(nX1​+X2​+...+Xn​​)

= D ( X 1 n ) + D ( X 2 n ) + . . . + D ( X n n ) = D({{{X_1}} \over n}) + D({{{X_2}} \over n}) + ... + D({{{X_n}} \over n}) =D(nX1​​)+D(nX2​​)+...+D(nXn​​)

= 1 n 2 D ( X 1 ) + 1 n 2 D ( X 2 ) + . . . + 1 n 2 D ( X n ) = {1 \over {{n^2}}}D({X_1}) + {1 \over {{n^2}}}D({X_2}) + ... + {1 \over {{n^2}}}D({X_n}) =n21​D(X1​)+n21​D(X2​)+...+n21​D(Xn​)

= 1 n 2 D ( X ) × n = D ( X ) n = σ 2 n = {1 \over {{n^2}}}D(X) \times n = {{D(X)} \over n} = {{{\sigma ^2}} \over n} =n21​D(X)×n=nD(X)​=nσ2​

對於 E ( S 2 ) = D ( X ) = σ 2 E({S^2}) = D(X) = {\sigma ^2} E(S2)=D(X)=σ2：

E ( S 2 ) = E ( ∑ i = 1 n ( X i − X ˉ ) 2 n − 1 ) E({S^2}) = E\left( {{{\sum\limits_{i = 1}^n {{{\left( {{X_i} - \bar X} \right)}^2}} } \over {n - 1}}} \right) E(S2)=E⎝⎜⎜⎛​n−1i=1∑n​(Xi​−Xˉ)2​⎠⎟⎟⎞​

= 1 n − 1 E ( ∑ i = 1 n ( X i − X ˉ ) 2 ) = {1 \over {n - 1}}E(\sum\limits_{i = 1}^n {{{\left( {{X_i} - \bar X} \right)}^2}} ) =n−11​E(i=1∑n​(Xi​−Xˉ)2)

= 1 n − 1 E ( ∑ i = 1 n ( X i 2 − 2 X i X ˉ + X ˉ 2 ) ) = {1 \over {n - 1}}E(\sum\limits_{i = 1}^n {\left( {X_i^2 - 2{X_i}\bar X + {{\bar X}^2}} \right)} ) =n−11​E(i=1∑n​(Xi2​−2Xi​Xˉ+Xˉ2))

= 1 n − 1 E ( ∑ i = 1 n X i 2 − 2 ∑ i = 1 n X i X ˉ + ∑ i = 1 n X ˉ 2 ) = {1 \over {n - 1}}E(\sum\limits_{i = 1}^n {X_i^2} - 2\sum\limits_{i = 1}^n {{X_i}\bar X} + \sum\limits_{i = 1}^n {{{\bar X}^2}} ) =n−11​E(i=1∑n​Xi2​−2i=1∑n​Xi​Xˉ+i=1∑n​Xˉ2)

= 1 n − 1 E ( ∑ i = 1 n X i 2 − 2 n X ˉ 2 + ∑ i = 1 n X ˉ 2 ) = {1 \over {n - 1}}E(\sum\limits_{i = 1}^n {X_i^2} - 2n{\bar X^2} + \sum\limits_{i = 1}^n {{{\bar X}^2}} ) =n−11​E(i=1∑n​Xi2​−2nXˉ2+i=1∑n​Xˉ2)

= 1 n − 1 E ( ∑ i = 1 n X i 2 − n X ˉ 2 ) = {1 \over {n - 1}}E(\sum\limits_{i = 1}^n {X_i^2} - n{\bar X^2}) =n−11​E(i=1∑n​Xi2​−nXˉ2)

= 1 n − 1 ( ∑ i = 1 n E ( X i 2 ) − n E ( X ˉ 2 ) ) = {1 \over {n - 1}}\left( {\sum\limits_{i = 1}^n {E(X_i^2)} - nE({{\bar X}^2})} \right) =n−11​(i=1∑n​E(Xi2​)−nE(Xˉ2))

由方差公式，有： E ( X 2 ) = D ( X ) + ( E ( X ) ) 2 E({X^2}) = D(X) + {(E(X))^2} E(X2)=D(X)+(E(X))2，代入即得上式：

= 1 n − 1 ( ∑ i = 1 n D ( X i ) + ( E ( X i ) ) 2 − n ( D ( X ˉ ) + ( E ( X ˉ ) ) 2 ) ) = {1 \over {n - 1}}\left( {\sum\limits_{i = 1}^n {D({X_i}) + (E({X_i})} {)^2} - n\left( {D(\bar X) + {{(E(\bar X))}^2}} \right)} \right) =n−11​(i=1∑n​D(Xi​)+(E(Xi​))2−n(D(Xˉ)+(E(Xˉ))2))

= 1 n − 1 ( ∑ i = 1 n σ 2 + μ 2 − n ( σ 2 n + μ 2 ) ) = {1 \over {n - 1}}\left( {\sum\limits_{i = 1}^n {{\sigma ^2} + {\mu ^2}} - n\left( {{{{\sigma ^2}} \over n} + {\mu ^2}} \right)} \right) =n−11​(i=1∑n​σ2+μ2−n(nσ2​+μ2))

= 1 n − 1 ( n σ 2 + n μ 2 − n ( σ 2 n + μ 2 ) ) = {1 \over {n - 1}}\left( {n{\sigma ^2} + n{\mu ^2} - n\left( {{{{\sigma ^2}} \over n} + {\mu ^2}} \right)} \right) =n−11​(nσ2+nμ2−n(nσ2​+μ2))

= σ 2 = {\sigma ^2} =σ2

相關文章