復(fù)習(xí)筆記Day121：卡爾曼濾波的推導(dǎo)

2023-07-28 01:46 作者:間宮_卓司 0人讀過 | 我要投稿

這篇專欄只涉及到卡爾曼濾波的一個(gè)比較原始的推導(dǎo)，比較現(xiàn)代的推導(dǎo)見

https://zhuanlan.zhihu.com/p/166342719

首先來介紹一下卡爾曼濾波要解決的問題：對(duì)于給定的系統(tǒng)

$%5Cbegin%7Bcases%7D%0A%09x%5Cleft(%20k%2B1%20%5Cright)%20%3D%5CvarPhi%20x%5Cleft(%20k%20%5Cright)%5C%5C%0A%09y%5Cleft(%20k%20%5Cright)%20%3D%5CvarTheta%20x%5Cleft(%20k%20%5Cright)%5C%5C%0A%5Cend%7Bcases%7D$

其中 $x(k)$ 為實(shí)際值， $y(k)$ 為觀測(cè)值，而現(xiàn)實(shí)中測(cè)量以及系統(tǒng)本身都是存在誤差的，也就是說，在這個(gè)系統(tǒng)中，初值不再是一個(gè)數(shù)，而是服從某個(gè)分布的隨機(jī)變量；觀測(cè)和迭代也不可能是絕對(duì)準(zhǔn)確的，要加上一個(gè)噪聲。在這樣的假設(shè)下，模型就變成了

$%5Cbegin%7Bcases%7D%0A%09x%5Cleft(%20k%2B1%20%5Cright)%20%3D%5CvarPhi%20x%5Cleft(%20k%20%5Cright)%20%2B%5Cxi%20%5Cleft(%20k%20%5Cright)%5C%5C%0A%09y%5Cleft(%20k%20%5Cright)%20%3D%5CvarTheta%20x%5Cleft(%20k%20%5Cright)%20%2B%5Ceta%20%5Cleft(%20k%20%5Cright)%5C%5C%0A%5Cend%7Bcases%7D$

其中假設(shè)了初值以及噪聲都是服從正態(tài)分布的。

因?yàn)榧尤肓嗽肼?，所以要絕對(duì)精確的得知 $x(k)$ 的值是不可能的，所以只能去求在已知信息下，關(guān)于 $x(k)$ 的最優(yōu)估計(jì) $%5Chat%7Bx%7D%5Cleft(%20k%7Ck%20%5Cright)%20$ ，其中后一個(gè)k代表前k步的信息作為已知信息

首先需要明確的是何為最優(yōu)，在我參考的教材中，對(duì)最優(yōu)的定義見復(fù)習(xí)筆記Day119，同樣在那篇文章里說明了，此時(shí)有 $x_0$ ，也就是說：在已知觀測(cè)值 $y%5Cleft(%201%20%5Cright)%20%2Cy%5Cleft(%202%20%5Cright)%20%2C%5Ccdots%20%2Cy%5Cleft(%20n%20%5Cright)%20$ 的情況下，對(duì) $x(k)$ 的最優(yōu)估計(jì)正是限制在這些觀測(cè)值下迭代條件期望。下面分幾步來給出計(jì)算這個(gè)值的遞推公式

注意這里的迭代關(guān)系都是線性的，故 $x(k)%2Cy(k)$ 都是服從某個(gè)多維正態(tài)分布的（注意它們是向量而不是一個(gè)數(shù)）

首先先來研究一下如果隨機(jī)變量 $X$ 的概率密度函數(shù)為 $f(x)$ ，隨機(jī)變量 $Y$ 的概率密度函數(shù)為 $f(y)$ ，它們的聯(lián)合分布的概率密度函數(shù)為 $f(x%2Cy)$ 。那么 $X$ 在 $Y$ 的限制下的條件概率密度函數(shù)如何？，也就是去計(jì)算 $f%5Cleft(%20x%7Cy%20%5Cright)%20%3D%5Cfrac%7Bf%5Cleft(%20x%2Cy%20%5Cright)%7D%7Bf%5Cleft(%20y%20%5Cright)%7D$

下面來做一些假設(shè)，設(shè) $X%5Csim%20N%5Cleft(%20m_x%2CR_x%20%5Cright)%2CY%5Csim%20N%5Cleft(%20m_y%2CR_y%20%5Cright)%20$ , $X%2CY$ 之間的協(xié)方差矩陣為 $R_%7Bxy%7D$ ，那么隨機(jī)變量 $C%3D(X%2CY)$ 的協(xié)方差矩陣就是 $R%3D%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09R_x%26%09%09R_%7Bxy%7D%5C%5C%0A%09R_%7Byx%7D%26%09%09R_y%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20$ ，于是就有下面的計(jì)算過程，計(jì)算過程其實(shí)不是很復(fù)雜，不過寫的比較詳細(xì)，所以看起來很長(zhǎng)。

記 $T%3D%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09I%26%09%09-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7D%5C%5C%0A%090%26%09%09I%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20$ ，則 $R_1%3DTR%3D%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09R_x-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7DR_%7Byx%7D%26%09%090%5C%5C%0A%09R_%7Byx%7D%26%09%09R_y%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20$ ，進(jìn)而

$R_%7B1%7D%5E%7B-1%7D%3D%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09%5Cleft(%20R_x-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7DR_%7Byx%7D%20%5Cright)%20%5E%7B-1%7D%26%09%090%5C%5C%0A%09-R_%7By%7D%5E%7B-1%7DR_%7Byx%7D%5Cleft(%20R_x-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7DR_%7Byx%7D%20%5Cright)%20%5E%7B-1%7D%26%09%09R_%7By%7D%5E%7B-1%7D%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20$ ，從后面的結(jié)果來看，這樣做是為了把 $y$ 單獨(dú)分離出來。

