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Bias and variance of nearest neighbours · Part 3 of 3

Data follow $y = f(x) + \varepsilon$ with independent noise of variance 4. A $k$ -nearest-neighbour regressor predicts at a point $x_0$ by averaging the $y$ values of its $k$ nearest training points.

Using $k = 32$ neighbours reaches points whose true values differ from $f(x_0)$ , adding a bias $b$ . Above what $|b|$ does $k = 32$ predict worse than $k = 8$ , to three decimal places?

Answer with a number. Fractions, powers and expressions like 23/6 or C(52,5) are read correctly in practice.

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The framework: bias, variance, capacity and dimensionality

Using k=32k = 32k=32 neighbours reaches points whose true values differ from f(x0)f(x_0)f(x0​), adding a bias bbb. Above what ∣b∣|b|∣b∣ does k=32k = 32k=32 predict worse than k=8k = 8k=8, to three decimal places?

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Using $k = 32$ neighbours reaches points whose true values differ from $f(x_0)$ , adding a bias $b$ . Above what $|b|$ does $k = 32$ predict worse than $k = 8$ , to three decimal places?