A derivation of the bias-variance decomposition of test error in machine learning.

analysis.ipynb

@bradyneal, I understand what @sdangi meant. My issue is that I don't see why the suggested change is correct. I could be wrong, though. Happy to look over the algebra if you or @sdangi are interested in providing it. Best, Peter

@bradyneal @sdangi : you are correct! Thanks! I've updated the jupyter notebook accordingly.

In section 'Reducible and irreducible error' why is E_e[2\epsilon (f - h)] equal to 0?

I agree that E_e[\epsilon f] = 0, but why E_e[\epsilon h] = 0? The hypothesis h is learned from some test set (X,Y) and Y depends on \epsilon and so the parameters of the learned model do. Thus one cannot write E_e[\epsilon h] = h E_e[\epsilon].

Could you explain this?

@rafgonsi : ... In performing the triple integral over $X$, $\cal{D}$ and $\epsilon$, I fix two variables ($X$ and $\cal{D}$) and vary the third ($\epsilon$). Since $X$ is fixed, so are $f$ and $h$, which may therefore be "pulled out" of the innermost integral over $\epsilon$.

An entirely analogous result to that outlined in this gist is obtained when one computes the error of an estimator of a parameter. Namely the mean square error of any estimator is equal to its variance plus (the square of) its bias. See section 7.7 at https://www.sciencedirect.com/science/article/pii/B9780123948113500071

In active machine learning, we assume that the learner is unbiased, and focus on algorithms that minimize the learner's variance, as shown in Cohn et al (1996): https://arxiv.org/abs/cs/9603104 (Eq. 4 is difficult to interpret precisely, though, in the absence of further reading).