從而

$%5Cbegin%7Baligned%7D%0A%09%26f(x%2Cy)%5C%5C%0A%09%26%3D(2%5Cpi%20)%5E%7B-%5Cfrac%7Bn_x%2Bn_y%7D%7B2%7D%7D(%5Cdet%20R)%5E%7B-%5Cfrac%7Bn_x%2Bn_y%7D%7B2%7D%7D%5Cexp%20%5Cleft%5C%7B%20-%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bl%7D%0A%09x-m_x%5C%5C%0A%09y-m_y%5C%5C%0A%5Cend%7Barray%7D%20%5Cright%5D%20%5E%7B%5Cmathrm%7BT%7D%7DR%5E%7B-1%7D%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bl%7D%0A%09x-m_x%5C%5C%0A%09y-m_y%5C%5C%0A%5Cend%7Barray%7D%20%5Cright%5D%20%5Cright%5C%7D%5C%5C%0A%09%26%3D(2%5Cpi%20)%5E%7B-%5Cfrac%7Bn_x%2Bn_y%7D%7B2%7D%7D(%5Cdet%20R)%5E%7B-%5Cfrac%7Bn_x%2Bn_y%7D%7B2%7D%7D%5Cexp%20%5Cleft%5C%7B%20-%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bl%7D%0A%09x-m_x%5C%5C%0A%09y-m_y%5C%5C%0A%5Cend%7Barray%7D%20%5Cright%5D%20%5E%7B%5Cmathrm%7BT%7D%7D%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09R_x%26%09%09R_%7Bxy%7D%5C%5C%0A%09R_%7Byx%7D%26%09%09R_y%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20%5E%7B-1%7D%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bl%7D%0A%09x-m_x%5C%5C%0A%09y-m_y%5C%5C%0A%5Cend%7Barray%7D%20%5Cright%5D%20%5Cright%5C%7D%5C%5C%0A%09%26%3D(2%5Cpi%20)%5E%7B-%5Cfrac%7Bn_x%2Bn_y%7D%7B2%7D%7D(%5Cdet%20R)%5E%7B-%5Cfrac%7Bn_x%2Bn_y%7D%7B2%7D%7D%5Cexp%20%5Cleft%5C%7B%20-%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bl%7D%0A%09x-m_x%5C%5C%0A%09y-m_y%5C%5C%0A%5Cend%7Barray%7D%20%5Cright%5D%20%5E%7B%5Cmathrm%7BT%7D%7D%5Cleft(%20T%5E%7B-1%7DR_1%20%5Cright)%20%5E%7B-1%7D%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bl%7D%0A%09x-m_x%5C%5C%0A%09y-m_y%5C%5C%0A%5Cend%7Barray%7D%20%5Cright%5D%20%5Cright%5C%7D%5C%5C%0A%09%26%3D(2%5Cpi%20)%5E%7B-%5Cfrac%7Bn_x%2Bn_y%7D%7B2%7D%7D(%5Cdet%20R)%5E%7B-%5Cfrac%7Bn_x%2Bn_y%7D%7B2%7D%7D%5Cexp%20%5Cleft%5C%7B%20-%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bl%7D%0A%09x-m_x%5C%5C%0A%09y-m_y%5C%5C%0A%5Cend%7Barray%7D%20%5Cright%5D%20%5E%7B%5Cmathrm%7BT%7D%7DR_%7B1%7D%5E%7B-1%7DT%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bl%7D%0A%09x-m_x%5C%5C%0A%09y-m_y%5C%5C%0A%5Cend%7Barray%7D%20%5Cright%5D%20%5Cright%5C%7D%5C%5C%0A%09%26%3D(2%5Cpi%20)%5E%7B-%5Cfrac%7Bn_x%2Bn_y%7D%7B2%7D%7D(%5Cdet%20R)%5E%7B-%5Cfrac%7Bn_x%2Bn_y%7D%7B2%7D%7D%5Cexp%20%5Cleft%5C%7B%20-%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bl%7D%0A%09x-m_x%5C%5C%0A%09y-m_y%5C%5C%0A%5Cend%7Barray%7D%20%5Cright%5D%20%5E%7B%5Cmathrm%7BT%7D%7D%20%5Cright.%5C%5C%0A%09%26%5Cleft.%20%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09%5Cleft(%20R_x-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7DR_%7Byx%7D%20%5Cright)%20%5E%7B-1%7D%26%09%090%5C%5C%0A%09-R_%7By%7D%5E%7B-1%7DR_%7Byx%7D%5Cleft(%20R_x-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7DR_%7Byx%7D%20%5Cright)%20%5E%7B-1%7D%26%09%09R_%7By%7D%5E%7B-1%7D%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09I%26%09%09-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7D%5C%5C%0A%090%26%09%09I%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bl%7D%0A%09x-m_x%5C%5C%0A%09y-m_y%5C%5C%0A%5Cend%7Barray%7D%20%5Cright%5D%20%5Cright%5C%7D%5C%5C%0A%09%26%3D(2%5Cpi%20)%5E%7B-%5Cfrac%7Bn_x%2Bn_y%7D%7B2%7D%7D(%5Cdet%20R)%5E%7B-%5Cfrac%7Bn_x%2Bn_y%7D%7B2%7D%7D%5Cexp%20%5Cleft%5C%7B%20-%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B%20%5Cleft(%20x-m_x%20%5Cright)%20%5ET%5Cleft(%20R_x-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7DR_%7Byx%7D%20%5Cright)%20%5E%7B-1%7D%20%5Cright.%20%5Cright.%5C%5C%0A%09%26%5Cleft.%20-%5Cleft(%20y-m_y%20%5Cright)%20%5ETR_%7By%7D%5E%7B-1%7DR_%7Byx%7D%5Cleft(%20R_x-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7DR_%7Byx%7D%20%5Cright)%20%5E%7B-1%7D%2C%5Cleft(%20y-m_y%20%5Cright)%20%5ETR_%7By%7D%5E%7B-1%7D%20%5Cright%5D%5C%5C%0A%09%26%5Cleft.%20%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bc%7D%0A%09%5Cleft(%20x-m_x%20%5Cright)%20-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7D%5Cleft(%20y-m_y%20%5Cright)%5C%5C%0A%09y-m_y%5C%5C%0A%5Cend%7Barray%7D%20%5Cright%5D%20%5Cright%5C%7D%5C%5C%0A%09%26%5Ctriangleq%20(2%5Cpi%20)%5E%7B-%5Cfrac%7Bn_x%7D%7B2%7D%7D(%5Cdet%20R)%5E%7B-%5Cfrac%7Bn_x%7D%7B2%7D%7D%5Cexp%20%5Cleft.%20%5Cleft%5C%7B%20-%5Cfrac%7B1%7D%7B2%7D%5Cleft(%20x-m_%7Bx%7Cy%7D%20%5Cright)%20%5ETR_%7Bx%7Cy%7D%5E%7B-1%7D%5Cleft(%20x-m_%7Bx%7Cy%7D%20%5Cright)%20%5Cright.%20%5Cright%5C%7D%20f%5Cleft(%20y%20%5Cright)%5C%5C%0A%5Cend%7Baligned%7D$

