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Derivation for MSE to Bias Squared Plus Variance

MSE(θ^m)=E((θ^m−θ)2)MSE(\hat{\theta}_m) = \mathbb{E}((\hat{\theta}_m - \theta)^2) =E(θ^m2+θ2−2θθ^m) = \mathbb{E}(\hat{\theta}_m^2 + \theta^2 - 2\theta\hat{\theta}_m) =E(θ^m2)+E(θ2)−E(2θθ^m) = \mathbb{E}(\hat{\theta}_m^2) + \mathbb{E}(\theta^2) - \mathbb{E}(2\theta\hat{\theta}_m) =E(θ^m2)+θ2−2θE(θ^m) = \mathbb{E}(\hat{\theta}_m^2) + \theta^2 - 2\theta\mathbb{E}(\hat{\theta}_m) =[E(θ^m2)+θ2−2θE(θ^m)]+[E2(θ^m)−E2(θ^m)] = [\mathbb{E}(\hat{\theta}_m^2) + \theta^2 - 2\theta\mathbb{E}(\hat{\theta}_m)] + [\mathbb{E}^2(\hat{\theta}_m) - \mathbb{E}^2(\hat{\theta}_m)] =[E2(θ^m)−2θE(θ^m)+θ2]+[E(θ^m2)−E2(θ^m)] = [\mathbb{E}^2(\hat{\theta}_m) - 2\theta\mathbb{E}(\hat{\theta}_m) + \theta^2] + [\mathbb{E}(\hat{\theta}_m^2) - \mathbb{E}^2(\hat{\theta}_m)] =(E(θ^m)−θ)2+E((θ^m−E(θ^m))2) = (\mathbb{E}(\hat{\theta}_m) - \theta)^2 + \mathbb{E}((\hat{\theta}_m - \mathbb{E}(\hat{\theta}_m))^2) =Bias2(θ^m)+Var(θ^m) = Bias^2(\hat{\theta}_m) + Var(\hat{\theta}_m)

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Updated 2021-05-24

Contributors are:

JM

Justin McIntosh
🏆 1

Who are from:

University of Toledo
University of Toledo
🏆 1

Tags

Data Science

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