h***t
发帖数: 2540 | 1 x(i) are iid from Normal(u,sigma^2), both u and sigma unknown but fixed.
x_star= arg max sum(i=1 to n) exp(-(x-x(i))^2/(2*sigma^2)
does x_star converge to u as n goes to infinity? Thanks |
B******5
发帖数: 4676 | 2 if sigma is unknown, why there is one in the argmax? |
h***t
发帖数: 2540 | 3 it is unknown but fixed, x is the arg max for.
【在 B******5 的大作中提到】
: if sigma is unknown, why there is one in the argmax?
|
D*******a
发帖数: 207 | 4 你又不对它求arg max，只要是unknown就不应该出现在表达式里，不管是不是fixed。 |
h***t
发帖数: 2540 | 5 这个是未知，为什么不可以写在表达式里？最多就是结果里包含SIGMA啊
【在 D*******a 的大作中提到】
: 你又不对它求arg max，只要是unknown就不应该出现在表达式里，不管是不是fixed。
|
D*******a
发帖数: 207 | 6
在统计的术语里面，这样的叫做“known”。
【在 h***t 的大作中提到】
: 这个是未知，为什么不可以写在表达式里？最多就是结果里包含SIGMA啊
|
B****n
发帖数: 11290 | 7 It should converge to u.
The reason is under regularity conditions, which you can check easily for
your special case, the argmax sum(i=1 to n) exp(-(x-x(i))^2/(2*sigma^2)->
argmax E( exp(-(x-X(i))^2/(2*sigma^2)), where X(i) is the ith random
variable and x(i) the ith observation.
【在 h***t 的大作中提到】
: x(i) are iid from Normal(u,sigma^2), both u and sigma unknown but fixed.
: x_star= arg max sum(i=1 to n) exp(-(x-x(i))^2/(2*sigma^2)
: does x_star converge to u as n goes to infinity? Thanks
|
