This question is regarding derivation of bias-variance decomposition as answered (which was accepted) in another thread. I am repeating the steps for this question:

\begin{align} \newcommand{\var}{{\rm var}} {\rm variance} &= \var \left( \frac{1}{k} \sum_i^k Y(x_i) \right) \\ &= \frac{1}{k^2} \sum_i^k \var \left( f(x_i) + \epsilon_i \right) \\ &= \frac{1}{k^2} \sum_i^k \var \left( f(x_i) \right) + \var \left( \epsilon_i \right) \\ &= \frac{1}{k^2} \sum_i^k \var \left( \epsilon_i \right) \\ &= \frac{1}{k^2} k \sigma_\epsilon^2 \\ &= \frac{\sigma^2_\epsilon}{k} \end{align}

Question 1: How did the author get $$\frac{1}{k^2}$$ in the 2nd step?


1 Answer 1


\begin{align} var(aX)&=E \left[ (aX-E\right[ {aX}\left])^2\right]\\ &=E\left[ (aX-aE\right[X\left])^2\right]\\ &=E\left[ (a(X-E\right[X\left]))^2\right]\\ &=E\left[a^2(X-E\right[X\left])^2\right]\\ &=a^2E\left[ (X-E\right[X\left])^2\right]\\ &=a^2var(X) \end{align} This depends on $E(aX)=aE(X)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.