其中最后一個(gè)等號(hào)的計(jì)算過程如下

$%5Cbegin%7Baligned%7D%0A%09%26%5Cleft(%20x-m_x%20%5Cright)%20%5ET%5Cleft(%20R_x-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7DR_%7Byx%7D%20%5Cright)%20%5E%7B-1%7D-%5Cleft(%20y-m_y%20%5Cright)%20%5ETR_%7By%7D%5E%7B-1%7DR_%7Byx%7D%5Cleft(%20R_x-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7DR_%7Byx%7D%20%5Cright)%20%5E%7B-1%7D%5C%5C%0A%09%26%3D%5Cleft(%20%5Cleft(%20x-m_x%20%5Cright)%20%5ET-%5Cleft(%20y-m_y%20%5Cright)%20%5ETR_%7By%7D%5E%7B-1%7DR_%7Byx%7D%20%5Cright)%20%5Cleft(%20R_x-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7DR_%7Byx%7D%20%5Cright)%20%5E%7B-1%7D%5C%5C%0A%09%26%3D%5Cleft(%20x-m_x-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7D%5Cleft(%20y-m_y%20%5Cright)%20%5Cright)%20%5ET%5Cleft(%20R_x-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7DR_%7Byx%7D%20%5Cright)%20%5E%7B-1%7D%5C%5C%0A%09%26%5Ctriangleq%20%5Cleft(%20x-m_%7Bx%7Cy%7D%20%5Cright)%20R_%7Bx%7Cy%7D%5E%7B-1%7D%5C%5C%0A%5Cend%7Baligned%7D$

也就是說

$%5Cfrac%7Bf%5Cleft(%20x%2Cy%20%5Cright)%7D%7Bf%5Cleft(%20y%20%5Cright)%7D%3D(2%5Cpi%20)%5E%7B-%5Cfrac%7Bn_x%7D%7B2%7D%7D(%5Cdet%20R)%5E%7B-%5Cfrac%7Bn_x%7D%7B2%7D%7D%5Cexp%20%5Cleft.%20%5Cleft%5C%7B%20-%5Cfrac%7B1%7D%7B2%7D%5Cleft(%20x-m_%7Bx%7Cy%7D%20%5Cright)%20%5ETR_%7Bx%7Cy%7D%5E%7B-1%7D%5Cleft(%20x-m_%7Bx%7Cy%7D%20%5Cright)%20%5Cright.%20%5Cright%5C%7D%20$

注意到這個(gè)式子中， $%5Cfrac%7Bf%5Cleft(%20x%2Cy%20%5Cright)%7D%7Bf%5Cleft(%20y%20%5Cright)%7D$ 同樣可以視為某個(gè)服從正態(tài)分布的隨機(jī)變量的概率密度函數(shù)

接下來來計(jì)算

記

$Z%5Csim%20N%5Cleft(%20m_%7Bx%7Cy%7D%2CR_%7Bx%7Cy%7D%5E%7B%7D%20%5Cright)%20%3DN%5Cleft(%20m_x-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7D%5Cleft(%20y-m_y%20%5Cright)%20%2CR_x-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7DR_%7Byx%7D%20%5Cright)%20$ ，注意這樣假設(shè)把y視為了常量，那么

$%5Cmathbb%7BE%7D%20%5Cleft(%20X%7Cy%20%5Cright)%20%3D%5Cmathbb%7BE%7D%20%5Cleft(%20z%20%5Cright)%20%3Dm_x-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7D%5Cleft(%20y-m_y%20%5Cright)%20$

注意這里把 $y$ 視為了一個(gè)數(shù)，而 $%5Cmathbb%7BE%7D%20%5Cleft(%20X%7Cy%20%5Cright)%20$ 是 $%5Cmathbb%7BE%7D%20%5Cleft(%20X%7CY%3Dy%20%5Cright)%20$ 的縮寫，所以如果把 $y$ 視為一個(gè)隨機(jī)變量，那么 $%5Cmathbb%7BE%7D%20%5Cleft(%20X%7Cy%20%5Cright)%20$ 也是一個(gè)隨機(jī)變量，所以干脆記成 $%5Cmathbb%7BE%7D%20%5Cleft(%20X%7CY%20%5Cright)%20$ （這塊不太能講清楚，湊合著看吧）。總之，這樣就得到了

$%5Cmathbb%7BE%7D%20%5Cleft(%20X%7CY%20%5Cright)%20%3Dm_x-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7D%5Cleft(%20Y-m_y%20%5Cright)%20$

接下來，為了獲得遞推式，來計(jì)算 $%5Cmathbb%7BE%7D%20%5Cleft(%20X%7Cy%2Cz%20%5Cright)%20$ ，其中 $X%2CY%2CZ$ 都是服從正態(tài)分布的。先考慮簡(jiǎn)單的情況，即 $Y%2CZ$ 是相互獨(dú)立的情況

$%5Cbegin%7Baligned%7D%0A%09%5Cmathbb%7BE%7D%20%5Cleft(%20X%7Cy%2Cz%20%5Cright)%20%26%3Dm_x-%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09R_%7Bxy%7D%26%09%09R_%7Bxz%7D%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09R_y%26%09%09%5C%5C%0A%09%26%09%09R_z%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20%5E%7B-1%7D%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bc%7D%0A%09y-m_y%5C%5C%0A%09z-m_z%5C%5C%0A%5Cend%7Barray%7D%20%5Cright%5D%5C%5C%0A%09%26%3Dm_x-%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09R_%7Bxy%7D%26%09%09R_%7Bxz%7D%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09R_%7By%7D%5E%7B-1%7D%26%09%09%5C%5C%0A%09%26%09%09R_%7Bz%7D%5E%7B-1%7D%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bc%7D%0A%09y-m_y%5C%5C%0A%09z-m_z%5C%5C%0A%5Cend%7Barray%7D%20%5Cright%5D%5C%5C%0A%09%26%3D%5Cleft(%20m_x-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7D%5Cleft(%20y-m_y%20%5Cright)%20%5Cright)%20%2B%5C%5C%0A%09%26%5Cleft(%20m_x-R_%7Bxy%7DR_%7Bz%7D%5E%7B-1%7D%5Cleft(%20z-m_z%20%5Cright)%20%5Cright)%20-m_x%5C%5C%0A%09%26%3D%5Cmathbb%7BE%7D%20%5Cleft(%20X%7Cy%20%5Cright)%20%2B%5Cmathbb%7BE%7D%20%5Cleft(%20X%7Cz%20%5Cright)%20-m_x%5C%5C%0A%5Cend%7Baligned%7D$

