Upper bound for probability when Bayes risk is zero

  • Thread starter GabrielN00
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GabrielN00

Homework Statement


Bayes' risk is ##L^*=0## for a classification problem. ##g_n(x)## is a classification rule (plug-in) such that ##g_n=0## is ##\eta_n(x)\leq 1/2## and ##g_n=1##$ otherwise. The function ##\eta##is given by ##\eta(x)=\mathbb{E}(Y|X=x)##. Then ##\mathbb{P}(g_n(X)\neq Y)\leq 4\mathbb{E}((\eta_n(X)-\eta(X))^2)##

Homework Equations


Bayes risk: For a loss function ·##L## and an estimator ##\hat{\theta}## the risk function is defined as the expected loss for that decisionis ##R:\Theta\times D \rightarrow \mathbb{R}, R(\theta, \hat{\theta}) = \mathbb{E_\theta}L(\theta,\hat{\theta})##.

Bayes risk is defined as the mean risk: ##r(\pi,\hat{\theta})=\int_\Theta R(\theta,\hat{\theta})\pi(\theta)d\theta##

The Attempt at a Solution



I see that if ##\eta_n\leq 1/2## then it reduces to show that ##\mathbb{P}(0\neq Y)\leq 4\mathbb{E}((\eta_n(X)-\eta(X))^2)##, but how do I get the bayesian risk involved. The right-hand of the inequality looks similar like the bayesian risk for the quadratic loss, but I can't see a way to use it - if it is related to the problem.
 

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