-
Notifications
You must be signed in to change notification settings - Fork 0
/
Copy pathR SST disassemble.Rmd
85 lines (76 loc) · 3.22 KB
/
R SST disassemble.Rmd
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
---
title: "R SST拆解證明"
author: "Ivan Lin"
date: "2020年4月18日"
output:
prettydoc::html_pretty:
theme: cayman
highlight: github
---
```{r setup, include=FALSE}
#這段並不會show出來
knitr::opts_chunk$set(echo = TRUE)
require(kableExtra)
require(car)
require(tidyverse)
require(prettydoc)
require(Metrics)
knitr::opts_chunk$set(echo = TRUE)
options(knitr.table.format = "html")
```
這個證明需要花點篇幅，因此獨立來寫。<br>
在做回歸時，我們能將總離均差平方和\(SS_{T}\)拆解成回歸離均差平方和\(SS_{regression}\)與誤差離均差平方和\(SS_{e}\)。<br>
之後在計算\(R^{2}\)時，會使用\(SS_{T}\)-\(SS_{e}\)來計算\(SS_{regression}\)<br>
但問題是，為什麼這個公式可以成立呢？<br>
\[SS_{t}=SS_{regression}+SS_{e}\]
\[\sum_{i=1}^{n} (y_{i}-\bar{y})^{2}=\sum_{i=1}^{n} (y_{i}-\hat{y_{i}})^{2}+\sum_{i=1}^{n} (\hat{y_{i}}-\bar{y})^{2}\]
<br>
這張圖修改自[Tommy線性回歸](https://medium.com/@chih.sheng.huang821/%E7%B7%9A%E6%80%A7%E5%9B%9E%E6%AD%B8-linear-regression-3a271a7453e)
<br>
可以參照
請看以下證明：
# 總離差平方和推導
\(y_{i}\)代表真實值；\(\bar{y}\)平均值；\(\hat{y_{i}}\)代表預測值；
\(i\in[1,n]\)<br>
\[(y_{i}-\bar{y})=(y_{i}-\hat{y_{i}})+(\hat{y_{i}}-\bar{y})\]
將兩邊平方並加總
\[\sum_{i=1}^{n} (y_{i}-\bar{y})^{2}=\sum_{i=1}^{n} (y_{i}-\hat{y_{i}})^{2}+\sum_{i=1}^{n} (\hat{y_{i}}-\bar{y})^{2}+\sum_{i=1}^{n}2(y_{i}-\hat{y_{i}})(\hat{y_{i}}-\bar{y})\]
之後我們來證明
\[\sum_{i=1}^{n}2(y_{i}-\hat{y_{i}})(\hat{y_{i}}-\bar{y})=0\]
就可以讓
\[SS_{t}=SS_{regression}+SS_{e}\]
以簡單回歸（一個自變數\(x_{i}\)）的狀況當作例子：
\[\hat{y_{i}}=\hat{a}+\hat{b}x_{i}\]
\[\bar{y}=\hat{a}+\hat{b}\bar{x}\]
而回歸模型中，斜率
\[\hat{b}=\frac{cov(x,y)}{s^2}=\frac{\sum_{i=1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sum_{i=1}^{n}(x_{i}-\bar{x})^2}\]
<br>
因此，
\[\hat{y_{i}}-\bar{y}=\hat{b}(x_{i}-\bar{x})\]
\[y_{i}-\hat{y_{i}}=(y_{i}-\bar{y})-(\hat{y_{i}}-\bar{y})=(y_{i}-\bar{y})-\hat{b}(x_{i}-\bar{x})\]
所以，
\[\sum_{i=1}^{n}2(y_{i}-\hat{y_{i}})(\hat{y_{i}}-\bar{y})=2\hat{b}\sum_{i=1}^{n}(y_{i}-\hat{y_{i}})(x_{i}-\bar{x})\]
\[=2\hat{b}\sum_{i=1}^{n}((y_{i}-\bar{y})-\hat{b}(x_{i}-\bar{x}))(x_{i}-\bar{x_{i}})\]
\[=2\hat{b}\sum_{i=1}^{n}((y_{i}-\bar{y})(x_{i}-\bar{x})-\hat{b}(x_{i}-\bar{x})^2)\]
\[=2\hat{b}\sum_{i=1}^{n}((y_{i}-\bar{y})(x_{i}-\bar{x})-\frac{\sum_{i=1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sum_{i=1}^{n}(x_{i}-\bar{x})^2}(x_{i}-\bar{x})^2)\]
\[=2\hat{b}(0)=0\]
# 在R中驗證
我們在R裡面創個簡單的數列與回歸來測試一下<br>
```{r prove sst}
y <- sort(round(runif(10,min=0,max=50))) #隨機生成Y數列
x <- 1:10
x_mean <- mean(x)
y_mean <- mean(y)
b_hat <- sum((x-x_mean)*(y-y_mean))/sum((x-x_mean)^2) ;b_hat #斜率
a_hat <- y_mean-b_hat*x_mean; a_hat #截距
lm(y~x)
y_estimate <- a_hat+b_hat*x
#SST=SSreg+SSe
sum((y-y_mean)^2)==sum((y_estimate-y_mean)^2)+sum((y-y_estimate)^2)
```
<br>
Bingo！之後會在另一篇文章來探討回歸的計算！<br>
***
如果內容有任何問題或錯誤，歡迎來信跟我聯絡唷！