k*******g
发帖数: 13 | 1 To predict y(t), we use two candidate vars x1(t) x2(t) separately and get
two linear regression models:
M1: y(t)=b1*x1(t)+b0
M2: y(t)=b1*x2(t)+b0
The coefficient of determination, R-squared, for M1 and M2 are 0.01 and 0.02
If we run a new regression model with both x1 and x2,
M: y(t)=b1*x1(t)+b2*x2(t)+b0
问：lower bound and upper bound of M's R-squared 是多少？这两个extreme case
分别会有什么问题？
多谢指教！ |
d******e
发帖数: 7844 | 2 0.02~1
02
case
【在 k*******g 的大作中提到】
: To predict y(t), we use two candidate vars x1(t) x2(t) separately and get
: two linear regression models:
: M1: y(t)=b1*x1(t)+b0
: M2: y(t)=b1*x2(t)+b0
: The coefficient of determination, R-squared, for M1 and M2 are 0.01 and 0.02
: If we run a new regression model with both x1 and x2,
: M: y(t)=b1*x1(t)+b2*x2(t)+b0
: 问：lower bound and upper bound of M's R-squared 是多少？这两个extreme case
: 分别会有什么问题？
: 多谢指教！
|
k*******g
发帖数: 13 | 3 Why 0.02~1? Could you please explain?
I think maybe one extreme case is that x1 and x2 are perfectly correlated.
But it makes M1 and M2 having same R-squared, which contradicts with their R
-squared values.
Is the other extreme case X1 and X2 have correlation 0? How does it make R-
squared of M to be 1?
Thanks!
【在 d******e 的大作中提到】
: 0.02~1
:
: 02
: case
|
k*******a
发帖数: 772 | 4 I think it is 0.02 - 0.03
x1 can explain 1% error and x2 can explain 2% error
so x1 together with x2 can explain at least 2% and at most 3% (when x1 and
x2 uncorrelated) of error |
a****g
发帖数: 8131 | 5 there is a case called suppressor
R
【在 k*******g 的大作中提到】
: Why 0.02~1? Could you please explain?
: I think maybe one extreme case is that x1 and x2 are perfectly correlated.
: But it makes M1 and M2 having same R-squared, which contradicts with their R
: -squared values.
: Is the other extreme case X1 and X2 have correlation 0? How does it make R-
: squared of M to be 1?
: Thanks!
|
k*******g
发帖数: 13 | 6 Thank you!
When will it equal 0.02? I think it is the case that the variation explained
by x1 has also been explained by x2, but when will it happen? Thanks!
【在 k*******a 的大作中提到】
: I think it is 0.02 - 0.03
: x1 can explain 1% error and x2 can explain 2% error
: so x1 together with x2 can explain at least 2% and at most 3% (when x1 and
: x2 uncorrelated) of error
|
d******e
发帖数: 7844 | 7 > y = rnorm(100)
> x1 = rnorm(100)
> cor(x1,y)
[1] -1.254646e-05
> x2 = x1 + 0.0000001*y
> cor(x2,y)
[1] -1.243022e-05
> cor(x2-x1,y)
[1] 1
自己试去。
【在 k*******a 的大作中提到】
: I think it is 0.02 - 0.03
: x1 can explain 1% error and x2 can explain 2% error
: so x1 together with x2 can explain at least 2% and at most 3% (when x1 and
: x2 uncorrelated) of error
|
d******e
发帖数: 7844 | 8 他告诉你的结论是错的。
explained
【在 k*******g 的大作中提到】
: Thank you!
: When will it equal 0.02? I think it is the case that the variation explained
: by x1 has also been explained by x2, but when will it happen? Thanks!
|
s*****n
发帖数: 2174 | 9 我给你个直观的解释。
y_n = a + b * x1_n + e
这个模型里面的R^2, 可以看成是n维空间中向量Y和向量X1
(1维线性子空间)之间夹角的a余弦平方
R_1^2 = Cos(a)^2
同理第二个模型里面的R^2, 可以看成是n维空间中向量Y和向量X2
(1维线性子空间)之间夹角的b余弦平方
R_2^2 = Cos(b)^2
现在你要把X1和X2一起放进来， 等于是要把向量Y向X1和X2生成
的2维线性子空间中投影。
这样的问题就成了初等立体几何的问题, 已知一向量y与一平面上
两个向量x1和x2的夹角分别为a和b, 问y与此平面夹角c的余弦。
c最小可以是0， 最大是min(a, b), 于是其余弦平方最大可以是1,
最小则是Cos(min(a,b))^2
就不画图了， 不明白的话， 用三根筷子摆一下就一目了然。
correlated.
their
R
make R-
【在 k*******g 的大作中提到】
: Why 0.02~1? Could you please explain?
: I think maybe one extreme case is that x1 and x2 are perfectly correlated.
: But it makes M1 and M2 having same R-squared, which contradicts with their R
: -squared values.
: Is the other extreme case X1 and X2 have correlation 0? How does it make R-
: squared of M to be 1?
: Thanks!
|
G**7
发帖数: 391 | 10 Please type in English to explain. Thank you. |
k****n
发帖数: 165 | 11 犀利。
【在 s*****n 的大作中提到】
: 我给你个直观的解释。
: y_n = a + b * x1_n + e
: 这个模型里面的R^2, 可以看成是n维空间中向量Y和向量X1
: (1维线性子空间)之间夹角的a余弦平方
: R_1^2 = Cos(a)^2
: 同理第二个模型里面的R^2, 可以看成是n维空间中向量Y和向量X2
: (1维线性子空间)之间夹角的b余弦平方
: R_2^2 = Cos(b)^2
: 现在你要把X1和X2一起放进来， 等于是要把向量Y向X1和X2生成
: 的2维线性子空间中投影。
|
b********8
发帖数: 3059 | 12 厉害
【在 s*****n 的大作中提到】
: 我给你个直观的解释。
: y_n = a + b * x1_n + e
: 这个模型里面的R^2, 可以看成是n维空间中向量Y和向量X1
: (1维线性子空间)之间夹角的a余弦平方
: R_1^2 = Cos(a)^2
: 同理第二个模型里面的R^2, 可以看成是n维空间中向量Y和向量X2
: (1维线性子空间)之间夹角的b余弦平方
: R_2^2 = Cos(b)^2
: 现在你要把X1和X2一起放进来， 等于是要把向量Y向X1和X2生成
: 的2维线性子空间中投影。
|
