# MSE Bias Variance tradeoff in estimating the variance of noise for MLE linear regression

I am just grasping the bias variance trade-off as it is explained by the MSE heuristic. We have that if $$y = f(x) + \epsilon$$ for $$\epsilon \sim N(0,\sigma^2)$$ we can show that

\begin{aligned} MSE(y, \hat y) &= Var(\epsilon) + Bias^2(\hat y) + Var(\hat y) \\ &= Var(\epsilon) + Var(\hat y) \qquad \text{ because we assume OLS is unbiased} \end{aligned}

I am confused because when we solve linear regression via MLE, we assume the same $$y = f(x) + \epsilon$$ for $$\epsilon$$ normally distributed with variance

$$Var(\epsilon) = MSE(y, \hat y)$$

What happened to the $$Var(\hat y)$$ term?

Thank you!

• Yes - I wrote that the bias term is 0. My question is why there is variance for noise ($\epsilon$) and variance for our estimator $\hat y$ Apr 30, 2020 at 21:15
• Hi: I still might not be understanding your question but in the unbiased case, the variance for the noise is the variance for the estimator $\hat{y}$. they're the same thing. . May 1, 2020 at 13:19