和上面一樣，也有 $%5Cmathbb%7BE%7D%20%5Cleft(%20X%7CY%2CZ%20%5Cright)%20%3D%5Cmathbb%7BE%7D%20%5Cleft(%20X%7CY%20%5Cright)%20%2B%5Cmathbb%7BE%7D%20%5Cleft(%20X%7CZ%20%5Cright)%20-m_x$

接下來討論 $Y%2CZ$ 相關(guān)的情況，實(shí)際上有結(jié)論 $%5Cmathbb%7BE%7D%20%5Cleft(%20X%7CY%2CZ%20%5Cright)%20%3D%5Cmathbb%7BE%7D%20%5Cleft(%20X%7CY%2C%5Cwidetilde%7BZ%7D%5Cleft(%20Y%20%5Cright)%20%5Cright)%20$ ，這里的 $%5Cwidetilde%7BZ%7D%5Cleft(%20Y%20%5Cright)%20$ 指的是 $Z$ 減去了Z在Y條件下的最優(yōu)預(yù)測(cè)值，即 $Z-%5Cmathbb%7BE%7D%20%5Cleft(%20Z%7CY%20%5Cright)%20$ 。這個(gè)結(jié)論的證明如下

$%5Cbegin%7Baligned%7D%0A%09%26%5Cmathbb%7BE%7D%20%5Cleft(%20X%7Cy%2Cz%20%5Cright)%5C%5C%0A%09%26%3Dm_x-%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09R_%7Bxy%7D%26%09%09R_%7Bxz%7D%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09R_y%26%09%09R_%7Byz%7D%5C%5C%0A%09R_%7Bzy%7D%26%09%09R_z%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20%5E%7B-1%7D%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bc%7D%0A%09y-m_y%5C%5C%0A%09z-m_z%5C%5C%0A%5Cend%7Barray%7D%20%5Cright%5D%5C%5C%0A%09%26%3Dm_x-%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09R_%7Bxy%7D%26%09%09R_%7Bxz%7D%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09R_y%26%09%09R_%7Byz%7D%5C%5C%0A%09R_%7Bzy%7D%26%09%09R_z%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20%5E%7B-1%7D%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bc%7D%0A%09y-m_y%5C%5C%0A%09z-m_z%5C%5C%0A%5Cend%7Barray%7D%20%5Cright%5D%5C%5C%0A%09%26%3Dm_x-%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09R_%7Bxy%7D%26%09%09R_%7Bxz%7D%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09%5Cleft(%20R_y-R_%7Byz%7DR_%7Bz%7D%5E%7B-1%7DR_%7Bzy%7D%20%5Cright)%20%5E%7B-1%7D%26%09%090%5C%5C%0A%09-R_%7Bz%7D%5E%7B-1%7DR_%7Bzy%7D%5Cleft(%20R_y-R_%7Byz%7DR_%7Bz%7D%5E%7B-1%7DR_%7Bzy%7D%20%5Cright)%20%5E%7B-1%7D%26%09%09R_%7Bz%7D%5E%7B-1%7D%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09I%26%09%09-R_%7Byz%7DR_%7Bz%7D%5E%7B-1%7D%5C%5C%0A%090%26%09%09I%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bc%7D%0A%09y-m_y%5C%5C%0A%09z-m_z%5C%5C%0A%5Cend%7Barray%7D%20%5Cright%5D%5C%5C%0A%09%26%3Dm_x-%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09R_%7Bxy%7D%26%09%09R_%7Bxz%7D%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09%5Cleft(%20R_y-R_%7Byz%7DR_%7Bz%7D%5E%7B-1%7DR_%7Bzy%7D%20%5Cright)%20%5E%7B-1%7D%26%09%090%5C%5C%0A%09-R_%7Bz%7D%5E%7B-1%7DR_%7Bzy%7D%5Cleft(%20R_y-R_%7Byz%7DR_%7Bz%7D%5E%7B-1%7DR_%7Bzy%7D%20%5Cright)%20%5E%7B-1%7D%26%09%09R_%7Bz%7D%5E%7B-1%7D%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bc%7D%0A%09y-m_y-R_%7Byz%7DR_%7Bz%7D%5E%7B-1%7D%5Cleft(%20z-m_z%20%5Cright)%5C%5C%0A%09z-m_z%5C%5C%0A%5Cend%7Barray%7D%20%5Cright%5D%5C%5C%0A%09%26%3Dm_x-%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09R_%7Bxy%7D%26%09%09R_%7Bxz%7D%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09R_%7Bx%7Cy%7D%5E%7B-1%7D%26%09%090%5C%5C%0A%09-R_%7Bz%7D%5E%7B-1%7DR_%7Bzy%7DR_%7Bx%7Cy%7D%5E%7B-1%7D%26%09%09R_%7Bz%7D%5E%7B-1%7D%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bc%7D%0A%09y-%5Cmathbb%7BE%7D%20%5Cleft(%20y%7Cz%20%5Cright)%5C%5C%0A%09z-m_z%5C%5C%0A%5Cend%7Barray%7D%20%5Cright%5D%5C%5C%0A%09%26%3Dm_x-%5Cleft%5B%20%5Cbegin%7Bmatrix%7D%0A%09R_%7Bxy%7D%26%09%09R_%7Bxz%7D%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%5D%20%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bc%7D%0A%09R_%7Bx%7Cy%7D%5E%7B-1%7D%5Cleft(%20y-%5Cmathbb%7BE%7D%20%5Cleft(%20y%7Cz%20%5Cright)%20%5Cright)%5C%5C%0A%09-R_%7Bz%7D%5E%7B-1%7DR_%7Bzy%7DR_%7Bx%7Cy%7D%5E%7B-1%7D%5Cmathbb%7BE%7D%20%5Cleft(%20y%7Cz%20%5Cright)%20%2BR_%7Bz%7D%5E%7B-1%7D%5Cleft(%20z-m_z%20%5Cright)%5C%5C%0A%5Cend%7Barray%7D%20%5Cright%5D%5C%5C%0A%09%26%3Dm_x-R_%7Bxy%7DR_%7Bx%7Cy%7D%5E%7B-1%7D%5Cleft(%20y-%5Cmathbb%7BE%7D%20%5Cleft(%20y%7Cz%20%5Cright)%20%5Cright)%20-R_%7Bxz%7D%5Cleft(%20-R_%7Bz%7D%5E%7B-1%7DR_%7Bzy%7DR_%7Bx%7Cy%7D%5E%7B-1%7D%5Cleft(%20y-%5Cmathbb%7BE%7D%20%5Cleft(%20y%7Cz%20%5Cright)%20%5Cright)%20%2BR_%7Bz%7D%5E%7B-1%7D%5Cleft(%20z-m_z%20%5Cright)%20%5Cright)%5C%5C%0A%09%26%3D%5Cleft(%20m_x-R_%7Bxz%7DR_%7Bz%7D%5E%7B-1%7D%5Cleft(%20z-m_z%20%5Cright)%20%5Cright)%20%2B%5Cleft(%20m_x-%5Cleft(%20R_%7Bxy%7D-R_%7Bxz%7DR_%7Bz%7D%5E%7B-1%7DR_%7Bzy%7D%20%5Cright)%20R_%7Bx%7Cy%7D%5E%7B-1%7D%5Cleft(%20y-%5Cmathbb%7BE%7D%20%5Cleft(%20y%7Cz%20%5Cright)%20%5Cright)%20%5Cright)%20-m_x%5C%5C%0A%5Cend%7Baligned%7D$

