復習筆記Day112：概率論知識總結(jié)(四)

2023-03-03 22:54 作者:間宮_卓司 0人讀過 | 我要投稿

在開始第六章之前先寫道題目吧

附錄 5.3.3 ( $%5Ctext%7BStieltjes%7D$ 積分的分部積分公式)如果 $F%2CG$ 是 $%5Ba%2Cb%5D$ 上右連續(xù)單調(diào)函數(shù)，那么

$F%5Cleft(%20b%20%5Cright)%20G%5Cleft(%20b%20%5Cright)%20-F%5Cleft(%20a%20%5Cright)%20G%5Cleft(%20a%20%5Cright)%20%3D%5Cint_%7B%5Cleft(%20a%2Cb%20%5Cright%5D%7D%5E%7B%7D%7BF%5Cleft(%20t-%20%5Cright)%20%5Cmathrm%7Bd%7DG%5Cleft(%20t%20%5Cright)%7D%2B%5Cint_%7B%5Cleft(%20a%2Cb%20%5Cright%5D%7D%5E%7B%7D%7BG%5Cleft(%20t%20%5Cright)%20%5Cmathrm%7Bd%7DF%5Cleft(%20t%20%5Cright)%7D$

112.1 證明：如果隨機變量 $%5Cxi$ 可積，則

(a)? $%5Cunderset%7By%5Crightarrow%20%2B%5Cinfty%7D%7B%5Clim%7Dy%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cxi%20%3Ey%20%5Cright)%20%3D0$

(b) $%5Cmathbb%7BE%7D%20%5Cxi%20%3D%5Cint_0%5E%7B%2B%5Cinfty%7D%7B%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cxi%20%3Ey%20%5Cright)%20%5Cmathrm%7Bd%7Dy%7D-%5Cint_0%5E%7B%2B%5Cinfty%7D%7B%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cxi%20%3C-y%20%5Cright)%20%5Cmathrm%7Bd%7Dy%7D$

不知道這題的正常證明方法是什么，我就寫一下我的方法吧···

先來證明一下(b)，我想到的是用離散型隨機變量去逼近這個隨機變量

第一步：先證明 $%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cxi%3B%5Cleft%5C%7B%200%3C%5Cxi%5Cle%20x%20%5Cright%5C%7D%20%5Cright)%20%3D%5Cint_0%5Ex%7B%5Cleft%5B%20F(x)%20-F%5Cleft(%20t%20%5Cright)%20%5Cright%5D%20%5Cmathrm%7Bd%7Dt%7D$ ，其中 $F(x)$ 是 $%5Cxi$ 的分布函數(shù)。證明方法如下：

類似于定理4.1.1中(3)的證明，構造函數(shù)

$%5Cphi%20_n%5Cleft(%20u%20%5Cright)%20%3Du1_%7B%5Cleft(%20-%5Cinfty%20%2C0%20%5Cright%5D%20%5Ccup%20%5Cleft(%20x%2C%2B%5Cinfty%20%5Cright)%7D%5Cleft(%20u%20%5Cright)%20%2B%5Csum_%7Bk%3D1%7D%5En%7B%5Cfrac%7Bx%5Cleft(%20k-1%20%5Cright)%7D%7Bn%7D1_%7B%5Cleft(%20%5Cfrac%7Bx%5Cleft(%20k-1%20%5Cright)%7D%7Bn%7D%2C%5Cfrac%7Bxk%7D%7Bn%7D%20%5Cright%5D%7D%7D%5Cleft(%20u%20%5Cright)%20$

并記 $%5Cxi_n%3D%5Cphi_n(%5Cxi)$ ，那么

$%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cxi_n%3D%5Cfrac%7Bxk%7D%7Bn%7D%20%5Cright)%20%3D%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cfrac%7Bx%5Cleft(%20k-1%20%5Cright)%7D%7Bn%7D%3C%5Cxi%5Cle%20%5Cfrac%7Bxk%7D%7Bn%7D%20%5Cright)%20%3DF%5Cleft(%20%5Cfrac%7Bkx%7D%7Bn%7D%20%5Cright)%20-F%5Cleft(%20%5Cfrac%7B%5Cleft(%20k-1%20%5Cright)%20x%7D%7Bn%7D%20%5Cright)%20%2Ck%3D1%2C2%2C%5Ccdots%20%2Cn$

