# Limit involving Bayes' risk

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##

## 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.