其中第三個(gè)等號(hào)和之前條件概率密度函數(shù)的推導(dǎo)思路是一樣的，第五個(gè)等號(hào)開始把 $y$ 視為了隨機(jī)變量，不過是隨機(jī)變量還是常量其實(shí)無所謂了吧···反正這里只是一個(gè)簡(jiǎn)化計(jì)算的符號(hào)

為了得到結(jié)論，只需要注意到

$%5Cbegin%7Baligned%7D%0A%09%26%5Cmathbb%7BE%7D%20%5Cleft(%20X%5Cleft(%20Y-m_y-R_%7Byz%7DR_%7Bz%7D%5E%7B-1%7D%5Cleft(%20Z-m_z%20%5Cright)%20%5Cright)%20%5ET%20%5Cright)%5C%5C%0A%09%26%3D%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cleft(%20X-m_x%20%5Cright)%20%5Cleft(%20Y-m_y-R_%7Byz%7DR_%7Bz%7D%5E%7B-1%7D%5Cleft(%20Z-m_z%20%5Cright)%20%5Cright)%20%5ET%20%5Cright)%5C%5C%0A%09%26%3DR_%7Bxy%7D-R_%7Bxz%7DR_%7Bz%7D%5E%7B-1%7DR_%7Bzy%7D%5C%5C%0A%5Cend%7Baligned%7D$

所以總而言之，上面的式子確實(shí)說明了

$%5Cmathbb%7BE%7D%20%5Cleft(%20X%7CY%2CZ%20%5Cright)%20%3D%5Cmathbb%7BE%7D%20%5Cleft(%20X%7CZ%20%5Cright)%20%2B%5Cmathbb%7BE%7D%20%5Cleft(%20X%7C%5Cwidetilde%7BY%7D%5Cleft(%20Z%20%5Cright)%20%5Cright)%20-m_x$

因?yàn)橥瑫r(shí)也有 $%5Cmathbb%7BE%7D%20%5Cleft(%20X%7CZ%2C%5Cwidetilde%7BY%7D%5Cleft(%20Z%20%5Cright)%20%5Cright)%20%3D%5Cmathbb%7BE%7D%20%5Cleft(%20X%7CZ%20%5Cright)%20%2B%5Cmathbb%7BE%7D%20%5Cleft(%20X%7C%5Cwidetilde%7BY%7D%5Cleft(%20Z%20%5Cright)%20%5Cright)%20-m_x$ ，而 $Z%2C%5Cwidetilde%7BY%7D%5Cleft(%20Z%20%5Cright)%20$ 之間是相互獨(dú)立的，所以上面的過程也可以看成把 $Y$ 與 $Z$ 無關(guān)的部分提取了出來再做利用先去的結(jié)論

有了這些結(jié)論，接下來就可以計(jì)算 $%5Chat%7Bx%7D%5Cleft(%20k%7Ck%20%5Cright)%20%3D%5Cmathbb%7BE%7D%20%5Cleft(%20x%5Cleft(%20k%20%5Cright)%20%7Cy%5Cleft(%201%20%5Cright)%20%2Cy%5Cleft(%202%20%5Cright)%20%2C%5Ccdots%20%2Cy%5Cleft(%20k%20%5Cright)%20%5Cright)%20$ 了

$%5Cbegin%7Baligned%7D%0A%09%26%5Cmathbb%7BE%7D%20%5Cleft(%20x%5Cleft(%20k%20%5Cright)%20%7Cy%5Cleft(%201%20%5Cright)%20%2Cy%5Cleft(%202%20%5Cright)%20%2C%5Ccdots%20%2Cy%5Cleft(%20k%20%5Cright)%20%5Cright)%5C%5C%0A%09%26%3D%5Cmathbb%7BE%7D%20%5Cleft(%20x%5Cleft(%20k%20%5Cright)%20%7Cy%5Cleft(%201%20%5Cright)%20%2Cy%5Cleft(%202%20%5Cright)%20%2C%5Ccdots%20%2Cy%5Cleft(%20k-1%20%5Cright)%20%5Cright)%20%2B%5Cmathbb%7BE%7D%20%5Cleft(%20x%5Cleft(%20k%20%5Cright)%20%7C%5Cwidetilde%7By%7D%5Cleft(%20k%7Ck%20%5Cright)%20%5Cright)%20-m_0%5C%5C%0A%09%26%3D%5Cmathbb%7BE%7D%20%5Cleft(%20%5CvarPhi%20x%5Cleft(%20k-1%20%5Cright)%20%2B%5Cxi%20%5Cleft(%20k-1%20%5Cright)%20%7Cy%5Cleft(%201%20%5Cright)%20%2C%5Ccdots%20%2Cy%5Cleft(%20k-1%20%5Cright)%20%5Cright)%5C%5C%0A%09%26%2B%5Cmathbb%7BE%7D%20%5Cleft(%20x%5Cleft(%20k%20%5Cright)%20%7C%5Cwidetilde%7By%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20%5Cright)%20-m_0%5C%5C%0A%09%26%3D%5CvarPhi%20%5Cmathbb%7BE%7D%20%5Cleft(%20x%5Cleft(%20k-1%20%5Cright)%20%7Cy%5Cleft(%201%20%5Cright)%20%2C%5Ccdots%20%2Cy%5Cleft(%20k-1%20%5Cright)%20%5Cright)%20%2B%5Cmathbb%7BE%7D%20%5Cleft(%20x%5Cleft(%20k%20%5Cright)%20%7C%5Cwidetilde%7By%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20%5Cright)%20-m_0%5C%5C%0A%09%26%3D%5CvarPhi%20%5Cmathbb%7BE%7D%20%5Cleft(%20x%5Cleft(%20k-1%20%5Cright)%20%7Cy%5Cleft(%201%20%5Cright)%20%2C%5Ccdots%20%2Cy%5Cleft(%20k-1%20%5Cright)%20%5Cright)%20%2BR_%7Bx%5Cleft(%20k%20%5Cright)%20%5Cwidetilde%7By%7D%5Cleft(%20k%7Ck-1%20%5Cright)%7DR_%7B%5Cwidetilde%7By%7D%5Cleft(%20k%7Ck-1%20%5Cright)%7D%5E%7B-1%7D%5Cwidetilde%7By%7D%5Cleft(%20k%7Ck-1%20%5Cright)%5C%5C%0A%5Cend%7Baligned%7D$