并且 $%5Cxi_n%5Crightarrow%5Cxi$

進一步可以計算出 $%5Cxi_n$ 的分布函數(shù)為

$F_n%3D1_%7B%5Cleft(%20-%5Cinfty%20%2C0%20%5Cright%5D%20%5Ccup%20%5Cleft(%20x%2C%2B%5Cinfty%20%5Cright)%7DF%2B%5Csum_%7Bk%3D1%7D%5En%7BF%5Cleft(%20%5Cfrac%7Bxk%7D%7Bn%7D%20%5Cright)%201_%7B%5Cleft(%20%5Cfrac%7Bx%5Cleft(%20k-1%20%5Cright)%7D%7Bn%7D%2C%5Cfrac%7Bxk%7D%7Bn%7D%20%5Cright%5D%7D%7D$

接下來，計算可得

$%5Cbegin%7Baligned%7D%0A%09%26%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cxi%20_n%2C%5Cleft%5C%7B%200%3C%5Cxi%20_n%5Cle%20x%20%5Cright%5C%7D%20%5Cright)%20%3D%5Cint_%7B%5Cleft(%200%2Cx%20%5Cright%5D%7D%5E%7B%7D%7Bt%5Cmathrm%7Bd%7DF_n%5Cleft(%20t%20%5Cright)%7D%5C%5C%0A%09%26%3D%5Csum_%7Bk%3D1%7D%5En%7B%5Cfrac%7Bkx%7D%7Bn%7D%5Cleft(%20%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cxi%20_n%5Cle%20%5Cfrac%7Bkx%7D%7Bn%7D%20%5Cright)%20-%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cxi%20_n%5Cle%20%5Cfrac%7B%5Cleft(%20k-1%20%5Cright)%20x%7D%7Bn%7D%20%5Cright)%20%5Cright)%7D%5C%5C%0A%09%26%3Dx%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cxi%20_n%5Cle%20x%20%5Cright)%20-%5Cfrac%7Bx%7D%7Bn%7D%5Csum_%7Bk%3D1%7D%5En%7B%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cxi%20_n%5Cle%20%5Cfrac%7Bkx%7D%7Bn%7D%20%5Cright)%7D%5C%5C%0A%09%26%3DxF%5Cleft(%20x%20%5Cright)%20-%5Csum_%7Bk%3D1%7D%5En%7B%5Cfrac%7Bx%7D%7Bn%7DF%5Cleft(%20%5Cfrac%7Bxk%7D%7Bn%7D%20%5Cright)%7D%5C%5C%0A%5Cend%7Baligned%7D$

所以

$%5Cunderset%7Bn%5Crightarrow%20%5Cinfty%7D%7B%5Clim%7D%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cxi%20_n%3B%5Cleft%5C%7B%200%3C%5Cxi%20_n%5Cle%20x%20%5Cright%5C%7D%20%5Cright)%20%3D%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cunderset%7Bn%5Crightarrow%20%5Cinfty%7D%7B%5Clim%7D%5Cxi%20_n%3B%5Cleft%5C%7B%200%3C%5Cxi%20_n%5Cle%20x%20%5Cright%5C%7D%20%5Cright)%20%3D%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cxi%20%3B%5Cleft%5C%7B%200%3C%5Cxi%20%5Cle%20x%20%5Cright%5C%7D%20%5Cright)%20$

結(jié)論得證

同理可證 $%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cxi%20%3B%5Cleft%5C%7B%20-y%5Cle%20%5Cxi%20%3C0%20%5Cright%5C%7D%20%5Cright)%20%3D%5Cint_0%5Ey%7B%5Cleft%5B%20F%5Cleft(%20-y%20%5Cright)%20-F%5Cleft(%20-t%20%5Cright)%20%5Cright%5D%20%5Cmathrm%7Bd%7Dt%7D$ ，其中 $y%5Cge0$

