I-paradox kaStein

Ngowe-1961 uJames noStein bapapasha iphepha elithi Estimation with Quadratic Loss . Thatha idata esasazwe ngokuqhelekileyo ngentsingiselo engaziwayo \(\mu\) kunye nokwahlukana \(1\) . Ukuba ngoku ukhetha ixabiso elingacwangciswanga \(x\) kule datha kwaye kufuneka uqikelele intsingiselo \(\mu\) ngokwesiseko soku, ngentuitively \(x\) luqikelelo olufanelekileyo lwe \(\mu\) (ekubeni unikezelo oluqhelekileyo lukhona, okukhethiweyo ngokungenamkhethe \(x\) mhlawumbi kufutshane \(\mu\) ).


Ngoku uvavanyo luyaphindwa - ngeli xesha kunye ezintathu ezizimeleyo, kwakhona ngokuqhelekileyo kusasazwa idatha iseti nganye ngokwahluka \(1\) kunye namaxabiso \(\mu_1\) , \(\mu_2\) , \(\mu_3\) . Emva kokufumana amaxabiso amathathu angaqhelekanga \(x_1\) , \(x_2\) kunye \(x_3\) , omnye uyaqikelela (esebenzisa inkqubo efanayo) \(\mu_1=x_1\) , \(\mu_2=x_2\) kunye \(\mu_3=x_3\) .

Isiphumo esimangalisayo sikaJames noStein kukuba kukho uqikelelo olungcono lwe- \( \left( \mu_1, \mu_2, \mu_3 \right) \) (o.k.t. indibaniselwano yeeseti ezintathu ezizimeleyo zedatha) kune \( \left( x_1, x_2, x_3 \right) \) . "Umqikelelo kaJames Stein" ungoko:

$$ \begin{pmatrix}\mu_1\\\mu_2\\\mu_3\end{pmatrix} = \left( 1-\frac{1}{x_1^2+x_2^2+x_3^2} \right) \begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} \neq \begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} $$

Intsingiselo yokutenxa isikwere esisikweri salo mqikelelo isoloko incinci kunentsingiselo yenxaso yesikwere \( E \left[ \left|| X - \mu \right||^2 \right] \) yoqikelelo oluqhelekileyo.

Kuyamangalisa kwaye mhlawumbi kuyaxaka ukuba umqikelelo kaJames-Stein atshintshe uqikelelo oluqhelekileyo (ngento ecuthekayo) ukuya kwimvelaphi kwaye ngaloo ndlela enika iziphumo ezingcono kuninzi lwamatyala. Oku kusebenza kwimilinganiselo \( \geq 3 \) , kodwa kungekhona kwimeko ye-dimensional.

Inkcazo enhle yejometri yokuba kutheni le nto isebenza inikezelwa nguBrown & Zao . Qaphela ukuba oku akuthethi ukuba unoqikelelo olungcono lweseti yedatha nganye - unoqikelelo olungcono olunomngcipheko omncinci odityanisiweyo .

Emva