其中第三個(gè)等號(hào)是因?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=%5Ceta(k-1)" alt="%5Ceta(k-1)">與 $y%5Cleft(%201%20%5Cright)%20%2C%5Ccdots%20%2Cy%5Cleft(%20k-1%20%5Cright)%20$ 都是無關(guān)的，下面就來計(jì)算一下第三個(gè)等號(hào)的第二項(xiàng)的各個(gè)系數(shù)，首先有

$%5Cbegin%7Baligned%7D%0A%09%26%5Cwidetilde%7By%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20%3Dy%5Cleft(%20k%20%5Cright)%20-%5Cmathbb%7BE%7D%20%5Cleft(%20y%5Cleft(%20k%20%5Cright)%20%7Cy%5Cleft(%201%20%5Cright)%20%2C%5Ccdots%20%2Cy%5Cleft(%20k-1%20%5Cright)%20%5Cright)%5C%5C%0A%09%26%3Dy%5Cleft(%20k%20%5Cright)%20-%5Cmathbb%7BE%7D%20%5Cleft(%20%5CvarTheta%20x%5Cleft(%20k%20%5Cright)%20%2B%5Cxi%20%5Cleft(%20k%20%5Cright)%20%7Cy%5Cleft(%201%20%5Cright)%20%2C%5Ccdots%20%2Cy%5Cleft(%20k-1%20%5Cright)%20%5Cright)%5C%5C%0A%09%26%3Dy%5Cleft(%20k%20%5Cright)%20-%5CvarTheta%20%5Cmathbb%7BE%7D%20%5Cleft(%20x%5Cleft(%20k%20%5Cright)%20%7Cy%5Cleft(%201%20%5Cright)%20%2C%5Ccdots%20%2Cy%5Cleft(%20k-1%20%5Cright)%20%5Cright)%5C%5C%0A%09%26%3D%5CvarTheta%20x%5Cleft(%20k%20%5Cright)%20%2B%5Ceta%20%5Cleft(%20k%20%5Cright)%20-%5CvarTheta%20%5Cmathbb%7BE%7D%20%5Cleft(%20x%5Cleft(%20k%20%5Cright)%20%7Cy%5Cleft(%201%20%5Cright)%20%2C%5Ccdots%20%2Cy%5Cleft(%20k-1%20%5Cright)%20%5Cright)%5C%5C%0A%09%26%3D%5CvarTheta%20%5Cwidetilde%7Bx%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20%2B%5Ceta%20%5Cleft(%20k%20%5Cright)%5C%5C%0A%5Cend%7Baligned%7D$

那么

$%5Cbegin%7Baligned%7D%0A%09%26R_%7Bx%5Cleft(%20k%20%5Cright)%20%5Cwidetilde%7By%7D%5Cleft(%20k%7Ck-1%20%5Cright)%7D%5C%5C%0A%09%26%3D%5Cmathbb%7BE%7D%20%5Cleft(%20x%5Cleft(%20k%20%5Cright)%20%5Cleft(%20%5Cwidetilde%7By%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20%5Cright)%20%5ET%20%5Cright)%20%3D%5C%5C%0A%09%26%3D%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cleft(%20x%5Cleft(%20k%20%5Cright)%20-m_%7Bx%5Cleft(%20k%20%5Cright)%7D%20%5Cright)%20%5Cleft(%20%5CvarTheta%20%5Cwidetilde%7Bx%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20%2B%5Ceta%20%5Cleft(%20k%20%5Cright)%20%5Cright)%20%5ET%20%5Cright)%5C%5C%0A%09%26%3D%5Cwidetilde%7Bx%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20%5Cwidetilde%7Bx%7D%5ET%5Cleft(%20k%7Ck-1%20%5Cright)%20%5CvarTheta%20%5ET%5C%5C%0A%5Cend%7Baligned%7D$

其中第三個(gè)等號(hào)是因?yàn)?/p>

$%5Cbegin%7Baligned%7D%0A%09%26%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cleft(%20x-%5Cmathbb%7BE%7D%20%5Cleft(%20x%7Cy%20%5Cright)%20%5Cright)%20%5Cmathbb%7BE%7D%20_%7B%7D%5E%7BT%7D%5Cleft(%20x%7Cy%20%5Cright)%20%5Cright)%5C%5C%0A%09%26%3D%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cleft(%20x-m_x%2BR_%7Bxy%7DR_%7By%7D%5E%7B-1%7D%5Cleft(%20y-m_y%20%5Cright)%20%5Cright)%20%5Cleft(%20m_x-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7D%5Cleft(%20y-m_y%20%5Cright)%20%5Cright)%20%5ET%20%5Cright)%5C%5C%0A%09%26%3D%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cleft(%20x-m_x%20%5Cright)%20%5Cleft(%20y-m_y%20%5Cright)%20%5ETR_%7By%7D%5E%7B-1%7DR_%7Byx%7D-R_%7Bxy%7DR_%7By%7D%5E%7B-1%7D%5Cleft(%20y-m_y%20%5Cright)%20%5Cleft(%20y-m_y%20%5Cright)%20%5ETR_%7By%7D%5E%7B-1%7DR_%7Byx%7D%20%5Cright)%5C%5C%0A%09%26%3D0%5C%5C%0A%5Cend%7Baligned%7D$