兩式相加就有

$%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cxi%20%3B%5Cleft%5C%7B%20-y%3C%5Cxi%20%3Cx%20%5Cright%5C%7D%20%5Cright)%20%3D%5Cint_0%5Ex%7B%5Cleft%5B%20F%5Cleft(%20x%20%5Cright)%20-F%5Cleft(%20t%20%5Cright)%20%5Cright%5D%20%5Cmathrm%7Bd%7Dt%7D%2B%5Cint_0%5Ey%7B%5Cleft%5B%20F%5Cleft(%20-y%20%5Cright)%20-F%5Cleft(%20-t%20%5Cright)%20%5Cright%5D%20%5Cmathrm%7Bd%7Dt%7D$

分別令 $x%2Cy%5Crightarrow%2B%5Cinfty$ ，可得

$%5Cmathbb%7BE%7D%20%5Cxi%20%3D%5Cint_0%5E%7B%2B%5Cinfty%7D%7B%5Cleft%5B%201-F%5Cleft(%20t%20%5Cright)%20%5Cright%5D%20%5Cmathrm%7Bd%7Dt%7D-%5Cint_0%5E%7B%2B%5Cinfty%7D%7BF%5Cleft(%20-y%20%5Cright)%20%5Cmathrm%7Bd%7Dt%7D%3D%5Cint_0%5E%7B%2B%5Cinfty%7D%7B%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cxi%20%3Ey%20%5Cright)%20%5Cmathrm%7Bd%7Dy%7D-%5Cint_0%5E%7B%2B%5Cinfty%7D%7B%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cxi%20%3C-y%20%5Cright)%20%5Cmathrm%7Bd%7Dy%7D$

另外這題也可以用@共與陽光擁抱哀醬提供的方法

這個方法比我自己想的方法要簡單很多

(a)依附錄 5.3.3，取 $F(x)%3Dx%2CG(x)%3DP(%5Cxi%5Cle%20x)%2Cb%5Cge0$

$%5Cbegin%7Baligned%7D%0A%09%5Cint_%7B%5Cleft(%200%2Cb%20%5Cright%5D%7D%5E%7B%7D%7By%5Cmathrm%7Bd%7DF%5Cleft(%20y%20%5Cright)%7D%26%3DbF%5Cleft(%20b%20%5Cright)%20-%5Cint_%7B%5Cleft(%200%2Cb%20%5Cright%5D%7D%5E%7B%7D%7BF%5Cleft(%20y%20%5Cright)%20%5Cmathrm%7Bd%7Dy%7D%5C%5C%0A%09%26%3D-b%5Cleft(%201-F%5Cleft(%20b%20%5Cright)%20%5Cright)%20%2B%5Cint_%7B%5Cleft(%200%2Cb%20%5Cright%5D%7D%5E%7B%7D%7B%5Cleft%5B%201-F%5Cleft(%20y%20%5Cright)%20%5Cright%5D%20%5Cmathrm%7Bd%7Dy%7D%5C%5C%0A%5Cend%7Baligned%7D$

從(b)可知

$%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cxi%20%3B%5Cleft%5C%7B%200%5Cle%20%5Cxi%20%3C%20%2B%5Cinfty%20%5Cright%5C%7D%20%5Cright)%20%3D%5Cint_0%5E%7B%2B%5Cinfty%7D%7B%5Cleft%5B%201-F%5Cleft(%20t%20%5Cright)%20%5Cright%5D%20%5Cmathrm%7Bd%7Dt%7D$

所以 $%5Cunderset%7By%5Crightarrow%20%2B%5Cinfty%7D%7B%5Clim%7Dy%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cxi%20%3Ey%20%5Cright)%20%3D0$

類似地可以證明 $%5Cunderset%7By%5Crightarrow%20-%5Cinfty%7D%7B%5Clim%7Dy%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cxi%20%3Cy%20%5Cright)%20%3D0$

