m******t
发帖数: 273 | 1 【 以下文字转载自 Statistics 讨论区 】
发信人: myregmit (myregmit), 信区: Statistics
标 题: data prediction by regression or better ways
发信站: BBS 未名空间站 (Fri Mar 7 17:24:34 2014, 美东)
I am working on data prediction.
Given data of a random variable X and Y, find out how to predict Y by X.
I know how to do it by linear regression, y = k x + b .
But, here, x is always non-negative and y is required to be non-negative.
Sometimes, b is not non-negative so that y < 0.
How to assure that b > 0 and also minimize the prediction error ?
Are there other better ways (not regression) to do the prediction ?
Any help would be appreciated. | B****n
发帖数: 11290 | 2 If you use matlab, the function nlinfit can fit nonlinear regression model
using least squares that allows restrictions for the parameters.
【在 m******t 的大作中提到】
: 【 以下文字转载自 Statistics 讨论区 】
: 发信人: myregmit (myregmit), 信区: Statistics
: 标 题: data prediction by regression or better ways
: 发信站: BBS 未名空间站 (Fri Mar 7 17:24:34 2014, 美东)
: I am working on data prediction.
: Given data of a random variable X and Y, find out how to predict Y by X.
: I know how to do it by linear regression, y = k x + b .
: But, here, x is always non-negative and y is required to be non-negative.
: Sometimes, b is not non-negative so that y < 0.
: How to assure that b > 0 and also minimize the prediction error ?
| m******t
发帖数: 273 | 3 【 以下文字转载自 Statistics 讨论区 】
发信人: myregmit (myregmit), 信区: Statistics
标 题: data prediction by regression or better ways
发信站: BBS 未名空间站 (Fri Mar 7 17:24:34 2014, 美东)
I am working on data prediction.
Given data of a random variable X and Y, find out how to predict Y by X.
I know how to do it by linear regression, y = k x + b .
But, here, x is always non-negative and y is required to be non-negative.
Sometimes, b is not non-negative so that y < 0.
How to assure that b > 0 and also minimize the prediction error ?
Are there other better ways (not regression) to do the prediction ?
Any help would be appreciated. | B****n
发帖数: 11290 | 4 If you use matlab, the function nlinfit can fit nonlinear regression model
using least squares that allows restrictions for the parameters.
【在 m******t 的大作中提到】
: 【 以下文字转载自 Statistics 讨论区 】
: 发信人: myregmit (myregmit), 信区: Statistics
: 标 题: data prediction by regression or better ways
: 发信站: BBS 未名空间站 (Fri Mar 7 17:24:34 2014, 美东)
: I am working on data prediction.
: Given data of a random variable X and Y, find out how to predict Y by X.
: I know how to do it by linear regression, y = k x + b .
: But, here, x is always non-negative and y is required to be non-negative.
: Sometimes, b is not non-negative so that y < 0.
: How to assure that b > 0 and also minimize the prediction error ?
| m******t
发帖数: 273 | 5 x and y are discrete and between 0 and 100,000.
Any help would be appreciated.
【在 m******t 的大作中提到】
: 【 以下文字转载自 Statistics 讨论区 】
: 发信人: myregmit (myregmit), 信区: Statistics
: 标 题: data prediction by regression or better ways
: 发信站: BBS 未名空间站 (Fri Mar 7 17:24:34 2014, 美东)
: I am working on data prediction.
: Given data of a random variable X and Y, find out how to predict Y by X.
: I know how to do it by linear regression, y = k x + b .
: But, here, x is always non-negative and y is required to be non-negative.
: Sometimes, b is not non-negative so that y < 0.
: How to assure that b > 0 and also minimize the prediction error ?
| m******t
发帖数: 273 | 6 This y and x scatter plot.
Any help would be appreciated.
【在 m******t 的大作中提到】
: x and y are discrete and between 0 and 100,000.
: Any help would be appreciated.
| m******t
发帖数: 273 | 7 The data are not counts and not number of success either.
They are numerical values calculated by black box functions.
Any help would be appreciated.
【在 m******t 的大作中提到】
: This y and x scatter plot.
: Any help would be appreciated.
| m******t
发帖数: 273 | 8 This is the log-norm scatter plot.
log with base as 10.
For the x =0 and y =0, we change it to be 1.
Any help would be appreciated.
【在 m******t 的大作中提到】
: The data are not counts and not number of success either.
: They are numerical values calculated by black box functions.
: Any help would be appreciated.
|
|