現(xiàn)在記 $%5Cwidetilde%7Bx%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20%5Cwidetilde%7Bx%7D%5ET%5Cleft(%20k%7Ck-1%20%5Cright)%20%3DP%5Cleft(%20k%7Ck-1%20%5Cright)%20$ ，則 $R_%7Bx%5Cleft(%20k%20%5Cright)%20%5Cwidetilde%7By%7D%5Cleft(%20k%7Ck-1%20%5Cright)%7D%3DP%5Cleft(%20k%7Ck-1%20%5Cright)%20%5CvarTheta%20%5ET$

同理

$%5Cbegin%7Baligned%7D%0A%09%26R_%7B%5Cwidetilde%7By%7D%5Cleft(%20k%20%5Cright)%7D%3D%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cleft(%20%5CvarTheta%20%5Cwidetilde%7Bx%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20%2B%5Ceta%20%5Cleft(%20k%20%5Cright)%20%5Cright)%20%5Cleft(%20%5CvarTheta%20%5Cwidetilde%7Bx%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20%2B%5Ceta%20%5Cleft(%20k%20%5Cright)%20%5Cright)%20%5ET%20%5Cright)%5C%5C%0A%09%26%3D%5CvarTheta%20P%5Cleft(%20k%7Ck-1%20%5Cright)%20%5CvarTheta%20%5ET%2BR_2%5C%5C%0A%5Cend%7Baligned%7D$

現(xiàn)在尚且需要得到 $P%5Cleft(%20k%7Ck-1%20%5Cright)%20$ 的遞推關(guān)系式，因?yàn)?/p>

$%5Cbegin%7Baligned%7D%0A%09%26%5Cwidetilde%7Bx%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20%3Dx%5Cleft(%20k%20%5Cright)%20-%5Cwidehat%7Bx%7D%5Cleft(%20k%7Ck-1%20%5Cright)%5C%5C%0A%09%26%3D%5CvarPhi%20x%5Cleft(%20k-1%20%5Cright)%20%2B%5Cxi%20%5Cleft(%20k-1%20%5Cright)%20-%5Cmathbb%7BE%7D%20%5Cleft(%20x%5Cleft(%20k%20%5Cright)%20%7Cy%5Cleft(%201%20%5Cright)%20%2C%5Ccdots%20y%5Cleft(%20n-1%20%5Cright)%20%5Cright)%5C%5C%0A%09%26%3D%5CvarPhi%20x%5Cleft(%20k-1%20%5Cright)%20%2B%5Cxi%20%5Cleft(%20k-1%20%5Cright)%20-%5Cmathbb%7BE%7D%20%5Cleft(%20x%5Cleft(%20k-1%20%5Cright)%20%2B%5Cxi%20%5Cleft(%20k-1%20%5Cright)%20%7Cy%5Cleft(%201%20%5Cright)%20%2C%5Ccdots%20y%5Cleft(%20n-1%20%5Cright)%20%5Cright)%5C%5C%0A%09%26%3D%5CvarPhi%20%5Ctilde%7Bx%7D%5Cleft(%20k-1%7Ck-1%20%5Cright)%20%2B%5Cxi%20%5Cleft(%20k-1%20%5Cright)%5C%5C%0A%5Cend%7Baligned%7D$

所以

$%5Cbegin%7Baligned%7D%0A%09%26P%5Cleft(%20k%7Ck-1%20%5Cright)%20%3D%5Cwidetilde%7Bx%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20%5Cwidetilde%7Bx%5ET%7D%5Cleft(%20k%7Ck-1%20%5Cright)%5C%5C%0A%09%26%3D%5Cleft(%20%5CvarPhi%20%5Ctilde%7Bx%7D%5Cleft(%20k-1%7Ck-1%20%5Cright)%20%2B%5Cxi%20%5Cleft(%20k-1%20%5Cright)%20%5Cright)%20%5Cleft(%20%5CvarPhi%20%5Ctilde%7Bx%7D%5Cleft(%20k-1%7Ck-1%20%5Cright)%20%2B%5Cxi%20%5Cleft(%20k-1%20%5Cright)%20%5Cright)%20%5ET%5C%5C%0A%09%26%3D%5CvarPhi%20%5Ctilde%7Bx%7D%5Cleft(%20k-1%7Ck-1%20%5Cright)%20%5Ctilde%7Bx%7D%5ET%5Cleft(%20k-1%7Ck-1%20%5Cright)%20%5CvarPhi%20%5ET%2BR_1%5C%5C%0A%09%26%5Ctriangleq%20%5CvarPhi%20P%5Cleft(%20k-1%7Ck-1%20%5Cright)%20%5CvarPhi%20%5ET%2BR_1%5C%5C%0A%5Cend%7Baligned%7D$

這樣一來，就得到了一個(gè)遞推關(guān)系：在已知 $%5Chat%7Bx%7D%5Cleft(%20k-1%7Ck-1%20%5Cright)%20$ 的情況下，可以計(jì)算出 $P%5Cleft(%20k-1%7Ck-1%20%5Cright)%20$ ，進(jìn)而可以計(jì)算出 $P%5Cleft(%20k%7Ck-1%20%5Cright)%20$ ，據(jù)此又可以算出 $R_%7Bx%5Cleft(%20k%20%5Cright)%20%5Cwidetilde%7By%7D%5Cleft(%20k%20%5Cright)%7D%2CR_%7B%5Cwidetilde%7By%7D%5Cleft(%20k%20%5Cright)%7D%5E%7B-1%7D$ ，此外根據(jù) $%5Chat%7Bx%7D%5Cleft(%20k-1%7Ck-1%20%5Cright)%20$ 還可以計(jì)算出 $%5Cwidetilde%7Bx%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20$ ，進(jìn)而算出 $%5Cwidetilde%7By%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20$ ，有了這些，就可以求出 $%5Chat%7Bx%7D%5Cleft(%20k%7Ck%20%5Cright)%20$ 了，這樣就完成了一次迭代

寫到這里其實(shí)已經(jīng)推導(dǎo)完了，不過和課本上的格式不太一樣，為了規(guī)范還是按課本上的來吧

記 $K%5Cleft(%20k%7Ck%20%5Cright)%20%3DR_%7Bx%5Cleft(%20k%20%5Cright)%20%5Cwidetilde%7By%7D%5Cleft(%20k%20%5Cright)%7DR_%7B%5Cwidetilde%7By%7D%5Cleft(%20k%20%5Cright)%7D%5E%7B-1%7D$ ，那么之前的遞推式就可以寫成