第六章隨機變量

§6.1 隨機向量及聯(lián)合分布

這節(jié)簡單介紹了一下隨機向量，想要進一步了解還是要參考別的概率論課本

協(xié)方差的定義是 $%5Cmathrm%7Bcov%7D%5Cleft(%20%5Cxi%20%2C%5Ceta%20%5Cright)%20%3A%3D%5Cmathbb%7BE%7D%20%5Cleft%5B%20%5Cleft(%20%5Cxi%20-%5Cmathbb%7BE%7D%20%5Cxi%20%5Cright)%20%5Cleft(%20%5Ceta%20-%5Cmathbb%7BE%7D%20%5Ceta%20%5Cright)%20%5Cright%5D%20%3D%5Cmathbb%7BE%7D%20%5Cxi%20%5Ceta%20-%5Cmathbb%7BE%7D%20%5Cxi%20%5Cmathbb%7BE%7D%20%5Ceta%20$

對于隨機向量 $X%3D%5Cleft(%20%5Cxi%20_1%2C%5Ccdots%20%2C%5Cxi%20_n%20%5Cright)%20$ ，定義其協(xié)方差矩陣為 $A%3D%5Cleft(%20%5Cmathrm%7Bcov%7D%5Cleft(%20%5Cxi%20_i%2C%5Cxi%20_j%20%5Cright)%20%5Cright)%20_%7Bn%5Ctimes%20n%7D$ ，也可以表示成 $A%3D%5Cmathbb%7BE%7D%20%5Cleft%5B%20X%5ETX%20%5Cright%5D%20-%5Cleft(%20%5Cmathbb%7BE%7D%20X%20%5Cright)%20%5ET%5Cleft(%20%5Cmathbb%7BE%7D%20X%20%5Cright)%20$

引理6.1.1 協(xié)方差矩陣是正定矩陣

只要把協(xié)方差運算看成是 $%5Cmathscr%7BF%7D$ 上的內(nèi)積，這個性質(zhì)應該很好理解

§6.2 均勻分布與正態(tài)分布

這節(jié)沒有任何的定理、引理、定義，按道理要直接跳過的，不過這本書的多維正態(tài)分布好像比其他（我讀過的）的概率論課本寫的要好一些（也可能是我之前沒有認真看），所以我大概寫一下

記 $f%5Cleft(%20x%20%5Cright)%20%3D%5Cexp%20%5Cleft%5C%7B%20-%5Cfrac%7B1%7D%7B2%7DxA%5E%7B-1%7Dx%5ET%20%5Cright%5C%7D%20$ ，則 $f$ 在 $%5Cmathbf%7BR%7D%5En$ 上可積當且僅當 $A$ 是正定矩陣，進一步計算可得 $%5Cint_%7Bx%5Cin%20%5Cmathbf%7BR%7D%5En%7D%7Bf%5Cleft(%20x%20%5Cright)%20%5Cmathrm%7Bd%7Dx%3D%5Csqrt%7B%5Cdet%20%5Cleft(%20A%20%5Cright)%7D%5Cleft(%20%5Csqrt%7B2%5Cpi%7D%20%5Cright)%20%5En%7D$

稱以 $f%5Cleft(%20x%20%5Cright)%20%3D%5Cfrac%7B%5Cexp%20%5Cleft%5C%7B%20-%5Cfrac%7B1%7D%7B2%7D%5Cleft(%20x-a%20%5Cright)%20A%5E%7B-1%7D%5Cleft(%20x-a%20%5Cright)%20%5ET%20%5Cright%5C%7D%7D%7B%5Csqrt%7B%5Cdet%20%5Cleft(%20A%20%5Cright)%7D%5Cleft(%20%5Csqrt%7B2%5Cpi%7D%20%5Cright)%20%5En%7D$ 為概率密度函數(shù)的隨機向量 $X$ 為服從參數(shù)為 $a%2CA$ 的隨機變量，記為 $X%5Csim%20N%5Cleft(%20a%2CA%20%5Cright)%20$ ，再經(jīng)過計算可得， $X$ 的協(xié)方差矩陣正是 $A$