m******y
发帖数: 266 | 1 譬如一个很简单的参数化的问题：E{(y-xb)exp(x)} >= 0. | e*******a
发帖数: 514 | 2 E{(y-xb)exp(x)}
=E[E{(y-xb)exp(x)}|X]
=E[exp(x)(E{y|X}-xb)]>= 0
Through sample analog,
1/n \sum_{i=1}^n[exp(x_i) (\sum{i=1}^n (y_i-x_i*b))]>=0.
I guess, matlab fminsearch will do the work.
This is my take on your question. Never read anything about partial
identification problem before. Low credibility in my answer.
Good luck.
【在 m******y 的大作中提到】
: 譬如一个很简单的参数化的问题：E{(y-xb)exp(x)} >= 0.
| m******y
发帖数: 266 | 3 你这想法不错，一开始我也是这么想的。但后来实际问题是，你很容易定义sample
sanlog of E[exp(x)(E{y|X}-
xb)]>= 0. 但是程序要run出一个set，而不是一个点的时候，计算问题就出来了。
【在 e*******a 的大作中提到】
: E{(y-xb)exp(x)}
: =E[E{(y-xb)exp(x)}|X]
: =E[exp(x)(E{y|X}-xb)]>= 0
: Through sample analog,
: 1/n \sum_{i=1}^n[exp(x_i) (\sum{i=1}^n (y_i-x_i*b))]>=0.
: I guess, matlab fminsearch will do the work.
: This is my take on your question. Never read anything about partial
: identification problem before. Low credibility in my answer.
: Good luck.
| e*******a
发帖数: 514 | 4 The sample analog (sa) is:
1/n \sum_{i=1}^n {exp(x_i)*[\sum_{j=1}^n y_j -x_i*\beta]}>=0,
define y_bar = \sum_{j=1}^n y_j,
the sa becomes:
1/n \sum_{i=1}^n {exp(x_i)*[y_bar -x_i*\beta]}>=0.
Let's start with a simple case, \beta is in R.
sa implies \beta <= (y_bar * \sum_{i=1}^n exp(x_i) )/(\sum_{i=1}^n x_i).
Here, you only need a calculator.
If \beta is in a compact subset of R^2. You could fix one of the coordinates
of \beta first, then it reduces to the previous case. So for your program,
you h
【在 m******y 的大作中提到】
: 你这想法不错，一开始我也是这么想的。但后来实际问题是，你很容易定义sample
: sanlog of E[exp(x)(E{y|X}-
: xb)]>= 0. 但是程序要run出一个set，而不是一个点的时候，计算问题就出来了。
| m******y
发帖数: 266 | 5 计算的问题在于，你不可能把参数空间都搜索一遍，因此我们最大化concave/convex的
目标函数时候，结合data，可以根据高斯牛顿法很“快”寻找出解。要把整个参数空间
都找一遍的办法是几乎不可行的。别的计算方法，譬如Han Hong他们的MCMC，也是希望
直接根据data直接找到解。
coordinates
,
【在 e*******a 的大作中提到】
: The sample analog (sa) is:
: 1/n \sum_{i=1}^n {exp(x_i)*[\sum_{j=1}^n y_j -x_i*\beta]}>=0,
: define y_bar = \sum_{j=1}^n y_j,
: the sa becomes:
: 1/n \sum_{i=1}^n {exp(x_i)*[y_bar -x_i*\beta]}>=0.
: Let's start with a simple case, \beta is in R.
: sa implies \beta <= (y_bar * \sum_{i=1}^n exp(x_i) )/(\sum_{i=1}^n x_i).
: Here, you only need a calculator.
: If \beta is in a compact subset of R^2. You could fix one of the coordinates
: of \beta first, then it reduces to the previous case. So for your program,
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