h***t
发帖数: 2540 | 1 x(i) are iid from Normal(u,sigma^2),
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 | m*e
发帖数: 146 | 2 x_star is MLE of u. it should converge to true value.
【在 h***t 的大作中提到】
: x(i) are iid from Normal(u,sigma^2),
: 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
| z****g
发帖数: 1978 | 3 It's sum not PI, or it's exp() not log of density. It thought it should be
the log-
liklihood function, although I doubt there is typo in the question
【在 m*e 的大作中提到】
: x_star is MLE of u. it should converge to true value.
| h***t
发帖数: 2540 | 4 there is no typo, indeed you can view it this way, x star is the gravity
center of the kernel function if proper constant added
【在 z****g 的大作中提到】
: It's sum not PI, or it's exp() not log of density. It thought it should be
: the log-
: liklihood function, although I doubt there is typo in the question
| z****g
发帖数: 1978 | 5 indeed, its a kernel function, or a mixture Gaussian distribution with
centers {x_i} and mixture probability 1/n. It is intuitive to say x^*
converge to u, however, hard to prove. | s***e
发帖数: 267 | 6 This is not the usual MLE if there is no typo. x_star will indeed converge
to u as n goes large. To see this
consistency result, you can look at the population equation and use an
symmetry argument.
【在 h***t 的大作中提到】
: x(i) are iid from Normal(u,sigma^2),
: 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
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