$%5Cbegin%7Baligned%7D%0A%09%26%5Cwidehat%7Bx%7D%5Cleft(%20k%7Ck%20%5Cright)%5C%5C%0A%09%26%3D%5CvarPhi%20%5Cwidehat%7Bx%7D%5Cleft(%20k-1%7Ck-1%20%5Cright)%20%2BK%5Cleft(%20k%7Ck%20%5Cright)%20%5Cleft(%20%5Cwidetilde%7By%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20%5Cright)%5C%5C%0A%09%26%3D%5CvarPhi%20%5Cwidehat%7Bx%7D%5Cleft(%20k-1%7Ck-1%20%5Cright)%20%2BK%5Cleft(%20k%7Ck%20%5Cright)%20%5Cleft(%20y%5Cleft(%20k%20%5Cright)%20-%5CvarTheta%20%5CvarPhi%20%5Cwidehat%7Bx%7D%5Cleft(%20k-1%7Ck-1%20%5Cright)%20%5Cright)%5C%5C%0A%5Cend%7Baligned%7D$

其中第二個(gè)等號(hào)由之前 $%5Cwidetilde%7By%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20$ 的推導(dǎo)的第三個(gè)等號(hào)可以得到

而

$K%5Cleft(%20k%7Ck%20%5Cright)%20%3DR_%7Bx%5Cleft(%20k%20%5Cright)%20%5Cwidetilde%7By%7D%5Cleft(%20k%20%5Cright)%7DR_%7B%5Cwidetilde%7By%7D%5Cleft(%20k%20%5Cright)%7D%5E%7B-1%7D%3DP%5Cleft(%20k%7Ck-1%20%5Cright)%20%5CvarTheta%20%5ET%5Cleft(%20%5CvarTheta%20P%5Cleft(%20k%7Ck-1%20%5Cright)%20%5CvarTheta%20%5ET%2BR_2%20%5Cright)%20%5E%7B-1%7D$

因?yàn)?/p>

$%5Cbegin%7Baligned%7D%0A%09%5Cwidehat%7Bx%7D%5Cleft(%20k%7Ck%20%5Cright)%20%26%3D%5Cwidehat%7Bx%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20%2BK%5Cleft(%20k%7Ck%20%5Cright)%20%5Cleft(%20%5Ctilde%7By%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20%5Cright)%5C%5C%0A%09%26%3D%5Cwidehat%7Bx%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20%2BK%5Cleft(%20k%7Ck%20%5Cright)%20%5Cleft(%20%5CvarTheta%20%5Ctilde%7Bx%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20%2B%5Ceta%20%5Cleft(%20k%20%5Cright)%20%5Cright)%5C%5C%0A%5Cend%7Baligned%7D$

故

$%5Cbegin%7Baligned%7D%0A%09%26%5Cwidetilde%7Bx%7D%5Cleft(%20k%7Ck%20%5Cright)%20%3Dx%5Cleft(%20k%20%5Cright)%20-%5Cwidehat%7Bx%7D%5Cleft(%20k%7Ck%20%5Cright)%5C%5C%0A%09%26%3Dx%5Cleft(%20k%20%5Cright)%20-%5Cleft(%20%5Cwidehat%7Bx%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20%2BK%5Cleft(%20k%7Ck%20%5Cright)%20%5Cleft(%20%5CvarTheta%20%5Ctilde%7Bx%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20%2B%5Ceta%20%5Cleft(%20k%20%5Cright)%20%5Cright)%20%5Cright)%5C%5C%0A%09%26%3D%5Cleft(%20I-K%5Cleft(%20k%7Ck%20%5Cright)%20%5CvarTheta%20%5Cright)%20%5Ctilde%7Bx%7D%5Cleft(%20k%7Ck-1%20%5Cright)%20-K%5Cleft(%20k%7Ck%20%5Cright)%20%5Ceta%20%5Cleft(%20k%20%5Cright)%5C%5C%0A%5Cend%7Baligned%7D$

所以

$%5Cbegin%7Baligned%7D%0A%09%26P(k%5Cmid%20k)%3D%5Cmathbb%7BE%7D%20%5Cleft(%20%5Ctilde%7Bx%7D(k%5Cmid%20k)%5Ctilde%7Bx%7D%5E%7B%5Cmathrm%7BT%7D%7D(k%5Cmid%20k)%20%5Cright)%5C%5C%0A%09%3D%26%5BI-K(k%5Cmid%20k)%5CTheta%20%5DP(k%5Cmid%20k-1)%5BI-K(k%5Cmid%20k)%5CTheta%20%5D%5E%7B%5Cmathrm%7BT%7D%7D%5C%5C%0A%09%26%2BK(k%5Cmid%20k)R_2K(k%5Cmid%20k)%5C%5C%0A%5Cend%7Baligned%7D$

進(jìn)一步化簡(jiǎn)可得

$P(k%5Cmid%20k)%3D%5BI-K(k%5Cmid%20k)%5CTheta%20%5DP(k%5Cmid%20k-1)$

（這兩個(gè)公式我直接抄書了，實(shí)在太復(fù)雜了）

總之這樣一來就可以得到計(jì)算 $%5Cwidehat%7Bx%7D%5Cleft(%20k%7Ck%20%5Cright)%20$ 的方法了，如下圖所示

這和我之前講的方法應(yīng)該是一樣的（大概吧），為了方便觀看再把相關(guān)的公式貼出來一下吧，其實(shí)相關(guān)的推導(dǎo)都在上面了

這個(gè)推導(dǎo)可以說是又臭又長(zhǎng)了，所以再次建議大家看看我在一開始發(fā)的那個(gè)鏈接上的推導(dǎo)，那個(gè)推導(dǎo)更接近本質(zhì)。不過不管怎么說，這個(gè)推導(dǎo)我覺得更加直接，沒有那么多彎彎繞繞，對(duì)于一個(gè)初學(xué)的人來說，雖然可能不會(huì)真的去把每一個(gè)公式都驗(yàn)證一遍（除非像我這么閑），但是也能很快的知道一個(gè)大致的思路，只能說各有所長(zhǎng)吧···

接下來如果我還有雅興的話，大概會(huì)試著應(yīng)用一下這個(gè)玩意吧，如果沒有雅興就沒了。

標(biāo)簽：控制論隨機(jī)過程高等數(shù)學(xué)線性代數(shù)概率論

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