Limit involving Bayes' risk

  • Thread starter GabrielN00
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  • #1
GabrielN00

Homework Statement


Prove ##\lim_{n\rightarrow +\infty}\frac{\mathbb{E}(L_n)-L^*}{\sqrt{\mathbb{E}( ( \eta_n(X)-\eta(X) )^2 )}}=0##

if ##\eta_n## verifies ##\lim_{n\rightarrow\infty} \mathbb{E}( ( \eta_n(X)-\eta(X) )^2 )=0##

Homework Equations




The Attempt at a Solution



The idea might be to use ##g_n(x)=1_{\{\eta_n(x)>1/2\}}## because now ##\mathbb{E}(L_n)-L^*=2\mathbb{E}(|n(X)-1/2|1_{\{g(X)\neq g^*(X)\}})##. Now I want to show that ##\mathbb{E}(L_n)-L^*=2\mathbb{E}(|n(X)-1/2|1_{\{g(X)\neq g^*(X)\}})## can be bounded, this is, I want to prove there is an ##\epsilon## such such that ## \mathbb{E}(|n(X)-1/2|1_{\{g(X)\neq g^*(X)\}})\leq \mathbb{E}(|n(X)-\eta_n(X)|1_{\{g(X)\neq g^*(X)\}}1_{|\eta(X)-1/2|\leq \epsilon}1_{\eta(X)\neq1/2}) + \mathbb{E}(|n(X)-\eta_n(X)|1_{\{g(X)\neq g^*(X)\}}1_{|\eta(X)-1/2|> \epsilon})##.

If the latter can proved, and it is bounded, taking limits would show the limit of the problem is zero.I want to prove that such ##\epsilon## exists.
 

Answers and Replies

  • #2
GabrielN00
I realize some details may be needed, that I didn't write in the first message. Here is some extra context, I would add it to the original post but I can't edit (or I cannot figure out how to do so). The object of the proof is to show that ##L_n-L^*## converges to ##0## faster than the error ##L_2##, the conditional expectation error. ##L^*## denotes Bayes' risk. The function ##\eta## is given by ##\eta(x)=\mathbb{E}(Y|X=x)## and ##L_n## is the empirical prediction error.
 

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