Exercise 7.13

Answers

(a): $w_{50}$ is generated using data in $D_{train}$ , which has nothing to do with $D_{val}$ , so $E_{val} (w_{50})$ is an unbiased estimate of $E_{out} (w_{50})$
(b): From formula 2.1, with $M = 1$ , $N = 50$ , $δ = 0.1$ , we have $\begin{array}{l} E_{out} (w_{50}) & \leq E_{val} (w_{50}) + \sqrt{\frac{1}{2 N} \ln \frac{2 M}{δ}} \\ = E_{val} (w_{50}) + \sqrt{\frac{1}{100} \ln \frac{2}{0.1}} \\ = 0.05 + 0.2995732273553991 \approx 0.35 . \end{array}$
(c): We can’t bound $E_{out}$ with $E_{train}$ because we don’t know the number the hypothesis used to create $w_{50}$ . But for validation set, we first fixed the $w_{50}$ from training, then calculate the $E_{val}$ , so there’s only one hypothesis.

niuers

2021-12-08 09:55