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!


1 Answer 1


Hi: In your second formula, the bias is assumed to be zero so there is no bias term. The variance term is the MSE

  • $\begingroup$ Yes - I wrote that the bias term is 0. My question is why there is variance for noise ($\epsilon$) and variance for our estimator $\hat y$ $\endgroup$
    – Jason
    Commented Apr 30, 2020 at 21:15
  • $\begingroup$ 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. . $\endgroup$
    – mlofton
    Commented May 1, 2020 at 13:19

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